9th Chemistry
9th Chemistry
9th Chemistry
TELANGANA, HYDERABAD.
ACADEMIC YEAR : 2020 – 21, LEVEL - 2
Learning outcomes
INTRODUCTION
● We have learnt a little information from a balanced chemical equation like physical state, temperature etc.
In previous class.
i. A chemical equation gives information about the reactants and products through their symbols and formulae.
iii. As molecular masses are expressed in ‘Unified Masses’ (U), the relative masses of reactants and products are known
from the equation.
iv. If the masses are expressed in grams then the equation also gives the molar ratios of reactants and products.
v. If gases are involved, we can equate their masses to their volumes and calculate the volumes or those gases liberated at
given condition of temperature and pressure using molar mass and molar volume relationship.
vi. Using molar mass and Avagadro’s number we can calculate the number of molecules and atoms of different substances
from the equation.
● Now let's learn all the information that can be known from a balanced chemical equation.
Sunlight
Ex. 6CO2+ 6 H2O C6H12O6 + 6O2
Chlorophyll
3. Physical State :
1. CO2 - Gas
2. H2O - Liquid
3. C6H12O6 - Solid
4. O2 - Gas
6. 6 moles of carbon dioxide reacts with 6 moles of water to form 1 mole of glucose and 6 moles of oxygen
gas.
8. 6 x 22.4 lit of CO2 reacts with 6 x 18 gm of H2O to form 180 gm of C6H12O6 and 6 x 22.4 lit. of oxygen
(STP)
2. 2 C(s) + O2 CO2g) + Q
State all the information provided by the above balanced chemical equation.
STATE COUNCIL OF EDUCATIONAL RESEARCH AND TRAINING
TELANGANA, HYDERABAD.
ACADEMIC YEAR : 2020 – 21, LEVEL - 2
CLASS: X MEDIUM : English SUBJECT : Physical Sciences
CHAPTER : Chemical Equations
TOPIC / CONCEPT : Mass – Mass relationship.
WORKSHEET NO : 16
Learning outcomes
Children can calculate numberical problems of mass – mass relationship based on balanced
on chemical equations.
INTRODUCTION
● A balanced chemical equation gives information of relative masses of reactants and products.
To understand these relation ships, Let’s revise the following concepts learnt in previous class.
MOLAR MASS:
No. of particles in one mole of a substance is called Avagadro’s Number. This is denoted by N.
One mole of any gas, at S.T.P, occupies a volume of 22.4 lit. This is called gram molar volume (G.M.V)
With the above equation, Calculate the amount of alluminium required to obtain 1120 kg of Iron.
2 X Al + [ 2 X Fe + 3 X O ] [ 2 X Al + 3 X O ] + 2 X Fe
2 X 27 + [ 2 X 56 + 3 X 16 ] [ 2 X 27 + 3 X 16 ] + 2 X 56
= 540 kg
➢ Calculate the mass of hydrogen (H2) liberated when 230 gm of Sodium reacts with excess of water at S.T.P.
SELF ASSESSMENT
2. C + O2 CO2
(12gm) (32 gm) (44 gm )
If 6 gm of carbon burns completely, How much mass of carbon di – oxide is liberated.
4. How much carbon – di – oxide is liberated when 50 gm of calcium carbonate reacts with excess of
hydrochloric acid.
[ Ca = 40 U , C = 12U , O = 16 U , H = 1 U , Cl = 35.5 U ]
CaCO3 (S) + 2 HCl (aq) CaCl2 (aq) + H2O (l) + CO2 (g)
MULTIPLE CHOICE QUESTIONS
5. STP refers to [ ]
A) 2 gm
B) 1 gm
C) 0.2 gm
D) 0.1 gm
A) 6.022 X 10 23
B) 6.022 X 10 22
C) 3.011 X 10 22
D) 3.011 X 10 23
8. 2H2 + O2 2H2O [ ]
(4 gm) (32 gm) (36 gm )
A) 18 gm
B) 36 gm
C) 9 gm
D) 72 gm
STATE COUNCIL OF EDUCATIONAL RESEARCH AND TRAINING
TELANGANA, HYDERABAD.
ACADEMIC YEAR : 2020 – 21, LEVEL - 2
CLASS: X MEDIUM : English SUBJECT : Physical Sciences
CHAPTER : Chemical Equations
TOPIC / CONCEPT : Mass – Volume, Mass -Molecule number,
Volume – Volume relationships WORKSHEET NO : 17
Key concepts Finding Mass – Volume, Mass – Molecule number , Volume – Volume relationships
based on balanced chemical equation.
Learning outcomes
Children can do calculations of mass – volume, volume – volume, mass – volume – molecule number
relationships based on balanced on chemical equations.
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● Calculate volume of hydrogen gas liberated when 230 gm of sodium reacts with excess of water at STP. Also
2 X Na + 2[2XH+1XO] 2 [ Na + O + H ] + 2XH
2 X 23 + 2 [ 2 X 1 + 1 X 16 ] 2 [ 23 + 16 + 1 ] + 2X1
46 U + 36 U 80 U + 2U
46 gm + 36 gm 80 gm + 2 gm
Mass – Mass relationship : As per equation mass of H2 liberated when 46 gm of Na reacts with excess water = 2 gm
10 1
230 X 2
Mass of H2 liberated when 230 gm of Na reacts with excess water = = 10 gm
46
23
1
Mass – Volume relationship
22.4 X 10
Volume of 10 gm of H2 at STP = = 112 lit.
2
=
6.022X10 23 X 10 = 5 X 6.022 X 10 23 molecule = 5 N
2
Volume – Volume relationship
● Ex: At STP , 44.8 lit. of oxygen is available for combustion of carbon. What volume of carbon-di-oxide is
44.8
= = 2 mol.
22.4
Volume of 1 mole of O2 gas at S.T.P = 22.4 lit
As per equation,
● 50 gm of calcium carbonate is treated excess hydro chloric acid. Calculate the volume and
SELF ASSESSMENT
∆
3. CaCO3(s) CaO(s) + CO2(g)
4. What is the volume of oxygen liberated when 49 gm of potassium chlorate is decomposed at S.T.P.
[ K = 49 U , Cl = 35.5 U , O = 16 U ]
∆
2KClO3 (S) 2KCl (s) + 3O2 (g)
MULTIPLE CHOICE QUESTIONS
A) 22.4 lit
B) 2.24 lit
C) 11.2 lit
D) 1.12 lit
A) 6.022 X 10 23
B) 3.011 X 10 23
C) 3.011 X 10 22
D) 6.022 X 10 22
A) 6.022 X 10 23
B) 6.022 X 10 22
C) 3.011 X 10 22
D) 3.011 X 10 23
STATE COUNCIL OF EDUCATIONAL RESEARCH AND TRAINING
TELANGANA, HYDERABAD
ACADEMIC YEAR 2020-21
LEVEL - 2
CLASS: 10 MEDIUM : ENGLISH SUBJECT: PHYSICAL
SCIENCE
Lesson : Acids – Bases -Salts Work sheet No:18
Topic: Introduction , Response of various substances with indicators.
Key Words : Acid, Base, Natural indicators, Synthetic indicators and olfactory indicators.
Learning Outcomes :
Children can give examples for acids, bases, natural indicators, synthetic indicators and olfactory
indicators
Children can explain the process of preparation of olfactory indicator.
Children can explain the changes in color when indicators or added to acids, bases.
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Acids and Bases :
A) Both 1, 2 are true B) Both 1, 2 are false C) 1 is true and 2 is false D) 1 is false and 2 is true.
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STATE COUNCIL OF EDUCATIONAL RESEARCH AND TRAINING
TELANGANA, HYDERABAD
ACADEMIC YEAR 2020-21
LEVEL - 2
CLASS: 10 MEDIUM : ENGLISH SUBJECT: PHYSICAL SCIENCE
Lesson : Acids – Bases -Salts Work sheet No:19
Learning Outcomes :
Children can explain the relations of acids and bases with metals with examples.
Children can explain neutralization reaction with an example
Repeat the above experiment with some other acids like H2SO4 , HNO3
From the above activities you can conclude that Hydrogen gas is evolved when acid reacts with
metal.
The reaction occurring between acid and base in the above activity can be written as
NaOH + HCl NaCl + H2O
* The reaction of an acid with a base to give a salt and water is known as a neutralization reaction.
* A neutralization reaction can be written as.
Assessment
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.
STATE COUNCIL OF EDUCATIONAL RESEARCH AND TRAINING
TELANGANA, HYDERABAD
ACADEMIC YEAR 2020-21
LEVEL - 2
CLASS: 10 MEDIUM : ENGLISH SUBJECT: PHYSICAL
SCIENCE
Lesson : Acids – Bases -Salts Work sheet No:20
Topic : 1. Reaction of acids with metal carbonates and metal hydrogen carbonates.
2.Reaction of acids and bases with Metal and Non-Metal oxides.
Key Words : Acid, Base, Metal carbonate, Metal Hydrogen carbonate, Metal oxide, Non-metal oxide.
Learning Outcomes:
Children can explain the reaction between acid and metal carbonates and metal hydrogen carbonates
with examples.
Children can explain the reaction of acid with metal oxide with an example.
Children can explain the reaction between base non-metal oxides with an example.
By this activity we can write the general reaction between a metal oxide and an acid as
➢ Thus we can conclude that the metal oxides are basic in nature like metal hydroxides.
The reaction is similar to the reaction of an acid with base. We can conclude that non- metal oxides
are acidic in nature.
.
Assessment
1. Explain the reaction of acids with metal carbonates and metal hydrogen carbonates with an example
for each.
2. Explain the reaction between acid and metal oxide with an example.
3. Explain the reaction between base and non-metal oxide with an example.
.
4. The gas evolved in the reactions between acid and metal carbonates or metal
hydrogen carbonates. ( )
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STATE COUNCIL OF EDUCATIONAL RESEARCH AND TRAINING
TELANGANA, HYDERABAD
ACADEMIC YEAR 2020-21
LEVEL - 2
CLASS: 10 MEDIUM : ENGLISH SUBJECT: PHYSICAL SCIENCE
Learning Outcomes:
Children can give the reason for the flow of electric current through acid solutions, alkali solutions..
Children can explain the ions present in acid solutions, alkali solutions.
In this activity, you will notice that the bulb glows only in acid solutions but not in glucose and
alcohol solutions.
Glowing of bulb indicates that there is flow of electric current through the solution.
Acid solutions have ions and the movement of these ions in solutions helps for flow of electric
current the solutions.
Acid solutions have ions and the movement of these ions in solution helps for flow of electric current
through the solution.
This suggests that acids produce hydrogen ions in solution, which are responsible for their acidic
properties.
In glucose and alcohol solution the bulb did not glow, indicating the absence of H+ ions in these
solutions.
. : Repeat the activity 7 using alkalis such as NaOH, Ca(OH)2 solutions etc.,
Note
instead of acid solutions.
You will notice the electric current pass through these alkali solutions also.
Ions present in alkalis are negative ions (anions).
They are [OH-] ions.
We can conclude that bases produce hydroxide ions (OH-) in solutions.
Assessment
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STATE COUNCIL OF EDUCATIONAL RESEARCH AND TRAINING
TELANGANA, HYDERABAD
ACADEMIC YEAR 2020-21
LEVEL - 2
CLASS: 10 MEDIUM : ENGLISH SUBJECT: PHYSICAL SCIENCE
Key Words: Hydrogen ion (H+), Hydronium ion (H3O+), Hydroxide ion(OH-)
Learning Outcomes:
Children can explain the process of dissociation of acids in the presence of water, with equations.
Explain the process of giving OH- ions when a base is dissolved in water, with an example.
Assessment
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STATE COUNCIL OF EDUCATIONAL RESEARCH AND TRAINING
TELANGANA, HYDERABAD
ACADEMIC YEAR 2020-21
LEVEL - 2
CLASS: 10 MEDIUM : ENGLISH SUBJECT: PHYSICAL SCIENCE
Key Words: Concentrated acid, concentrated base, dilution, diluted acid, diluted base, strong acid,
strong base, weak acid, weak base, universal indicator.
Learning Outcomes:
If we allow electric current to pass through diluted CH3COOH (acetic acid) and diluted HCl
(Hydrochloric acid) separately as in activity 7 we will notice that bulb glows brightly in HCl solution
while the intensity of the bulb is low in acetic acid solution.
This indicates that there are more ions (H3O+ ions) in HCl solution than acetic acid solution.
Therefore HCl is strong acid. Where as acetic acid has fewer H3O+ ions and hence it is weak acid. .
If we carry out the above experiment by taking bases like diluted NaOH (sodium hydroxide) and
diluted NH4OH (Ammonium hydroxide), we will notice that NaOH is strong base and NH4OH is weak
base.
The universal indicator can also be used to know the strength of acid or base.
The universal indicator shows different colors at different concentrations of hydrogen ions in a
solution.
Assessment
1. What may happen if we add water to a concentrated acid? State the reason.
3. What do you mean by a strong acid or base? Explain in your own words.
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STATE COUNCIL OF EDUCATIONAL RESEARCH AND TRAINING
TELANGANA, HYDERABAD.
READINESS PROGRAMME LEVEL - 2
CLASS: X MEDIUM : English SUBJECT : Physical Sciences
Name of the Chapter : Acids, Bases and Salts
TOPIC / CONCEPT : PH Value, Neutral Solution
WORKSHEET NO : 24
Learning outcomes
Children can tell the range of PH values for acids, bases and neutral solutions.
Children can identify acids, bases and neutral solutions by testing with PH paper.
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PH SCALE:
PH value of a solution is simply a number which indicates the acidic or basic nature of a solution.
The PH of neutral solutions is 7.
As the PH value increases from 7 to 14, it represents a decrease in H3O+ ion concentration or an increase in
Compare the colour with the colours in figure given in this worksheet. Identify approximate PH value.
You will notice that all types of fruits juices, vegitable juices are acidic in nature.
Some salt solutions are acidic in nature and some are basic.
The strength of acid or base depends on the concentration of H3O+ ions or OH- ions produced in solution.
HCl and CH3COOH of some concentration will produce different concentration of hydrogen ions.
Acids that give more H3O+ ions are said to be strong. Acids that give fewer H3O+ ions are said to be weak.
Bases that give more OH- ions are said to be strong. Bases that give fewer OH- ions are said to be weak.
SELF ASSESSMENT
2. Give two examples for each acidic, basic and neutral solutions.
A) 0 B) 6 C) 7 D) 14
A) Solution with PH =1
B) Solution with PH =4
C) Solution with PH =7
D) Solution with PH = 14
A) Solution with PH =3
B) Solution with PH =7
C) Solution with PH =9
D) Solution with PH = 13
STATE COUNCIL OF EDUCATIONAL RESEARCH AND TRAINING
TELANGANA, HYDERABAD.
READINESS PROGRAMME LEVEL - 2
CLASS: X MEDIUM : English SUBJECT : Physical Sciences
Name of the Chapter : Acids, Bases and Salts
TOPIC / CONCEPT : Daily life applications of PH
WORKSHEET NO : 25
Learning outcomes
1. ACID RAIN :
Acid rain lowers the PH of water sources like rivers, the survival of aquatic life in such rivers becomes
difficult.
2. TOOTH DECAY :
Tooth decay starts when the PH of the mouth is lower than 5.5.
Tooth enamel, the hardest substance of the body is corroded when the PH in the mouth is bvelow 5.5.
Using tooth pastes, which are generally basic neutralise the excess acid and prevent tooth decay.
During indigestion the stomach produce too much Hcl and this causes paid and irritation.
To get rid of this pain, people use bases called antacids like milk of magnesia, a mild base ( Magnesium
Hydroxide ).
4. PH OF THE SOIL :
If the PH of soil is very low farmers would treat the soil of their fields with quick like or calcium carbonate.
Honey – bee sting leaves an acid in our body which causes pain and irritation.
Use of a mild base like baking soda on the stung area gives relief.
Stinging hair of leaves of nettle plant, inject methanoic acid ( Formic acid ) causing burning pain.
A traditional remedy for this is rubbing the area with the leaf of the dock plant.
SELF ASSESSMENT
3. State the remedy to get rid of the pain due to indigestion problem.
A) 1 B) 5 C) <5.5 D) >5.5
Key concepts Family of salts, Acidic salt, basic salt, Neutral salt.
Learning outcomes
FAMILY OF SALTS:
We can identify the acid and base from which a salt is obtained.
For example, NaCl and Na2SO4 belong to the family of Sodium (Na) salts.
Similarly, NaCl and Kcl belong to the family of chloride (Cl) salts.
PH OF SALTS:
● You will find some salts with acidic nature, some salts with basic nature and some are neutral.
● Salts of a strong acid and a strong base are neutral and the PH value is 7.
● The salts of a strong acid and weak base are acidic and the PH value is less than 7.
● The salts of a strong base and weak acid are basic in nature and the PH value is more than 7.
● If the acid and base both are weak, the PH of the salt depends on the relative strengths of acid and base.
SELF ASSESSMENT
1. Give an example for a family of salts.
2. Why some salts are acidic in nature and some are basic ? State the reason.
2. Give two examples for each acidic, basic and neutral solutions.
A) 0 B) 6 C) 7 D) 14
A) Solution with PH =1
B) Solution with PH =4
C) Solution with PH =7
D) Solution with PH = 14
A) Solution with PH =3
B) Solution with PH =7
C) Solution with PH =9
D) Solution with PH = 13
STATE COUNCIL OF EDUCATIONAL RESEARCH AND TRAINING
TELANGANA, HYDERABAD.
READINESS PROGRAMME LEVEL - 2
CLASS: X MEDIUM : English SUBJECT : Physical Sciences
Name of the Chapter : Acids, Bases and Salts
TOPIC / CONCEPT : Common Salts
WORKSHEET NO : 27
Key concepts Common salts, rock salt, brine, sodium hydroxide, bleaching powder,
chloro – alkali process
Learning outcomes
Salts are the ionic compounds which are produced by the neutralization of acid with base.
Deposits of solid are also found in several parts of the world. This is called rock salt.
The common salt is an important raw material for various materials of daily use such as sodium hydroxide,
When electricity is passed through an aqueous solution of sodium chloride ( called Brine ), it decomposes to
The process is called the chloro – alkali process – because of the products formed chloro for chlorine and
Bleaching powder is produced by the action of chlorine on dry slaked lime [ Ca(OH)2 ].
SELF ASSESSMENT
2. Name any two material which can be produced using common salt.
4. Why the preperation of NaOH from NaCl is called chloro – alkali process?
8. Raw material for the materials like sodium hydroxide, bleaching powder is [ ]
Learning outcomes
Children can explain the process of preparation of Baking soda and write the equation for the chemical
reaction.
Children can explain the process of preparation of Washing soda and write the equation for the chemical
reaction.
Children states the uses of Baking soda, washing soda.
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Baking powder mainly contains baking soda. It produces CO2 which rises through bubling dough into cake
or bread.
Baking soda is also an ingrediant in antacids. Being alkaline, it neutralizes excess acid in the stomach and
provides relief.
SELF ASSESSMENT
1. Explain the process of preparation of baking soda, and write the chemical equation.
2. Write the chemical equation for the reaction takes place when baking soda is heated during cooking.
3. Explain the process of preparation of washing soda, and write the chemical equation.
8. An ingredient in antacids… [ ]
Learning outcomes
crystallisation.
When the crystals are moistened with water, the blue colour
reappears.
Water of crystallisation is the fixed number of water molecules present in one formula unit of salt.
You know that one formula unit of washing soda contains 10 molecules of water. ( Na2CO3.10H2O )
➔ On careful heating of gypsum ( CaSO4.2 H2O ) at 373 K it loses water molecules partiallyt to
➔ Doctors use plaster of paris as plaster of supporting fractured bones in the right position.
➔ Plaster of paris is a white powder and on mixing with water, it sets into hard solid mass due to the
formation of gypsum.
➔ Two formula units of CaSO4 share one molecule of water. That is why the formula of plaster of
➔ Plaster of paris is used for making toys, materials for decoration and for making surfaces smooth.
SELF ASSESSMENT
7. The material which doctors use for supporting factured bones in the right position.. [ ]
❖ Describes the image of an object seen through empty cylindrical shaped transparent vessels such as a glass
tumbler.
❖ Differentiates between the images of an Object seen through empty glass tumbler and glass tumbler filled
with water.
❖ Draws ray diagrams to show the path of light when light ray travels from one medium to another medium.
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CONTENT :
● You know that when light travels from one medium into another medium, it bends at the surface separating
● Does it happen only at the plane surface ? Does it also happen at curved surfaces ?
● In the first case, when the vessel is empty, the image of an arrow is diminished.
● The reason is, light from the arrow refracts at the curved interface, moves through the glass and enters into
air then it again undergoes refraction on the opposite curved surface of the vessel and comes out into air.
● In this way light travels through two media and comes out of the vessel and forms a diminished image.
● In the second case, when the vessel is filled with water, there is a curved interface between air and water.
● We assume that the refractive indices of both water and glass are the same.
● Thus, light from the arrow enters through the curved surface, moves through water, comes out of the glass
● From the above activity, we conclude that refraction takes place at curved surfaces in the same way as we
● The centre of the sphere, of which the curved surface is a part, is called the
● Any line drawn from the centre of curvature to a point on the curved surface
● The direction of the normal changes from one point to another point on the curved surface.
● The centre of the curved surface is called the pole (P) of the curved surface.
● The line that joins the centre of curvature and the pole is called principal axis.
● A ray will bend towards the normal if it travels from a rarer medium to a denser medium.
● A ray will bend away from the normal if it travels from a denser medium to a rarer medium.
SELF ASSESSMENT
1. Draw the curved surface and mark centre of curvature, pole and principal axis.
2. Can you draw two normals from one point on the curved surface ? Explain.
3. Draw a diagram to show that normals drawn at different points on the curved surface are in different
directions.
4. State the similarities in the method of refraction at plane surface and refraction at curved surface in the form
of table.
a. The shape of the object appears to be inverted when seen through an empty cylindrical shaped glass
beaker
b. The shape of the object appears to be inverted when seen through a cylindrical shaped glass beaker
A) a is true B) b is true C) Both a and b are true D) Both a and b are false.
6. Identify the shape of this object when seen through an empty cylindrical glass beaker.
[ ]
A) B) C) D)
7. Imagine the shape of this object when seen through cylindrical glass beaker filled
Learning outcomes :
★ Draws ray diagrams to show the refracted light ray, when the incident light ray parallel to the principal axis
★ Identifies the difference, when light travels from one medium to another medium and undergoes refraction
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CONCEPT :
❖ Refraction of a light ray that travels along the principal axis / a light ray that travels through the centre of
curvature.
❖ The ray which travels along the normal drawn to the surface does not
deviate from its path. Hence both rays mentioned in the adjacent figure.,
CASE (1) :
A ray travelling parallel to the principal axis strikes a convex surface and
passes from a rarer medium to denser medium the refracted ray reaches a particular
CASE (2) :
A ray travelling parallel to the principal axis strikes a convex surface and
passes from a denser medium to rarer medium (see adjacent figure). The refracted ray
moves away from the principal axis. When you extend the refracted ray backwards, it
CASE (3) :
A ray travelling parallel to the principal axis strikes a concave surface and
passes from a denser medium to rarer medium ( see adjacent figure). The refracted
CASE (4) :
A ray travelling parallel to the principal axis strikes a concave surface and
passes from a rarer medium to denser medium ( see adjacent figure ). The refracted
ray moves away from the principal axis. When you extend the refracted ray
FOCUS :
The point where refracted ray intersects the axis in all the above cases is called the focal point.
*** In case of concave and convex mirrors, the focus is the midpoint between the pole and center of curvature.
In any situation focus does not change. But in case of lenses made of transparent material, the focus changes
depending upon the medium surrounding the lens. The point where refracted ray intersect the principal axis is
SELF ASSESSMENT
1. Draw ray diagrams to show the process of refraction of light at different curved surfaces.
2. Draw ray diagrams to show the situation when an incident light ray at a transparent curved surface does not
3. A light ray parallel to the principal axis, when incident on a transparent curved surface, undergoes
refraction and the point where the refracted light ray intersects the principal axis is called [ ]
4. Identify the false statement, for an incident light ray parallel to the principal axis.
i) Travels from rarer medium to denser medium through a concave surface, after refraction bends
ii) Travels from denser medium to rarer medium through a convex surface, after refraction bends
away from principal axis and when produced backward intersects the principal axis at a point.
❖ Draws ray diagrams to who the image formation, when light rays from an object fall opn curved surface.
❖ Deduces formula for refraction at plane surfaces using the curved surface formula.
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CONTENT :
Consider a curved surface separating two media of refractive indices n1 and n2 .
( n1 is considered as rarer medium and n2 is considered as denser medium)
★ Two incident light rays from object O fall on curved surface, undergo refraction as shown in figure.
★ In figure PO is the object distance which we denote as u, PI is the image distance which we denote as v, PC
is the radius of curvature which we denote as R , n1 , n2 are refractive indices of two media.
★ (See pages 60,61,62 of the 10th class English medium book for
derivation)
formula.
Sign conventions :
❖ Distances measured along the direction of the incident light rayt are taken as positive.
❖ Distances measured opposite to the direction of the incident light ray are taken as negative.
❖ The heights measured vertically above from the points on the axis are taken as positive.
❖ The heights measured vertically down from the points on the axis are taken as negative.
SELF ASSESSMENT
1. Write the curved surface formula and explain the terms in it?
3. A man standing near a water body, observes a rock immersed inside the water. Can he see the rock in its
4. A glass is half - filled with water. Diagram shows how long the spoon
5. If a bird flying in the sky looks vertically down and sees a fish in water. How does the fish appear to the
bird ? [ ]
7. A man inside the water sees a fish in water. Will the fish appear to man in its actual position ? Will the man
appear to fish in his actual position ? Imagine the situation and answer.
LEARNING OUTCOMES :
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CONTENT :
By observing the above example, we conclude that image distance V is independent of the refractive
index of the material of the sphere. That means, the apparent position of the dot is the same as its actual
LENSES :
A lens is formed when a transparent material is bounded by two surfaces of which one (or ) both
TYPES OF LENSES ;
Lenses can be of various tyupes. Some typical lenses along with their names are shown in the figure
below.
● A lens may have two spherical surfaces bulging outwards. See fig.6(a). It is thick at the middle as
● A lens bounded by a plane surface on one side and spherical surface on the other side is called plano -
● A lens bounded by a concave surface on one side and convex surface on the other side is called a
● A lens bounded by two spherical surfaces curved inwards is called Bi-concave lens. See fig.6(d). It is
● A lens bounded by a plane surface on one side and concave surface on the other side is called
● A lens bounded by a convex surface on one side and a concave surface on the other side is called
● Bi - convex, plano - convex, concavo - convex lenses are thick in the middle and thinner at the edges.
These lenses are called converging lenses. The reason is when parallel beams of light rays are incident
on these lenses, refraction takes place and refracted light rays meet at a point.
● Bi - concave, plano-concave, convexo-concave lenses are thin at the middle and thicker at the edges.
These lenses are called diverging lenses. The reason is, when parallel beam of light rays are incident
on these lenses, refraction takes place and refracted light rays appear to diverge and when produced
SELF ASSESSMENT
1. A transparent sphere of radius 6ch is kept in air. An object is placed at 12 cm distance from
the surface of the sphere and its real image is formed at the same distance from the second
surface of the sphere. Find the refractive index of the material of the sphere ? ( Ans: 1.5)
2. Mohan tells “When we take two similar transparent spheres made of different material having
a small, opaque dot at its centre, we find the dot appears to be at different positions for two
different spheres when observed from outside” Is he correct ? Justify your answer.
3. An object is placed at 4cm distance from the surface of a transparent sphere made of material
of refractive index 2.5, and the real image of the object is formed at the same distance from the
second surface of the sphere. Find the diameter of the sphere? ( Ans 12 cm)
4. A light ray parallel to the principal axis is incident on Bi - convex lens. Show the process of
refraction with the help of ray - diagram.
1) Lens which is thick at the middle and thin at the edges. a) Bi - convex lens
2) Lens which is bound by two spherical surfaces bent towards b) Concavo - convex lens
the same, Thin at the middle and thicker at the edges. c) Convexo - concave lens
● Identifies the focal point of the lens and measures focal length.
● Draws ray diagrams to show behaviour of light rays when they are incident on a lens.
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CONTENT :
★ The centre of the sphere which contains the part of the curved
★ If a lens contains two curved surfaces then their centres of curvature are denoted as C1 and C2
★ The distance between the centre of curvature and curved surface is called radius of curvature R.
The process of refraction in case of a parallel beam of light incident on a lens is shown in below figure.
❖ A parallel beam of light incident on a lens converges to a point as shown in the first figure.
❖ A parallel beam of light incident on a lens, after refraction seems to emanate from a point on the principal
❖ The point of convergence (or) the point from which rays seem to emanate is called focal point or focus (F)
❖ The distance between the focal point and optic centre is called the focal length of lens denoted by ‘f’.
For drawing ray diagrams related to lenses, we represent convex lens with a symbol and concave lens with a
symbol .
To know the formation of an image by lenses, let us see the behaviour of light rays when they meet a lens.
hile drawing ray diagrams, we consider the lens as a single surface element because we assume that the
NOTE : W
thickness of the lens is very small and show the net refraction at only one of the surfaces.
Bhhaviour of certain light rays when they are incident on lens :
From the above figures we conclude that, Any ray passing along the principal axis is undeviated.
The points C1, C2 are not centre of curvatures. They are the points at ‘2f’ distances from the optic centre.
From the above figures we conclude that, Any ray passing through the optic centre is also undeviated.
SELF ASSESSMENT
1. Explain the process of identifying the focus of a diverging lens, with the help of ray diagram.
4. A Bi-Convex lens is made from two spheres of different radii, focus is marked from both spherical surfaces
and focal length is measured on both sides. Will focal length be equal on both sides? Imagine and write an
answer.
5. The line joining the two centres of curvature of a lens is called ... [ ]
● Draws ray diagrams to explain the behaviour of light rays travelling parallel to the principal Axis and
incident on a lens.
● Draws ray diagrams to explain the behaviour of light rays passing through focus and incident on the lens.
● Identifies a point on a focal plane, where parallel rays, maing an angle with principal axis, fall on a lens and
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CONTENT:-
We know that light rays passing along the principal axis and light rays passing through the optic centre are
Let us know the behaviour of certain light rays when they are incident on a lens.
From the above figures, we conclude that light rays passing parallel to the principal axis converge at the focus or
From the above figures, we conclude that the light rays passing through the focus will take a path parallel to
principal axs after refraction. Light rays obey the principle of least time.
What happens when parallel rays of light fall on a lens making
Before knowing about such light rays, let us define a focal plane.
FOCAL PLANE : - Focal plane is the plane perpendicular to the
principal axis at the focus.
When parallel rays, making an angle with principal axis, fall on a lens, the rays converge at a point or
appear to diverge from a point lying on a focal plane.
Observe the following figures.
SELF ASSESSMENT
1. What do you mean by the term Focal plane? write in your own words.
2. How do you identify the focal point of a Bi-concave lens? Explain with the help of a ray diagram.
3. Explain the behaviour of light rays passing through focus and incident on Bi-convex lenses.
A) Bi-convex lens
B) Bi-concave lens
C) Concavo convex
D) Convexo concave
STATE COUNCIL OF EDUCATIONAL RESEARCH AND TRAINING
TELANGANA, HYDERABAD
ACADEMIC YEAR 2020-21
LEVEL - 2
CLASS: 10 MEDIUM: ENGLISH SUBJECT: PHYSICAL SCIENCE
Concept:
1. Rules to draw Ray diagrams for image formation by lenses.
2. Image formation by a convex lens for various positions of object and to describe the nature of image
formed.
Learning Outcomes:
Explains the rules for drawing Ray diagrams for image formation by lenses
Draws ray diagram to explain image formation by a convex lens when the object is at infinity.
Describes the nature of image formed by convex lens when an object is placed beyond the centre of curvature
on the principal axis and when an object is placed at the centre of curvature draws Ray diagrams.
…………………………………………………………………………………………
Rules to draw Ray diagrams for image formation by lenses:
2. Describe the nature of an image formed by an object placed beyond the centre of curvature on the
principal axis of a convex lens.
3. Draw a ray diagram and write the characteristics of an image formed when an object is placed at
centre of curvature of a convex lens.
4. Draw a ray diagram to show the image formation by a convex lens, when an object is placed beyond
the centre of curvature on the principal axis with the help of incident light ray parallel to the principal
axis and light ray passing through focus.
5. If an object is placed beyond the centre of curvature on the principal axis of a convex lens, then the
height of an image formed, when compared to the height of an object is. ( )
6. Which characteristics of an image are not true, when an object is placed beyond the centre of
curvature on the principal axis of a convex lens? ( )
------------------------------- * * * * * ------------------------------
STATE COUNCIL OF EDUCATIONAL RESEARCH AND TRAINING
TELANGANA, HYDERABAD
ACADEMIC YEAR 2020-21
LEVEL - 2
CLASS: 10 MEDIUM: ENGLISH SUBJECT: PHYSICAL SCIENCE
Lesson: Refraction of light at curved surfaces Work sheet No: 37
Concept
Image formation by a convex lens for various positions of the object and to study the characteristics of the
images formed.
Learning Outcomes:
Explains image formation by a convex lens for various positions of the object with the help of ray
diagrams.
Describes the characteristics of the images formed
Differentiate between the images formed by convex lens and concave lens.
Content
We have learnt about the characteristics of the image formation when the object is placed at
infinity, beyond centre of curvature and at centre of curvature of a bi-convex lens.
When an object is placed between centre of curvature (C2) and focus (F2), we will get an image which is
the real inverted and magnified the image will form beyond C. See figure 15.
Situation 5:- Object located at the focal point:
When an object is placed at the focus (F2) the image will be at infinity. See figure 16. As the image is
formed at infinite distance, we cannot discuss the size and nature of the image.
Situation 6:-Object placed between focal point and optic centre:
If we place an object between focus and optic centre, we will get an image which is virtual, erect and
magnified. From the ray diagram shown in figure 17, you will notice that image is formed on the same
side of the lens. From the above situation of image formation, we understand two things.
As the image formed is virtual, we can see it with our eyes. In all other cases the images is real which
we can't see with our eyes but can be viewed if the image is captured on the screen
A magnified virtual image is formed on the same side of the lens, where the object is placed thus the
image you are seeing through a lens is not real, it is a virtual image of the object.
When the object is placed at a distance less than the focal length of the lens, we get a virtual magnified
image. This particular behavior of convex lens helps to construct a microscope, which gives magnified
image of a small object.
Let us draw ray diagram for an object placed between C1 and F1 for a concave lens.
From figure 18, we observe that a virtual, erect and diminished image is formed between optic centre and
focus (F1)
Try to draw ray diagrams for other position of object for concave lens, as you have done in case of
convex lens. You will notice that irrespective of the position of object, on the principal axis, you will
get same characteristics of the image as shown in figure 18 for all cases.
Assessment
1) Write similarities/differences between the images formed by convex lens and concave lens?
2) Prepare a label to show the characteristics of an image formed when the object is placed at different
positions on the principal axis of a convex lens?
3) When you place an object at the focal point of a concave lens, what will be the characteristics of image
formed imagines and write.
4) Draw ray diagrams for an object placed beyond the centre of curvature (C1) and object placed at the
centre of curvature (C1) on the principal axis of a concave lens.
5) Choose the correct answer ( )
i convex lens can form both real and virtual images.
ii. Concave lens can form only virtual images.
A.) i) is true ii) is false B) i) is true ii) is true C).i) is false ii) is true D) i) is false ii) is false
6) The position of an object to obtain a virtual image using a convex lens should be ( )
A) At Infinity B) at centre of curvature
C) Centre of between centre of curvature and focus D) between focus and optic centre
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STATE COUNCIL OF EDUCATIONAL RESEARCH AND TRAINING
TELANGANA, HYDERABAD
ACADEMIC YEAR 2020-21
LEVEL - 2
CLASS: 10 MEDIUM: ENGLISH SUBJECT: PHYSICAL SCIENCE
Lesson: Refraction of light at curved surfaces Work sheet No: 38
Learning Outcomes:
Draws refracted light ray for any given incident light ray on convex or concave lens.
………………………………………………………………………………………
Content:
Example:
Draw a ray diagram to locate the position of image when a point sources is placed on optical axis MN of
a convex lens, in such a way that it is beyond focal point F 2. See figure E(4)
Solution:
Draw a perpendicular line to principal axis passing through the focus F 1. This line represents focal plane.
Draw a ray from point source (S) in any direction to meet lens at point P 1
Now draw another line parallel to sp1 such that it passes through optic centre P
Extend this line till it intersects the focal plane at F 0.
Now draw a line passing from point P 1 to pass through the point F0, such that it meets principal axis
at a point I
I is the point image for the point source (S)
Example:
Complete the ray diagrams to show the paths of rays after refraction through the lenses shown in the
figure E 5 (a) and E5 (b)?
Solution:
1. In figure E 5 (a), we see a convex lens. So, following the steps mentioned in the above example, we
get the path of light rays after refraction as shown in figure E 5(c)
fig-E5(d)
ASSESSMENT
1. As shown in figure below, if an object O is placed on the principal axis of a concave lens, find the
position of the image formed with the help of ray diagram .write down the steps involved in the method.
O F1 F2
P
F2 P F1 F1 F2
F1 F2 F2 F1
F2 F1
4. For concave lens the focal plane is represented by a line drawn perpendicular to the principal Axis at a
point ( )
A) Optic centre B) Focal point towards the object C) Centre of curvature towards the object
D) Focal point on the other side of the object.
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STATE COUNCIL OF EDUCATIONAL RESEARCH AND TRAINING
TELANGANA, HYDERABAD
ACADEMIC YEAR 2020-21
LEVEL - 2
CLASS: 10 MEDIUM :ENGLISH
ENGLISH SUBJECT: PHYSICAL SCIENCE
Concept:
1. Observing the types of images obtained in the ray diagrams of convex lens by performing a lab
activity.
2. Measuring the object distance and image distance from the lens.
3. Lens formula:
Learning Outcomes:
Proves that characteristics
stics of images formed through rray
ay diagrams are correct by performing an activity.
Solves problems using lens formula.
………………………………………………………………………….........................
…………………………………………………………………………
Content:
Perform the lab activity from the above activity
activity, we observe the following points.
We cannot get an image on the screen for every object distance.
To get an image, the object distance should be greater than the focal length of the lens.
When we do not get an image on the screen screen, you can see the image with your eye directly through
the lens.
You will see a magnified image on the same side where we kept the object.
This is the virtual image of the object which we cannot capture on the screen.
This virtual image is formed between fo focus and optic centre. Thuss the image distance in this case
is less than the focal length of the lens.
Could you find focal length of the lens from the values of object distance (u) and image distance
(v) recorded in table 1
Relation between u, v and f
Observe the figure 19
= -
This equation is called lens formula. It can be used for any lens while using this equation we should use
sign convention
Assessment
1.Magnification
Concept
Learning Outcomes:
outcomes solves problems using lens formula and magnification formula
…………………………………………………………………………
………………………………………………………………………….........................
Content:
Let
et us discuss the size of the image formed by a lens
OI
F1 I
ho
F2 P hi
u v I I
fig-20
Fig.20
= => =
=
Magnification m = =
Using lens formula and magnification formula letus solve some problems.
Example:
An electrical lamp and a screen are placed on the , in a line at a distance of 1 m
In what positions of convex lens of focal length of f= 21 cm will the the image of lamp be sharp?
Solution:
The distance between the lamp and the screen is 100 cm and` x’ be the distance between the lamp and the lens
from figure E(6) we have
u =- x
v=100-x
f=21
= -
We get
= +
After solving this equation,we get
X2-70x-30x+2100
X(x-70)-30(x-70)=0
⇒x(x-70)-30(x-70)=0
c(x-700 (x-30)=0
∴x=70 or30
We can place the bulb at adistance of 70cm or 30cm to get a sharp image
Example:
An object of height 2cm is placed on the principal axis at a distance of 45cm from aconvex lens of focal
length 15cm .find the position and the height of the image formed.
Solution:
= -
= -( )
⇒ = +
⇒ = - = =
Magnification =
u = 45 cm
v = 22.5 cm
ho = 2 cm
.
hi = = -1 cm
Assessment
Proves that the focal length of lens depends up on the surrounding medium in which it is kept by
performing an experiment
Explains the terms in lens Makers formula
Solves problems using lens maker's formula
--------------------------------------------------------------------------------------------------------------------
CONTENT:
Let us see on what factors the focal length of the lens depends
Perform activity 2
From the above activity we observe that focal length of lens in water has increased when compared to
focal length of the lens in air.
Thus we conclude that, the focal length of the lens depends upon the surrounding medium in which it is
kept
Lens maker's formula
- )
n – Refractive index of lens in the air
f – Focal length of the lens
R1, R2 – radii of the curvature
Note: If the refractive index of the convex lens is more than the refractive index of the surrounding
medium then it is behaves like a converging lens. If the refractive index of the convex lens is less
than the refractive index of the surrounding medium then it behaves like a diverging lens.
Example: Air bubble in water behaves like a diverging lens
Let us see an example for lens maker formula
Example
What is the focal length of double concave lens kept in air with two spherical surfaces of radii R 1 =30 cm
and R2 = 60 cm. Take refractive index of lens as n = 1.5
Solution
From figure E (7), using sign convention we get
R1 = 30cm
R2=60cm
n=1.5
Using lens maker's formula,
- )
- )
1. Write the procedure and the precautions to be taken in an experiment to verify that focal length of
lens depends on the surrounding medium.
2. Write the lens maker's formula and explain the terms in it.
3. When can a converging lens act as a diverging lens? Explain.
4. Find the radii of curvature of a concavo- convex convergent lens made of glass with the refractive
index n = 1.5 having focal length of 20% one of the radii of curvature is double the other
5. Find the refractive index of the glass which is symmetrical convergent lens if its focal length is equal
to the radius of curvature of its surface (note): A symmetrical convergent lens is the one that has
both radii of curvature equal)
6. Which of the following is the lens maker's formula ( )
A) + )
B) - )
C) - )
D) + )
7. Focal length of the Plano - convex lens is ______when its radius of curvature of the surface is R and n is
the refractive index of the lens ( )
A) f=r B) f= C) f= D) f=
…………………………………..*****………………………………………..
STATE COUNCIL OF EDUCATIONAL RESEARCH AND TRAINING
TELANGANA, HYDERABAD
ACADEMIC YEAR 2020-21
Learning Outcomes:
finds least distance of distinct vision for any e person by performing an activity
Gives reasons for variations in least distance of distinct vision from person to person
---------------------------------------------------------------------------------------------------------------------------------
Content:
The human eye function on the principle of sensation of vision. Let us do the following
activities to know about some interacting facts about our vision.
Activity 1.
Take a text book and hold it with your hands in front of you at a certain distance. Now try to read the
contents on the page. Slowly move the book towards your eye till it is very close to your eyes.
What changes do you notice?
You may see that printed letters on the page of the textbook appear blurred or you feel strain in the
eye.
Now slowly more the book backwards to a position where you can see clear printed letters without
straining your eye. Ask your friend to measure the distance between your eye and textbook at this
position. Note down its value.
Repeat the activity with other friends and note down the distance is for distinct vision in each case.
Find the average of all these distances of clear vision.
2. Angle of vision
The maximum angle which is able to see the whole object is called angle of vision.
The angle of vision for a healthy human being is about 60 degree
It varies from person to person with age.
Assessment
1. What what do you mean by least distance of distinct vision? Write in your own words.
2. What do you mean by angle of vision? write in your own words.
3. There is a large variation seen in the values of least distance of distinct vision in case of in children and
old age people. give reason
4. Write the procedure to find the angle of vision
5. For healthy adult the value of least distance of distinct vision is ( )
A) 20cm B) 25cm C) 15cm D) 30cm
7. For children below 10 years of age the least distance of distinct vision will be ( )
A) 25cm B) 1 to 2 m C ) 7 cm to 8 cm D)20 cm
8. For old age people the least distance of distinct vision will be ( )
A) 25cm B) 1 mt to 2 mt. C) 7 cm to 8 cm D). 20 cm
-*************
STATE COUNCIL OF EDUCATIONAL RESEARCH AND TRAINING
TELANGANA, HYDERABAD
ACADEMIC YEAR 2020-21
Learning Outcomes
Content : To know the reason for the variation of values of least distance of distinct vision and
angle of vision with person and age ,we need to understand the structure of eye and its
functioning.
The human eye is one of the most important sense organs. It enables as to use the
object and colors around us.
Assessment
1. Which part of the Human eye is called pupil?
2. What is the function of rods present in the retina?
3. What is the function of iris inhuman eye?
4. Describe briefly the structure of human eye with the help of neat labeled diagram.
-*************
STATE COUNCIL OF EDUCATIONAL RESEARCH AND TRAINING
TELANGANA, HYDERABAD
ACADEMIC YEAR 2020-21
LEVEL - 2
CLASS: 10 MEDIUM :ENGLISH SUBJECT: PHYSICAL SCIENCE
Lesson: Human eye and colorful world Work sheet No: 44
1. Accommodation of lens
Concept
Learning Outcomes:
Explains the relationship between the accommodation of lens and the ability of eye lens to see
objects placed at various positions in front of eye
…………………………………………………………………………
………………………………………………………………………….........................
Content:
Generally, when we use a convex lens, if focal length is constant, whenever there is change in
object distance, there will be a chance in image distance. In case of eye lens, for any position of
object in front of the eye, the image is formed on retina at distance of 205 cm from eye lens. i.e.
for different positions of object, the image distance is fixed and is about 2.5 cm . This is possible
only when there is change in focal length of eyeye lens.
The focal length of the lens depends on the material by which it has made and radii of curvature
(Thickness) of lens
Eye lens is able to change its shape and hence there will be a change in radii of curvature. Thus
,it cais possible to cha
change focal length of eye lens to get same image distance for various
positions of object in front of eye
The sillier muscle to which eye lens is attached helps the eye lens is attached helps the eye lens
to change its focal length by changing the radii of curvature
urvature of the eye lens
Case 1:
When the image is focused on a distance object , the ciliary muscles are relaxed so that the focal length of eye
lens has its maximum value which is equal to its distance from the retina .The parallel rays coming into the eye
are then focused on to the retina and we see the objects
clearly.
u= - ∞; v = 2.5 cm (image distance which is equalto distance between eye-lens and retina)
f max = 2.5 cm
Case 2:
When the eye is focused on a closer object (not less than 25 cm, which is the least distance of distinct vision).
The ciliary muscles are strained and length of the eye lens decreases. The ciliary muscles adjust the focal length
in such a way that the image is formed on retina and we see the object clearly.
Consider an object placed at a distance of 25 cm from our eye. In this situation, eye has
minimum focal length
Note: ciliary muscles cannot strain beyond a limit and hence if the object is brought too close to
eyes , the focal length cannot be adjusted to form an image on retina . Thus there should be a
minimum distance for distinct vision of object which is equal to 25 cm (Least Distance of distinct
vision)
Assessment
1. What do you understand by Accommodation of lens .Write in your own words
2. To Have maximum focal length for an eye lens , at what distance should an object be placed in front
of an eye
3. Explain the situation for an eye lens to have minimum focal length
8. If the focal length of the eye lens has minimum value, then the object distance will be ( )
A) 2.5 cm B) 2.27 cm C ) 25 cm D) A and B
-----------------------------------------------------------******--------------------------------- ------------------------
STATE COUNCIL OF EDUCATIONAL RESEARCH AND TRAINING
TELANGANA, HYDERABAD
ACADEMIC YEAR 2020-21
LEVEL - 2
CLASS: 10 MEDIUM :ENGLISH SUBJECT: PHYSICAL SCIENCE
Lesson: Human eye and colorful world Work sheet No: 45
Learning Outcomes:
Explains the common defects of vision arising due to accommodation defects of the eye lens
………………………………………………………………………….........................
Content:
Human eye helps us to see carious objects around us by adjusting the focal length of eye lens between 2.27cm
to 2.5 cm and forming a clear image on retina. Sometimes the eye may gradually lose its ability for
accommodation the person cannot see an object clearly and comfortably.The vision becomes blurred due to
accommodation defect of the lens.There are mainly three types of defect vision.
They are
(i) Myopia
(ii) Hypermetropia
(iii) Presbyopia
Myopia:
Some people cannot see objects at long distances but can see nearby objects clearly. This type of
defect in vision is called “Myopia “
It is also called “Nearsightedness”
For these People the maximum focal length of eye lens is less than 2.5 cm
In such cases the rays coming from distant objects , after refraction through the eye-lens, forms
an image in front of the retina as shown in fig 5(a) and 5(b)
A healthy person can see objects at all distances more than 25 cm clearly but a person
with myopia can see objects clearly up to a certain distance
Let the extreme point from where an object appears clearly to a person with myopia be
“M”
In general, a healthy person (with no eye defects) can see clerly closer objects (at a distance more than
least distinct vision) and farther objects (as far as objects at infinite distance)
1. How many types of defects of vision are there due to accommodation defects of eye lens? What are they?
2. Explain the eye defect Myopia?
3. Give reasons for getting the eye- defect Myopia
4. A person suffering from Myopia, cannot see the farther objects . Show with the help of diagram, Objects at
what distance they cannot see clearly
B) Can see objects placed between far point and least distance of distinct vision
6. A healthy person ( )
C ) Can see clearly any object placed between infinity and least distance of distinct vision
D) A and B
--------------------------------------------------------------*****----------------------------------------------------------------
STATE COUNCIL OF EDUCATIONAL RESEARCH AND TRAINING
TELANGANA, HYDERABAD
ACADEMIC YEAR 2020-21
Concept: Correction of Myopia, Finding Focal length of Bi-concave lens used for
correcting Myopia.
Learning Outcomes:
Gives reasons for the use of Bi-concave les for correcting Myopia.
Explains the method of finding focal length of Bi-concave lens use for correcting Myopia.
…………………………………………………………………………………………
Content
People with this defect Myopia cannot see objects beyond far point.
The eye lens can form clear image on the retina , when an object is placed between far point and
point of least distance of distinct vision.
Using a lens , if we are able to bring the image of the object kept beyond far point, between the
far point ( M ) and the point of least distance of distinct vision ( L ) , this image acts as an
object for the eye lens.
This can be made possible by the use of Bi-concave lens.
Image :
Now let us know how to decide the focal length of the lens to be used to correct Myopia.
In general, for an object at infinity, the Bi-concave lens forms virtual image at its focus.
For this reason, to correct one’s Myopia we select a Bi-concave lens which forms an image at
the far point from an object at infinity.
This image acts like an object for the eye lens. Hence the final image is formed on the retina.
= –
= + = [ f = -D]
Here ‘f’ is negative showing that it is concave lens.
Example :
A Person is unable to see objects beyond 4 m distinctly. What should be the focal length of Bi-
concave lens that must be used to correct his Myopia.
Solution :
= –
= –
= +
= f = -4mts.
f = -4
To correct Myopia, one should use Bi- concave lens of focal length equal to Maximum distance up
to which he can see clearly.
Assessment
1. Where will the image of an object be formed by the lens used for correcting Myopia ?
4. How can you decide the focal length of the lens to be used to correct Myopia ? Deduce Mathematically?
6. Keeping in view Myopia, chose the correct answer from the following. ( )
a) F = -D b) f = +D c) f = 1/-D d) f = 1/+D
a) Forms real image b) forms virtual image c) forms the other side of the lens you see the image
is formed before the object d) a & b.
------------------------------- * * * * * ------------------------------
STATE COUNCIL OF EDUCATIONAL RESEARCH AND TRAINING
TELANGANA, HYDERABAD
ACADEMIC YEAR 2020-21
TOPIC: Hypermetropia
Learning Outcomes :
Gives reasons for type of defect in vision called Hypermetropia.
------------------------------------------------------------------------------------------------------------------------
Content: For a healthy person , the maximum focal length of the eye lens will be 25 cm and minimum
focal length of the eye lens will be 2.27 cm. Such person can see far objects as well as near objects clearly
and distinctly.
If the minimum focal length of the eye lens is greater than 2.27 cm , then it would mean that ciliary muscles
are unable to adjust focal length to form an image on the retina. Ciliary muscles cannot strain beyond a limit
to decrease the focal length of eye lens. In this situation , the objects at near distance cannot be seen. This
type of defect in vision is called Hypermetropia.
Hypermetropia
Hypermetropia is also known as farsightedness.
A person with hypermetropia can see far objects clearly and distinctly but cannot see objects at near
distance.
In Hypermetropia, the rays coming from a nearby object after refraction at eye lens, forms an image
beyond the retina as shown in figure 6 (a).
Figure 6 (a)
Let the point of least distance at which the eye lens forms a clear image on the retina for a person with
Hypermetropia be ‘H’ . See figure 6 (b)
If an object is at ‘H’ or beyond ‘H’ , the eye can form its image on retina . See Figure 6 (a) and 6 ( c)
If the object is between ‘ H’ and point of least distance of distinct vision ‘L’ , then it cannot form an image.
See figure 6 (a)
Figure 6 (a)
Near Point :
The point of minimum distance at which the eye lens can form an image on the retina is called near point
‘H’. The people with defect of hypermetropia cannot see objects placed between the near point ‘H’ and the
point of least distance of distinct vision ‘L’ .
Assessment
1. What is hypermetropia? Write in your own words.
2. Draw a ray – diagram to show objects placed at what distance from an eye can be seen by a person
suffering from hypermetropia
4. Draw a ray diagrams to show objects placed at what distance from an eye cannot be seen by a person
suffering from hypermetropia.
6. For a person with hypermetropia, the image of an object placed at 25 cm distance from the eye
( )
A) forms on retina B) forms behind retina C) forms in front of retina D) A & C
------------------------------- * * * * * ------------------------------
STATE COUNCIL OF EDUCATIONAL RESEARCH AND TRAINING
TELANGANA, HYDERABAD
ACADEMIC YEAR 2020-21
Topic : Correction of Hypermetropia, finding focal length of bi-convex lens used for correcting
Hypermetropia.
LEARNING OUTCOMES
Gives reasons for the use of Bi-convex lens for correcting Hypermetropia.
Explains the method of finding focal length of Bi-convex lens used for correcting
Hypermetropia.
…………………………………………………………………………………………
Content
‘f ‘ is measured in cm
We know that d >25 cm, then ‘f’ becomes Positive.
That is we need to use biconvex lens to correct the defect of hypermetropia.
Eg : A person can see only objects placed beyond 33.33 cm from the eye clearly.
Name the type of defect in vision , the person is suffering from . What should be the focal length of the lens
that must be used to correct the defect in vision ?
Solution : -
Since the person cannot see objects placed at a distance less than 33.33 cm from the eye, he must be suffering
from hypermetropia. Biconvex lens can be used to correct his defective vision.
f = +100cm
Positive sign indicates Bi-convex lens.
Assessment
1. Where will the image of an object be formed by the lens used for correcting Hypermetropia ?
4. How can you decide the focal length of the lens to be used to correct Hypermetropia ? Deduce
Mathematically ?
6. Keeping in view Hypermetropia, chose the correct answer from the following. ( )
a.
b.
c.
d.
7. A person suffering from Hypermetropia, uses a lens whose focal length is +100 m. ( )
c) forms image behind the object when seen form the other side of the lens
d) b & c.
------------------------------- * * * * * ------------------------------
LEARNING OUTCOMES :
● Gives reasons for getting vision defect presbyopia, explains the type of lens to be used for correcting
presbyopia.
● Determines the type of lens and finds its focal length based on the value of power of lens.
@@@
Presbyopia: -
★ Presbyopia is a vision defect when the ability of accommodation of the eye usually decreases with
ageing.
★ For most people the near point gradually recedes away. They find it difficult to see nearby objects
★ This happens due to gradual weakening of ciliary muscles and diminishing flexibility of the eye lens.
★ Sometimes a person may suffer from both myopia and hypermetropia with ageing. This type of
★ Its upper portion consists of the concave lens and lower portion consists of the convex lens.
If you go to an eye hospital, to get tested for vision defects, the doctor gives you a prescription
that contains some information regarding the type of lens indicating their power, to be used to
correct vision.
The power of a lens determines the type of lens to be used and its focal length.
POWER OF LENS : -
The degree of convergence or divergence of light rats that can be achieved by a lens is expressed in
The reciprocal of focal length is called power of lens. Let ‘f’ be the focal length of lens,
Power of lens
Ex : - 1) Doctor advised to use +2D lens. What is its focal length ? Mention the type of lens.
using
The lens has focal length + 50 cm. ‘+’ sign of focal length indicates Bi-convex lens. Hence, the
Ex : - 2) Doctor advised to use -4D lens. What is its focal length ? Mention the type of lens.
The lens has focal length -25 cm. ‘-’ sign indicates Bi-concave lens.
SELF ASSESSMENT
3. A person is suffering from both Myopia and hypermetropia. What type of lens should be used to
4. Sita goes to an eye hospital and gets tested for vision defects. The Doctor prescribes her the use of -3D
lens. What type of defect in vision sita is suffering from? Mention the type of lens and find the focal
A) Metre B) Dioptre
C) Centimetre D) A and C
A) Bi-convex lens
B) Bi-concave lens
C) Bi-focal lens
D) A and B
7. Power of lens p = [ ]
● Defines angle of prism, Angle of incidence, Incident ray, Angle of deviation, Emergent ray, Angle of
@@@
PRISM : -
A prism is a transparent medium reparation from the surrounding medium by at least two
plane surfaces which are inclined at a certain angle in such a way that, light incident on one of the
★ To understand the behaviour of light when it is incident on the plane of a prism and passes into the
★ Let us consider that triangle PQR represents the outline of the prism where it rests on its triangular
base.
★ Let us assume that a light ray is incident on the plane surface PQ of a prism at M, as shown in above
figure.
★ The angle between the incident ray and normal is called angle of incidence (i1).
★ The ray is refracted at M. It moves through the prism and meets the other plane surface at N and
★ The ray which comes out of the surface PR at N is called emergent ray.
★ The angle between the emergent ray and normal is called angle of emergence (i2).
★ The angle between the plane surfaces PQ and PR is called the angle of prism or refracting angle of
prism (A).
★ The angle between the incident ray and emergent ray is called angle of deviation (d).
SELF ASSESSMENT
A) i1
B) i2
C) A
D) d
A) i1
B) i2
C) A
D) d
LEARNING OUTCOMES :
@@@
To study the refraction of light through a triangular prism, do an activity given in page
Repeat the above activity for various angles of incidence [ i = 30o,40o,50o….] and find the
★ In the above table observe the values of angles of incidence (i1) and angles of deviation (d)
★ You will notice that the angle of deviation decreases first and then increases with increase in angle of
incidence.
★ Draw a graph, taking angle of incidence (i1) along X- axis and angle of deviation (d) along Y-axis
using a suitable scale, markl points on a graph paper for every pair of angles. Join the points to
★ Draw a tangent line to the curve, parallel to X-axis, at the lowest point of the graph.
★ The point where this live cuts the Y-axis gives the angle of minimum deviation. It is denoted by D.
★ Draw a parallel line to the Y-axis through the point where the tangent touches the graph. This line
meets the X-axis at a point showing angle of incidence corresponding to the minimum deviation.
★ With this angle of incidence, repeat the above activity and find corresponding angles of emergence.
★ You will notice that angle of incidence will be equal to angle of emergence (i1= i2) in this case.
★ We can find the refractive index of a prism, by substituting the values of Angle of Prism (A) and
Solution:
2. List the material required to perform the lab activity for finding the refractive index of a prism?
3. A prism with angle A = 60o produces an angle of minimum deviation of 30o. Find the refractive
C) i2 =
d D) i1 > i2
6. Angles to be taken along the X, Y-axes for the graph to determine the
A) i1 along
X-axis, i2 along Y-axis
B) i2 along
X-axis, i1 along Y-axis
● Gives reason for greater deviation of violet colour than red colour, during the process of dispersion of
light.
@@@
❖ By doing activity 4 also, you will observe white light falling on a mirror partially immersed in water,
split into different colours. In this activity, the arrangement of mirror in water behaves like a prism.
❖ In activity 3, you will also observe that red light is less deviated than blue light.
DISPERSION OF LIGHT:
FREQUENCY:
The frequency of light is the property of the source and it is equal to the number of waves leaving the
★ Thus coloured light passing through any transparent medium retains its colour.
★ While refraction occurs at the interface, the number of waves that are incident on the interface in a
second must be equal to the number of waves passing through any point taken in another medium.
★ This means that the frequency of the light wave remains unaltered while its wavelength changes
★ We know that the relation between the speed of wave (v), wave length (λ), and frequency (𝝂) is
v = 𝝂λ
★ For refraction at any interface, the speed of wave (v) is proportional to wavelength (λ).
★ Speed of the wave increases with increase in wavelength of light and vice versa.
★ We noticed that white light passes through a prism, it splits into seven colours.
★ If we compare the wavelengths of seven colours in VIBGYOR, red colour has the longest wavelength
★ As violet colour has the shortest wavelength, its refractive index is high and hence it suffers more
deviation.
SELF ASSESSMENT
2. List out the materials required to do an activity using plane mirrors, to show splitting of white light
3. Write an activity showing splitting of white light into various colours using a prism.
4. Write the precautions to be taken in the activity, using a plane mirror to show dispersion of light.
5. Which of the following quantities does not change during refraction? [ ]
A) Red colour has more wavelength, violet colour has less wavelength
B) Red colour has more speed, violet colour has less speed.
C) Red colour hasless refractive index, violet colour has more refractive index.
D) All the above.
8. The relation between speed of wave (v), wave length (λ), and frequency (𝝂) is [ ]
LEARNING OUTCOMES :
Rainbow is a good example of dispersion of light. Sometimes when it rains, we can see a
Do activity 5 given in page no.98 of your textbook and observe the results.
➔ Are the sun rays coming back to your eyes from the wall or from water drops?
➔ The beautiful colours of the rainbow are due to dispersion of the sunlight by millions of tiny
water droplets.
➔ At this first refraction, the white light is dispersed into its spectrum of colours, violet being deviated
➔ Reaching the opposite side of the drop, each colour is reflected back into the drop because of total
internal reflection.
➔ Arriving at the first surface of the drop, each colour is again refracted into air.
➔ At the second refraction the angle between red and violet rays further increases when compared to
➔ The angle between the incoming and outgoing rays can be anything between 0o and about 42o We
observe bright rainbows when the angle between incoming and outgoing rays is near the maximum
angle of 42o.
➔ The red colour will be seen when the angle between a beam of sunlight and light sent back by a drop
is 42o.
➔ The colour violet will be seen when the angle between a sunbeam and light sent back by a drop is 40o.
➔ If you look at an angle between 40o and 42o you will observe the remaining colours of VIBGYOR.
➔ Although each drop disperses a full spectrum of colours, an observer is in a position to see only a
single colour from any one drop depending upon its position.
➔ If violet light from a single drop reaches the eye of an observer, red light from the same drop can’t
➔ The rainbow you see is actually a three dimensional cone with the tip at your eye as shown in figure.
➔ All the drops that disperse the light towards you lie in the shape of the cone - a cone of different
layers.
➔ The drops that disperse red colour to your eye are on the outermost layer of the cone.
➔ Similarly, the violet colour cone becomes the innermost cone. Remaining colour cones can be seen
3. What happens to the rays of sunlight on entering the water drop? Show with the help of diagrams.
5. The colour red will be seen when the angle between a beam of sunlight and light sent back
by a drop is [ ]
6. The colour violet will be seen when the angle between a beam of sunlight and light sent back
by a drop [ ]
is
B) 7 B) 2 C) 1 D) 3
8. The sequence of phenomena that occurs inside the water drop during dispersion of sunlight
[ ]
➢ Atoms or molecules which are exposed to light absorb light energy and emit some part of the light
➢ The effect of light on a molecule or an atom depends on the size of atom or molecule.
➢ If the size of an atom or molecule is small, it will be affected by higher frequency light.
➢ If the size of an atom or molecule is a bit, it will be affected by lower frequency light.
➢ This atom or molecule responds to the light whenever the size of the atom or molecule is comparable
➢ Due to these vibrations, the atom re-emits a certain fraction of absorbed energy in all directions with
different intensities.
➢ The re-emitted light is called scattered light.
SCATTERING OF LIGHT:
The process of re-emission of absorbed light in all directions with different intensities by atoms or
➔ The angle between the incident light and a direction in which intensity of scattered light is observed is
➔ For an angle of scattering less than 90° or more than 90°, the intensity of scattered light decreases.
★ Our atmosphere contains different types of molecules and atoms. The reason for blue sky is
due to the molecules N2 and O2. The sizes of these molecules are comparable to the wavelength
of blue light. These molecules act as scattering centres for scattering of blue light.
★ When you look at the sky in a direction perpendicular to the direction of the sun rays, you can
★ If opur angle of view is changed, the intensity of blue colour also changes.
★ On a hot day, due to rise in temperature water vapour enters into the atmosphere which leads
★ The size of water molecules is greater than the size of N2 and O2.Thus, water molecules act as a
scattering centre for other frequencies which are lower than the frequency of blue light.
★ Hence, on a hot day, all such colours of other frequencies reach your eye and the sky appears
white.
● The light from the sun needs to travel more distance in the atmosphere during sunrise and sunset to
reach your eye. In morning and evening times, during sunrise and sunset, except red colour light all
● Since scattering of red light is very small, it reaches you. As a result the sun appears red in colour
● During noon hours, the distance to be travelled by the sun rays in the atmosphere is less than that
compared to morning and evening hours. Therefore all colours reach your eye without much
● Do Activity 6, to demonstrate scattering of light and observe the results. Refer Activity 6 in page
● Due to chemical reaction between Hypo and sulphuric acid, sulphur precipitates.
● At the beginning, the size of grains is small and almost comparable to the wavelength of blue light.
● As the reaction progresses, the size of sulphur gains increases due to precipitation.
● As the size of grains increases, their size becomes comparable to wavelengths of other colours. As a
result of this, they act as scattering centres for other colours and hence they appear white in colour.
● Thus, sulphur grains appear blue in colour at the beginning and slowly their colour becomes white as
3. If the phenomenon of scattering of light were not there, then how would the sky appear. Imagine
4. The intensity of blue colour is not uniform throughout the sky. Give reason based on the concept of
scattering of light.
7. The process of re-emission of absorbed light in all directions with different intensities by atoms or
C) Equal to 90o
D) A and B
STATE COUNCIL OF EDUCATIONAL RESEARCH & TRAINING
TELANGANA, HYDERABAD.
ACADEMIC YEAR: 2020-21, LEVEL-2
Class: X Medium: English Subject: physical science
Name of the lesson: Structure of atom
Name of the topic/concept: Wave nature of light. WORKSHEET: 55
CONCEPTS:
1. Wave nature of light.
LEARNING OUTCOMES:
1. The student explains the wave nature of light.
2. The student gives reasons why visible light is claimed as an electromagnetic wave.
INTRODUCTION:
➢ We discussed the existence of subatomic particles in the previous class. You might have
acquired fundamental ideas about various atomic models proposed by J.J. Thomson,
Rutherford and Neils Bohr.
Activity-1: Try to prepare a new model of an atom by taking advice from your teacher. You can
also discuss among your friends. During this process some questions may arise in your mind …..
1. Can subatomic particles be arranged in any other way?
2. Do all atoms have the same subatomic particles?
3. Why is an atom of one element different from an atom of other element?
4. How are the electrons distributed in the space of an atom?
To answer them, we need to understand Nature of light.
SPECTRUM:
➢ You might have observed the formation of a rainbow. There are seven colours in a
rainbow namely; violet, indigo, blue, green, yellow, orange and red.
➢ We can find the colours spreading continuously. We can also find the intensity of each
colour varies from one point to another.
➢ This is a familiar example of visible spectrum. The range of wavelengths covering red
colour to violet colour is called the visible spectrum.
CONCEPTS:
1. Electromagnetic spectrum.
LEARNING OUTCOMES:
1. The student explains the concept of the electromagnetic spectrum.
2. The student explains Planck’s proposal and its significance.
ELECTROMAGNETIC SPECTRUM:
➢ Electromagnetic waves can have a wide variety of wavelengths .The entire range of
wavelengths is known as the electromagnetic spectrum .
➢ The electromagnetic spectrum consists of a continuous range of wavelengths. This
ranges from gamma rays to radio waves. Here Gamma rays are of shorter wavelength
whereas radio waves are of longer wavelength.
➢ There are several waves in the electromagnetic spectrum. Namely radio waves,
microwaves, infrared waves, visible light, UV rays, X-rays and gamma rays.
➢ But our eyes are sensitive only to visible light.
Let us recall an incident we generally come across.
➢ When an iron rod is heated, some of the heat energy is emitted as light.
➢ Initially the emitted light is red in colour (Lower energy corresponding to higher
wavelength ).
➢ As the rod is heated further, it starts glowing in colours namely orange, yellow, blue
(higher energy corresponding to lower wavelength).
➢ The rod may even turn white (all visible wavelengths) if the temperature is high
enough.
➢ That means the energy emitted from a material body increases with increase in heat
energy.
➢ we may not observe other colours meanwhile one colour is emitted. Because when the
temperature is high enough all colours are emitted but due to high intensity of one
emitted colour others cannot be observed.
➢ It is believed (about instance mentioned above) previously that energy is emitted
continuously
➢ A famous scientist Max planck disputed with the continuous energy tradition of
electromagnetic energy.
➢ He assumed that the energy is always emitted in multiples of hν. For example, 1hν
2hν, 3hν,........ nhν.
➢ That means the energy ‘E’ for a certain frequency ‘ν’ (of the radiation absorbed or
emitted) can be represented by the equation,
E=hν
Here, h is Planck’s constant and h=6.626 X 10-34 Js
The significance of Planck’s proposal
➢ electromagnetic energy can be gained or lost in discrete values (1hν 2hν, 3hν,........
nhν). But not in a continuous manner.
➢ Emission or absorption of light spectrum is a collection or group of wavelengths.
ACTIVITY-2:
➢ Take a pinch of cupric chloride in a watch glass and make a paste with concentrated
hydrochloric acid.
➢ Take this paste on a platinum loop and introduce it into a non-luminous flame.
➢ As a result cupric chloride produces a green colour flame.
➢ If we carry out similar activity with strontium chloride it produces a crimson red
colour flame.
➢ Similarly sodium vapours produce yellow light in street lamps.
➢ Scientists found that each element emits its own characteristic colour.
➢ These colours correspond to certain discrete wavelengths of light and are called line
spectra.
➢ The lines in atomic spectra can be used to identify unknown atoms, just like
fingerprints are used to identify people. .
ASSESSMENT:
I) Answer the following questions
1. Explain electromagnetic spectrum.
2. Write down the significance of Planck's proposal.
II) Choose the correct answer
1. The value of planck’s constant
A) 6.626 X 10-34 J/s. B) 6.626 X 10-27 Js.
C) 6.626 X 10-34 Js. D) 6.626 X 10-27 J/s.
2. The electromagnetic wave of shortest wavelength among given below
A) X-rays. B) UV rays.
C) Gamma rays. D) Hulk rays.
STATE COUNCIL OF EDUCATIONAL RESEARCH & TRAINING
TELANGANA, HYDERABAD.
ACADEMIC YEAR: 2020-21, LEVEL-2
Class: X Medium: English Subject: physical science
Name of the lesson: Structure of atom
Name of the topic/concept: Bohr’s model of hydrogen atom WORKSHEET: 57
CONCEPTS:
1. Bohr’s model of hydrogen atom and its limitations.
LEARNING OUTCOMES:
1. The student explains proposals of Bohr's model of hydrogen atom.
2. The student explains the limitations of Bohr’s model of hydrogen atom.
INTRODUCTION:
➢ The atomic model proposed by Rutherford could not explain the
reason for stability of atoms.
➢ To overcome this, Niels Bohr came up with a thought that
electrons can be found only in certain energy levels or regions
around the nucleus.
➢ Let us try to understand Bohr’s model by using line spectra of
hydrogen.
BOHR’S MODEL OF HYDROGEN ATOM:
The proposals are,
➢ Electrons in an atom occupy stationary orbits (states) of fixed energy at different
distances from the nucleus.
➢ When an electron jumps from a lower energy state (ground state) to higher energy states
(excited state) it absorbs energy.
➢ Energy is emitted when such a jump occurs from higher energy state to lower energy
state.
➢ The energies of an electron in an atom can have only certain values E1 E2 E3… that means
the energy is quantized.
➢ The states corresponding to these energies are called stationary states and the possible
values of the energy are called energy levels.
➢ The lowest energy state of an electron in an atom is known as its ground state.The
electron moves to a higher energy level whenever it gains energy. Such a higher energy
level is called as excited stage.
➢ The electron loses the energy and gets back to its ground state and energy is emitted in
the form of electromagnetic radiation if the wavelength is within the visible region it is
visible as an emission line.
Bohr’s model is successful model as far as line spectra of hydrogen
Spectrum of hydrogen
LIMITATIONS OF BOHR’S MODEL:
➢ The line spectrum of an hydrogen atom , when observed through a high resolution
spectroscope, appears as a group of finer lines. Bohr’s model failed to account for
splitting of line spectra.
ASSESSMENT:
I) Answer the following questions
1. Write down the proposals of Bohr’s atomic model.
2. Write down the limitations of Bohr model.
3. Explain the following terms.
A) Ground state B) Excited state.
II) Choose the correct answer.
1. The lowest energy state of electron is called as
A) Excited stage B) Energy state
C) Ground state D) Mid state
2. Bohr’s model explains line spectra of
A) Hydrogen B) Helium
C) Oxygen D) Nitrogen
STATE COUNCIL OF EDUCATIONAL RESEARCH & TRAINING
TELANGANA, HYDERABAD.
ACADEMIC YEAR: 2020-21, LEVEL-2
Class: X Medium: English Subject: physical science
Name of the lesson: Structure of atom
Name of the topic/concept: Bohr -Sommerfeld model of atom. WORKSHEET: 58
CONCEPTS:
1. Bohr -Sommerfeld model of atom.
LEARNING OUTCOMES:
1. The student explains proposals of the Bohr-Sommerfeld model of atom.
INTRODUCTION:
➢ Bohr's model might be successful to explain all the line spectra of hydrogen atom but
failed to account for splitting of line spectra.
➢ Sommerfeld made an attempt to account for the structure of line spectra which is also
known as fine spectra.
➢ Let us try to understand how Sommerfeld modified Bohr's model.
SOMMERFELD MODIFICATIONS TO BOHR’S MODEL :
➢ He added elliptical orbits to circular orbits proposed by Bohr. The elliptical orbits exist in
atoms such that the nucleus of the atom is at one of the principal foci of them.
➢ He was guided by the factthat,“Periodic motion under the influence of a central force
will lead to elliptical orbits with the force at one of the foci.”
➢ He retained first orbit as it is. It is represented byn=1 and l=0.
➢ He added one elliptical orbit to Bohr's second orbit. Bohr's second orbit is represented by
n=2, l=0and elliptical orbit is represented byl=1.
➢ He also added two elliptical orbits to Bohr's third orbit. Bohr's third orbit is represented
byn=3, l=0 and two elliptical orbits are represented byl=1, l=2.
➢ He added the elliptical orbits in a similar way to other orbits. For example, n=4
CONCEPTS:
1. Quantum mechanical model of atom.
LEARNING OUTCOMES:
1. The student explains proposals of the Quantum mechanical model of atoms.
QUANTUM MECHANICAL MODEL OF AN ATOM:
➢ Bohr atomic model has contributed to its best to know more about the structure of atoms.
Sommerfeld made a tremendous attempt to explain fine spectra.
➢ But they seem to have restricted electrons that revolve in orbits.
➢ The shortcomings of these models and following questions were at discussions.
1) Why is the electron in an atom restricted to revolve around the nucleus at a certain
distance?
2) Do the electrons really follow defined paths around the nucleus?
3) Is it possible to find the exact position of the electron?
4) What is the velocity of the electron?
➢ Some scientific investigations were going on at the same time.These were about to
address them.
➢ If the electron revolves around the nucleus in defined paths or orbits then the exact
position of the electron at various times would be known.Moreover electrons are invisible
to naked eye.
➢ We have to take the help of suitable light (of shorter wavelength) to find the position and
velocity of the electron.
➢ This light interacts with the electron and disturbs the motion of it. Hence the position and
the velocity cannot be measured accurately at a time.
➢ So now it is clear that electrons do not follow a definite path in an atom. It is also not
possible to pinpoint an electron in an atom.
➢ Under these circumstances Erwin Schrodinger developed a model of an atom to
understand the properties of electrons.
It is known as the quantum mechanical model of atom. According to this model of an atom,
➢ The electrons are thought to exist in a particular region of space around the nucleus at a
given instant of time instead of in orbits as proposed by Bohr.
➢ The region of space around the nucleus where the probability of finding the electron is
maximum is called an orbital.
➢ In a given space around the nucleus certain orbitals can only exist.
ASSESSMENT:
I) Answer the following questions.
1. What is an orbital? How is it different from Bohr's orbit?
2. Explain the circumstances that lead to the development of the quantum mechanical model
of atom?
II) Choose the correct answer.
1. The region of space around the nucleus where the probability of finding the electron is
maximum is
A)Orbit B)orbital
C)shell D)field
STATE COUNCIL OF EDUCATIONAL RESEARCH & TRAINING
TELANGANA, HYDERABAD.
ACADEMIC YEAR: 2020-21, LEVEL-2
Class: X Medium: English Subject: physical science
Name of the lesson: Structure of atom
Name of the topic/concept: Quantum numbers-I WORKSHEET: 60
CONCEPTS:
1. Principal Quantum number.
2. Angular-momentum quantum number.
LEARNING OUTCOMES:
1. The student explains quantum numbers.
2. The student differentiates the principal quantum number and angular quantum number.
QUANTUM NUMBERS:
➢ According to the quantum mechanical model of atom,the electrons exist in a particular
region of space around the nucleus at an instant of time. This region of space is called an
orbital.
➢ Each electron in an atom is described by a set of three numbers. These numbers are called
quantum numbers and denoted by n, l, ml.
➢ These indicate the probability of finding the electron in the space around the nucleus.
They describe the space around the nucleus where electrons are found and their energies.
➢ There are different quantum numbers as given below,
1) The principal quantum number(n)
2) The angular- momentum quantum number ( l )
3) The magnetic quantum number ( ml )
4) The spin quantum number(ms )
We now discuss the significance of quantum numbers .
THE PRINCIPAL QUANTUM NUMBER (n):
➢ This is related to the size and energy of the main shell. It is denoted by ‘n’.
➢ Main shells are represented by K,L,M,...... and ‘n’ has positive integer values of
1,2,3,........
➢ The shells become bigger and of higher energy as the value of ‘n’ is increased. The
bigger the shell, the farther it is from the nucleus.
➢ For example,
1) When n=1, the shell is K-shell and there is only one subshell with l=0.it is
designated by ‘1s’ orbital.
2) When n=2, the shell is L-shell and there are two subshells with l=0 and l=1. They
are designated by ‘2s’ ,’2p’ orbitals respectively.
ASSESSMENT:
I) Answer the following questions.
1. Explain the significance of principal quantum number and angular-momentum quantum
numbers in predicting the position of an electron in an atom.
2. Which electronic shell is of higher energy in K and L?
3. Differentiate the principal quantum number from the angular-momentum quantum
number.
II) Choose the correct answer.
1. The value of ‘n’ for M-shell
A) 3 B) 2
C) 1 D) 0
2. The ‘l’ values for n=4
A) 0,1,2 ,3 B) 1,2,3,4
C) 1,2,3 D) 2,3,4
STATE COUNCIL OF EDUCATIONAL RESEARCH & TRAINING
TELANGANA, HYDERABAD.
ACADEMIC YEAR: 2020-21, LEVEL-2
Class: X Medium: English Subject: physical science
Name of the lesson: Structure of atom
Name of the topic/concept: Quantum numbers-II WORKSHEET: 61
CONCEPTS:
1. The magnetic quantum number
2. The spin quantum number.
LEARNING OUTCOMES:
1. The student explains quantum numbers.
2. The student differentiates the magnetic quantum number and spin quantum number.
QUANTUM NUMBERS:
➢ We have discussed two quantum numbers namely, the principal quantum number and the
angular-momentum quantum number earlier.
➢ The principal quantum number (n) describes the size and energy of the main shell
meanwhile the angular-momentum quantum number ( l ) represents the subshell.
➢ Now we try to learn about the remaining quantum numbers.
1) The magnetic quantum number.
2) The spin quantum number
THE MAGNETIC QUANTUM NUMBER (ml):
➢ The magnetic quantum number describes the orientation of the orbital in space relative to
the other orbitals in the atom. It is denoted by ml .
➢ It has integer values between -l and l including zero where ‘l’ represents
angular-momentum quantum number.
➢ Thus for a certain value of ‘l’ there are ( 2l+1) integer values of ‘ml’.
-l, (-l+1), …..-1, 0, 1,.....(+l+1), l.
➢ When l=0, (2l+1)=1 and there is only one value (0) of ml,
thus we have only one orbital i.e. s-orbital. s-orbital is spherical in shape.
➢ When l=1, (2l+1)=3, that means ml has three values, namely, -1,0, and 1 or three
p- orbitals, with different orientations along x, y, z axes. These are labelled as pX, pY, and
pz. p-orbital is dumbbell-shaped.
➢ When l=2, (2l+1)=5, that means ml has five values, namely,-2, -1,0, 1, and 2 or five
, dyz , dxz
d-orbitals, with different orientations along x, y, z axes. These are labelled as dxy
, dx2-y2 , and dz2 . d-orbital is double dumbbell-shaped.
➢ The number of ‘ml ’ values indicates the number of orbitals in subshell with a particular
‘l’ value.
➢ Orbitals in the subshell (belonging to the same shell) possess the same energy. These are
called degenerated orbitals.
➢ Each subshell holds a maximum of twice as many electrons as the number of orbitals in
the subshell.
THE SPIN QUANTUM NUMBER (mS):
➢ This quantum number refers to the two possible orientations of the spin of an
electron,one clockwise and the other anticlockwise spin.
➢ These are represented by +½ and -½ .
➢ The importance of the spin quantum number is seen when electrons occupy specific
orbitals in multy-electron atoms.
ASSESSMENT:
I) Answer the following questions.
1. Explain the values of magnetic quantum numbers for a given value of l’ with an
example.
2. Differentiate the magnetic quantum number from the spin quantum number.
II) Choose the correct answer.
1. The maximum number of electrons that can occupy f-subshell
A) 2 B) 6
C) 14 D) 10
2. Number of orbitals in d-subshell
A) 3 B) 1
C) 5 D) 7
STATE COUNCIL OF EDUCATIONAL RESEARCH & TRAINING
TELANGANA, HYDERABAD.
ACADEMIC YEAR: 2020-21, LEVEL-2
Class: X Medium: English Subject: physical science
Name of the lesson: Structure of atom
Name of the topic/concept: Electronic configuration WORKSHEET: 62
CONCEPTS:
1. Electronic configuration.
LEARNING OUTCOMES:
1. The student explains electronic configuration in their own words.
2. The student can tell the names of the rules useful to describe the electronic configuration.
INTRODUCTION:
➢ We have discussed various models of atoms and quantum numbers so far.
➢ Atomic models provided information about the place where the nucleus exists ,about the
region of space where electrons are found.
➢ Quantum numbers gave an idea of shells, subshells which are occupied by electrons and
their energy levels in the atom.
➢ Now we try to understand how electrons are distributed in the atom.
ELECTRONIC CONFIGURATION:
➢ The distribution of electrons in shells,subshells and orbital in an atom is known as
electronic configuration.
➢ The shorthand notation nl ,consists of the principal energy level (‘n’ value), the letter
x
representing sub energy level ( ‘l’ value), and the number of electrons ( x ) in the subshell
(written as a superscript) , is used to show electronic configuration.
➢ Let us consider the hydrogen ( H ) atom to understand the arrangement of electrons, as it
contains only one electron.
➢ The electronic configuration of hydrogen is‘1s1’. This means there exists one electron in
s-orbital in K-shell(n=1).
➢ The electronic configuration can also be represented by showing the spin of the electron.
➢ For the electron in H ,as we have seen, the set of quantum numbers is:
n=1,l=0,ml =0,m
s=½
or -½ .
➢ We must know the electronic configuration of the atom for multi-electron atoms.
➢ The distribution of electrons in various atomic orbitals provides an understanding of the
electronic behaviour of the atom and, in turn, its reactivity.
➢ We need to know three principles to describe the electronic configuration for more than
one electron in an atom. Those are the Pauli exclusion principle, Aufbau principle and
Hund’s rule. We discuss them later.
ASSESSMENT:
I) Answer the following questions.
1. Explain the electronic configuration in your own words.
2. Name the three principles useful in describing electronic configuration of more electrons
in an atom.
II) Choose the correct answer.
1. The letter/set of letters which represents the number of electrons in notation nl x is
A) n B) x
C) l D) nl x
2. The letter/set of letters which represents the subshell in notation nl x i s
A) n B) x
C) l D) nl x
STATE COUNCIL OF EDUCATIONAL RESEARCH & TRAINING
TELANGANA, HYDERABAD.
ACADEMIC YEAR: 2020-21, LEVEL-2
Class: X Medium: English Subject: physical science
Name of the lesson: Structure of atom
Name of the topic/concept: Aufbau principle. WORKSHEET: 63
CONCEPTS:
1. The Aufbau principle.
LEARNING OUTCOMES:
1. The student explains the Aufbau principle with an example.
2. The student can communicate the filling order of atomic orbitals with a neat diagram.
INTRODUCTION:
➢ We know that electronic configuration is nothing but distribution of electrons in shells
and subshells.
➢ We also know that three rules are to be followed to describe the electronic configuration
for multi-electron atoms.
➢ We now discuss one important principle to be well aware of in order to describe
electronic configuration, popularly known as the Aufbau principle.
➢ The German word ‘Aufbau’ means ‘building up’.
AUFBAU PRINCIPLE:
➢ We know that as we pass from one element to another one of the next higher atomic
number, one electron is added every time to the atom.
➢ We also know that the maximum number of electrons in any shell is ‘2n2’, where ‘n’ is
the principal quantum number.
➢ The maximum number of electrons in a subshell is equal to 2(2l+1) where l=0,1,2,3….
➢ Thus these subshells s,p,d and f can have a maximum of 2,6,10, and 14 electrons
respectively.
➢ The electronic configuration in the ground state can be built up by placing electrons in
the lowest energy orbitals available until the total number of electrons (equal to the
atomic number) are added. This is called the Aufbau principle.
➢ Thus orbitals are filled in the order of increasing energy.
Two general rules help us to predict electronic configurations.
1. Electrons are assigned to orbitals in order of increasing value of (n+l).
2. For subshells with the same value of (n+l), electrons are assigned first to the
subshell with lower ‘n’ value.
➢ The following diagram shows the increasing value of (n+l). This is known as
Moeller chart and it provides the filling order of atomic orbitals.
CONCEPTS:
1. The Pauli exclusion principle.
2. Hund’s rule.
LEARNING OUTCOMES:
1. The student explains the Pauli exclusion principle with an example.
2. The student explains Hund's rule with an example.
INTRODUCTION:
➢ We have discussed the Aufbau principle which is essential to describe the electronic
configuration.
➢ According to this rule, the electronic configuration in the ground state can be built up by
placing electrons in the lowest energy orbitals available until the total number of
electrons (equal to the atomic number) are added.
➢ Let us learn another two namely,
1) The pauli exclusion principle,
2) Hund's rule.
THE PAULI EXCLUSION PRINCIPLE:
➢ The Pauli exclusion principle states that, “No two electrons of the same atom can have all
four quantum numbers the same”.
➢ Let us consider the helium (He) atom. Helium atom has two electrons.
➢ The first electron occupies ‘1s’ orbital. The second electron joins the first one in the
‘1s’ orbital so the electron configuration of the ground state of ‘He’ is ‘1s 2’.
➢ According to principle, the two electrons should not have the same quantum numbers.
➢ Hence ‘ms’ must be different even though n,l, and ml are the same for two electrons of
helium. Moreover the spins must be paired in the helium atom.
➢ Electrons with paired spins are denoted by ‘↑↓’ .One electron has ms=+½, the other has
ms=-½
. They have antiparallel spins.
➢ The major consequence of the exclusion principle involves orbital occupancy. Since only
two values of ms are allowed, an orbital can hold only two electrons and they must have
opposite spins.
➢ Hence,the electronic configuration of helium atom is:
HUND'S RULE :
➢ According to this rule electron pairing in orbitals starts only when all available empty
orbitals of the same energy (degenerate orbitals) are singly occupied.
➢ Let’s try to understand with an example.The electronic configuration of Boron ‘B’(Z=5)
is 1s2 2s2 2p1 and of Carbon ‘C’ (Z=6) is 1s2 2s2 2p2.
➢ The first four go into the 1s a nd 2s orbitals. The next two electrons in carbon go into
separate 2p orbitals with both electrons having the same spin.
➢ Note that the unpaired electrons in the 2p orbitals are shown with parallel spins.
ACTIVITY 3:
➢ Complete the electronic configuration of elements given in the following table.
ASSESSMENT:
I) Answer the following questions.
1. Explain the Pauli exclusion principle.
2. Explain the Hund's rule with suitable examples.
II) Choose the correct answer.
1. The rule/principle that talks about pairing of electrons in degenerate orbitals.
A) Configuration rule B) Exclusion principle
C) Hund's rule D) Aufbau principle.
2. The statement “No two electrons of the same atom can have all four quantum numbers
the same”. Is known as
A) Configuration rule B) Exclusion principle
C) Hund's rule D) Aufbau principle.
STATE COUNCIL OF EDUCATIONAL RESEARCH AND TRAINING
TELANGANA, HYDERABAD
ACADEMIC YEAR 2020-21
LEVEL - 2
Class: X Medium: English Subject: Physical Sciences
Topic/Concept: Classification of Elements-The Periodic Table/Dobereiner’s Triads Work sheet No:65
Concepts Identified:
Dobereiner’s Triads
Learning outcomes:
Explains Dobereiner’s Law of Triads.
Gives examples related to Dobereiner’s Triads.
Applies the average formula to check the triads.
Introduction:
Look at the adjacent figure.
Do you know why he arranged the fruits in that way?
It is easy
to identify the same kind of fruits.
to tell the rate according to their size and type.
If he don’t arrange them in a particular manner, it is
difficult to him and also us to select the needy fruits. So
they should be classified.
From this example we understand that for any system involving several things, a particular order of
arrangement of those things is essential.
Robert Boyle (1661) defined an element as any substance that cannot be decomposed
into a further simple substance by a physical or chemical change.
By this activity we conclude that the atomic weight of the middle element in each triad is close to the
average (or arithmetic mean) of atomic weights of the other two elements.
Dobereiner’s attempts encouraged others to classify the known elements.
Each theory was true for some elements but failed for the remaining ones.
But they become the stepping stones for the modern periodic law.
And first of these steps was laid by Dobereiner. So we should honor him for this purpose.
Limitations:
All the known elements could not be arranged in the form of triads
For very low mass or for very high mass elements, the law was not holding good.
Assessment
I. Explain “Dobereiner’s Law of triads”
II. N, P and As have similar properties. Check whether the average of atomic masses of N(14) and
As(74.9) is equal to the atomic mass of P(31). Could we call it as a Dobereiner triad? Why? Give
reason.
III. See the elements and their atomic masses (given in brackets). Classify them as Dobereiner’s triads.
Ba(137.3) Br(79.9) Ca(40.1) Cl(35.5) I(126.9) Sr(87.6)
IV. Choose the correct answer
1. In which order do we need to put Barium, Calcium & Strontium for Dobereiner's Triads to apply? ( )
a) calcium, strontium, barium b) barium, strontium, calcium
c) calcium, barium, strontium d) strontium, barium, calcium
2. Identify the group which is not a Dobereiner triad ( )
a. Li, Na, K b. Be, Mg, Cr c. Ca, Sr, Ba d. Cl, Br, I
3. On what basis were the elements arranged in Dobereiner's triads? ( )
a) atomic number b) atomic mass c) number of protons d) number of electrons
4. Given that A, B, C is a triad. Atomic masses of A and C are 32,125 respectively. Then what is the
atomic mass of B is? a) 157 b) 93 c)78.5 d)75.5 ( )
5. If two elements of a Dobereiner triad are Ca and Sr, then the third element of the triad is: ( )
a) Cs b) Mg c) Ba d) Na
STATE COUNCIL OF EDUCATIONAL RESEARCH AND TRAINING
TELANGANA, HYDERABAD
ACADEMIC YEAR 2020-21
LEVEL - 2
Class: X Medium: English Subject: Physical Sciences
Lesson:Classification of Elements-The Periodic Table Work sheet No: 66
Concepts Identified:
Newlands’ law of Octaves
Learning outcomes:
Explains Newlands’ law of Octaves.
Gives the reasons for rejection of Newlands’ law of Octaves.
Introduction:
We have already known that the first scientist to classify the elements was Dobereiner
According to him, When the elements were arranged in the order of increasing atomic
weights, the relative atomic weight of the middle element in each triad was close to the
average(or arithmetic mean) of atomic weights of the other two elements.
But he could identify only a few such triads and so the law could not gain importance.
The law failed for very low mass or for very high mass elements.
The attempts of Dobereiner encouraged other chemists to correlate the properties of
elements with their atomic weights.
Assessment
Concepts Identified:
Mendeleeff’s Periodic Table
Learning outcomes:
Explains salient features of Mendeleeff’s Periodic Table.
Explains limitations of Mendeleeff’s Periodic Table.
Introduction:
Concepts Identified:
Modern Periodic Table - Introduction
Learning outcomes:
Explains Modern Periodic Law.
Explains the various blocks of Modern Periodic Table.
Introduction:
We have already known the Mendeleeff’s Classifications of elements.
According to him, ‘The physical and chemical properties of the elements
are periodic functions of their atomic weights.’
The Modern periodic table of elements is based on Mendeleev’s Dimitri Mendeleeff
observations.
It is the extension of the original Mendeleeff’s periodic table known as
short form of the table and this modern table is called the long form of the periodic
table
The only difference is instead of being arranged by atomic weights, the modern table is
arranged by atomic number (Z).
Modern Periodic Table:
Henry Moseley, in 1913, found that atomic number is more
fundamental property of an element than its atomic weight.
Atomic number helps in arranging the elements
according to their electronic configuration as it is indirectly
equal to the number of electrons of a neutral atom.
Atomic number of an element is equal to the number
Henry Moseley
of protons in an atom of the element since number of protons
equal to the number of electrons in a neutral atom.
So, Atomic number of an element (Z) = Number of protons present in the atom = number of
electrons present in the neutral atom.
This atomic number concept forced the periodic law to be changed.
The periodic law is changed from atomic weight concept to atomic number concept and now it is
called the modern periodic law.
The Modern periodic law states that “the properties of the elements are periodic functions of their
atomic numbers.”
This law may be stated as “The physical and chemical properties of elements are the periodic functions
of the electronic configurations of their atoms.”
The modern
periodic table has
18 vertical columns
known as groups
and 7 horizontal
rows known as
periods.
The elements
with similar outer
shell (valence shell)
electronic
configurations in
their atoms are in the same column called group.
Depending on the differentiating electron’s sub-shell, the elements are classified as ‘s’ ‘p’, ‘d’ and
‘f’ block elements.
Let us observe the electronic configurations of the following elements. The last coming electron is
underlined.
Sodium: 11Na 1s2 2s2 2p6 3s1
Aluminium : 13Al 1s2 2s2 2p6 3s2 3p1
Scandium : 21Sc 1s2 2s2 2p6 3s2 3p6 4s2 3d1
Cerium : 58Ce 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 5d1 4f2
Sodium (Na) gets its differentiating electron into 3s level. Therefore ‘Na’ is an s-block element.
Aluminium (Al) gets its differentiating electron into ‘p’ sub shell and it is a p-block element.
Scandium (Sc) gets its differentiating electron into‘d’ sub shell. Therefore it is a d-block element.
Cerium (Ce) gets its new coming electron into ‘f’ sub shell; hence it is an f-block element.
Assessment
Concepts Identified:
Modern Periodic Table - Groups
Learning outcomes:
Explains about the groups in the Modern Periodic table.
Explains the element families in the periodic table.
Introduction:
Moseley
We have already known the Modern periodic law and the
classification of elements based on this law given by Moseley in 1913 .
According to him, ‘The physical and chemical properties of the elements
are periodic functions of the electronic configurations of their atoms.’
We know that the modern periodic table has 18 vertical columns known as
groups and 7 horizontal rows known as periods.
Depending on the differentiating electron’s sub-shell, the elements are
classified as ‘s’ ‘p’, ‘d’ and ‘f’ block elements.
Now let us understand about the groups in detail…..
Groups:
There are eighteen
groups in long form
of periodic table.
The elements
with similar outer
shell (valence shell)
electronic
configurations in their
atoms are in the
group.
They are
represented by using
Roman numeral I through VIII with letters A and B in traditional notation.
According to latest recommendation of the IUPAC, these groups are represented by Arabic numerals 1
through 18 with no A and B designations.
We use the latest system with the traditional heading following in parenthesis.
Ex: Group 2 (II A); Group 16 (VI A)
Group of elements is also called element family or chemical family.
For example Group 1 (IA) has from Li to Fr with outer shell electronic configuration ns 1 and is called
Alkali metal family.
Let us do the activity-2 given in the Text book for better understanding ……
Activity:2
Some main group elements of s-block and p-block have family names as given in the following table.
Observe the long form of the periodic table and complete the table with proper information.
Assessment
Concepts Identified:
Modern Periodic Table - Periods
Learning outcomes:
Explains about the Periods in the Modern Periodic table.
Introduction:
We have already known about the groups in the Modern periodic table Proposed by Moseley.
We know that the modern periodic table has 18 vertical columns known as groups and 7 horizontal rows
known as periods.
The elements with similar outer shell (valence shell) electronic configurations in their atoms are in the
group.
Now let us understand about the periods in detail…..
Periods:
The horizontal
rows in the periodic
table are called
periods.
There are seven
periods in the modern
periodic table.
These periods are
represented by Arabic
numerals 1 to 7.
The number of
main shells present in
the atom of particular element decides to which period it belongs. For example, hydrogen (H) and helium
(He) atoms contain only one main shell (K). Therefore they belong to period-1.
The number of elements in period depends on how electrons are filled into various shells. Each period
starts with a new main shell ‘s’ sub shell and ends when the main shell is filled with respect to the ‘s’ and
‘p’sub shells (except the first period).
The first period starts with K-shell. The first main shell (K) contains only one sub-shell, the (1s). This
first period contains only two elements.
Second period starts with the 2nd main shell (L). Eight types of configurations are possible in this shell
(L) like 2s1 and 2s2 and 2p1 to 2p6. L-shell has two sub shells, namely, 2s and 2p. This second period
contains 8 elements Li, Be, B, C, N O, F and Ne in the order given.
Third period starts with third main shell (M). This shell (M) has 3 sub shells, namely, 3s, 3p and 3d, but
while electrons are being filled into the shell ‘3d’ gets electrons only after ‘4s’ is filled. Therefore, the 3 rd
period contains again 8 elements, which includes two s-block elements (Na, Mg) and six p-block elements
(Al to Ar).
Fourth period starts with fourth main shell (N). This shell (N) has four sub-shells namely 4s, 4p, 4d and
4f and this fourth period contains 18 elements.
Again the fifth period contains 18 elements i.e. from 37Rb to 54Xe.
There are thirty two elements in the Sixth period from 55Cs to 86Rn.
‘4f’ elements are called Lanthanoids or lanthanides.
There are thirty two elements in the Seventh period from 87Fr to 118Og.
The 5f elements are called Actinoids or as Actinides. They are from 90Th to 103Lr.
The f-block elements known as lanthanides and actinides are shown separately at the bottom of the
periodic table.
Assessment
Concepts Identified:
Modern Periodic Table - Metals and non Metals
Learning outcomes:
Gives examples for Metals and non Metals.
Introduction:
We have already known about the groups and periods in the Modern periodic table Proposed
by Moseley.
We know that the modern periodic table has 18 vertical columns known as groups and 7 horizontal rows
known as periods.
Now let us understand about Metals, Semi Metals and non Metals in detail…..
Metals and non Metals:
The elements
with three or less
electrons in the outer
shell are considered
to be metals and those
with five or more
electrons in the outer
shell are considered
to be non metals.
We may find some
exceptions to this. ‘d’
block elements (3rd
group to 12th group) are metals and they are also known as transition metals and the metallic character of d-
block elements decreases gradually from left to right in periodic table.
Lanthanoids and actinoids actually belong to 3rd group (III B) which is within the transition elements:
hence they are called the inner transition elements.
Metalloids or semi-metals are elements which have properties that are intermediate between the
properties of metals and non metals.
They possess properties like metals but brittle like non metals. They are generally semi conductors.
Eg: B, Si, As, Ge.
All elements in s-block are metals, whereas in p-block (except 18 th group) there are metals, non metals
and metalloids.
In the periodic table you will notice a staircase like separation.
The elements to the left of this separation are metals and to the right are non-metals.
The elements on staircase (or) very near to it like B, Si, As, Ge etc., are metalloids.
Assessment
Concepts Identified:
Modern Periodic Table - Valence
Learning outcomes:
Explains about the valency of an element in their own words.
Explains about the valency of an element with respect to hydrogen and oxygen.
Explains the trends of valence in Groups and in periods
Introduction:
We have already known about the groups and periods in the Modern periodic table Proposed
by Moseley.
We know that the modern periodic table has 18 vertical columns known as groups and 7 horizontal rows
known as periods.
And also we know that the modern periodic table is organized on the basis of the electronic
configuration of the atoms of the elements. Physical and chemical properties of elements are related to
their electronic configurations particularly the outer most shell configurations.
The atoms of the elements in a group possess similar electronic configurations.
And the electronic configuration of valence shell of any two elements in a given period is not same.
Now let us understand about valence and its trend in Groups and in Periods in detail…..
Valence and its trend in Groups and in Periods:
The Valence (or)
valency of an element
was defined as the
combining power of
an element with
respect to hydrogen,
oxygen or indirectly
any other element
through hydrogen and
oxygen.
Valence of an element with respect to hydrogen is the number of hydrogen atoms with which one atom
of that element chemically combines.
Valence of an element with respect to oxygen is twice the number of oxygen atoms with which one
atom of that element combines.
In general, the valence of an element with respect to hydrogen is its traditional group number. That
means the valence of Group IA elements is 1 and Group IIA is 2.
If the element is in the group V or above, its valence is 8– group number.
For example, chlorine valence is 8-7 = 1.
In general, each period starts with valency 1 for 1st group elements, increases upto 4 with respect to the
group number and then decreases from 4 to 3 to 2 to 1 to zero in the following groups (This is applicable
only for main group elements i.e., ‘s’ and ‘p’ block elements with respect to hydrogen.)
Now a days the valence of an elment is generally taken as the number of valence shell (outer most shell)
electrons in its atom.
Assessment
Concepts Identified:
Modern Periodic Table - Atomic radius
Learning outcomes:
Explains Atomic radius in their own words.
Explains the trends of Atomic radius in Groups and in periods
Explains about the atomic radius of cation and anion.
Introduction:
We have already known about the groups and periods in the Modern periodic table Proposed
by Moseley.
We know that the modern periodic table has 18 vertical columns known as groups and 7 horizontal rows
known as periods.
We have already known about the valence and its trend in groups and periods.
Now let us understand about Atomic radius and its trends in the periodic table in
detail…..
Atomic radius and its trend in the Modern Periodic Table:
Atomic radius of an element may be difined as the distance from the centre of the nucleus of the atom to
its outermost shell.
Atomic radius of an element is not possible to measure in its isolated state.
However, we can measure the distance between the nuclei of adjacent atoms in a solid. From this we
can estimate the size of the atom by assigning half of this distance to the radius of
each atom.
Another way of estimating the size of an atom is to measure the
distance between the two atoms in covalent molecules.
The size of a chlorine atom is estimated by measuring the length of the
covalent bond between two chlorine atoms in a Cl2 molecule. Half of this
distance is taken as atomic radius which is called as the covalent radius of chlorine atom.
Atomic radius is measured in ‘pm’ (pico meter) units. 1 pm = 10-12m
Variation of atomic radii in group:
Atomic radii increase from top to bottom in a group
(column) of the periodic table. As we go down in a group,
the atomic number of the element increases.
As the number of shells in a group from top to bottom
increases, the distance between the nucleus and the outer shell
of the atom increases.
Hence the atomic size increases with atomic number when
we go down the group.
Variation of atomic radii in Period:
Atomic radii of elements decrease across a period from left to right.
As we go to right, electrons enter into the same main shell or even inner shell in case of ‘d’ block and
‘f’ block elements.
Therefore, there should be no change in distance between nucleus and outer shell but nuclear charge
increases because of the increase in the atomic number of elements in period.
Hence, the nuclear attraction on the outer shell electrons increases.As a result the size of the atom
decreases.
Do the atom of an element and its ion have same size?
Example -1: Assume that sodium (Na) atom has lost an electron and formed a cation of sodium (Na+).
As proton number is more than electrons in it, the nucleus of Na + ion attracts outer shell electrons with
strong nuclear force.
As a result the Na+ ion shrinks in size. Therefore, the size of Na+ ion is less than ‘Na’ atom.
The size of any positive ion is less than the size of its neutral atom.
Example – 2: Assume that chlorine (Cl) atom has gained an electron to form anion of chlorine (Cl –) i.e.
chloride ion.
As the number of electrons is more than the number of protons in it, the nuclear attraction is less in Cl –
ion when compared with chlorine atom. Therefore the size of the chlorine (Cl) atom is less size than Cl- ion.
The size of any negetive ion is more than the size of its neutral atom.
Assessment
I. What is atomic radius? Explain in your own words
II. Why does the atomic radius increase going down a group? Explain with the reason.
III. Do Al and Al+ have same size? Why?
IV. Choose the correct answer
1. Which has the maximum atomic radius? ( )
a) 13Al b) 14Si c) 15P d) 12Mg
2. Which one of the following indicates the correct order of atomic size of 2nd period elements? ( )
A) Be > F > C > Ne B) Be < C < F < Ne C) Be > C > F > Ne D) F < Ne < Be < C
STATE COUNCIL OF EDUCATIONAL RESEARCH AND TRAINING
TELANGANA, HYDERABAD
ACADEMIC YEAR 2020-21, LEVEL - 2
Class: X Medium: English Subject: Physical Sciences
Name of the Chapter: Classification of Elements-The Periodic Table
Name of the Topic: Modern Periodic Table – Ionization Energy & Electron Affinity Work sheet No: 74
Concepts Identified:
Modern Periodic Table - Ionization Energy & Electron Affinity
Learning outcomes:
Explains Ionization Energy & Electron Affinity in their own words.
Explains the factors which influence the Ionization Energy & Electron Affinity of an atom.
Introduction:
We have already known about the Atomic radius and its trends in the Modern periodic table
Proposed by Moseley.
Now let us understand about Ionization Energy and its trends in the periodic table in
detail…..
Ionization Energy and its trend in the Modern Periodic Table:
The energy required to remove an electron from the outer most orbit or shell of a neutral gaseous atom
is called ionization energy.
The energy required to remove the first electron is called its first ionization energy.
The energy required to remove an electron from uni-positive ion of the element is called the 2 nd
ionization energy of that element and so on.
M(g) + IE1 → M+(g) + e- (IE1= first ionization energy)
M+(g) + IE2 → M+2(g) + e- (IE2= second ionization energy)
Ionization energy is expressed in kJ/mol
Ionization energy is also called the ionization potential but when we use the term ionization potential, it
is better to write the unit eV/atom.
The factors which influence the ionization energy:
Nuclear charge: More the nuclear charge more is
the ionization energy.
Screening effect or sheilding effect: More the
screening effect, less is the ionization energy.
Penetration power of the orbitals: orbitals belonging to the same main shell have diffrent piercing
There fore, it is easier to remove 4f electron than 4s. Between 4Be 1s2 2s2 and 5B 1s2 2s2 2p1, the element
B has less ionization energy due to less penetration power of ‘2p’ compared to‘2s’.
Stable configuration: it is easier to remove one electron from 8O (1s22s22p4) than 7N (1s22s22p3). This is
Atomic radius: More the atomic radius, less is the ionization energy.
Ionization energy decreases as we go down in a group and generally increases from left to right in
a period.
Metals have very low electron gain enthalpy values and alkaline earth metals have even positive values.
All the factors which influence the ionization energy would also influence the electron gain enthalpy.
Assessment