Class XI – NCERT – Chemistry Chapter 2

Structure of Atom

NCERT Exercise

Question 1:
(i) Calculate the number of electrons which will together weigh one gram.
(ii) Calculate the mass and charge of one mole of electrons.

Solution 1:
–31
(i) Mass of one electron = 9.10939×10 kg
–31
Number of electrons that weigh 9.10939 × 10 kg = 1
–3
Number of electrons that will weigh 1 g = (1 × 10 kg)
1
31
(1103 kg )
9.10939 10 kg
–3 + 31
= 0.1098 × 10
= 0.1098 × 1028
= 1.098 × 1027
–31
(ii) Mass of one electron = 9.10939 × 10 kg
23 –31
Mass of one mole of electron = (6.022 × 10 ) × (9.10939 ×10 kg)
–7
= 5.48 × 10 kg
–19
Charge on one electron = 1.6022 × 10 coulomb
–19 23
Charge on one mole of electron = (1.6022 × 10 C) (6.022 × 10 )
4
= 9.65 × 10 C

Question 2:
(i) Calculate the total number of electrons present in one mole of methane.
14
(ii) Find (a) the total number and (b) the total mass of neutrons in 7 mg of C. (Assume that mass of a
–27
neutron = 1.675 × 10 kg).
(iii) Find (a) the total number and (b) the total mass of protons in 34 mg of NH 3 at STP. Will the answer
change if the temperature and pressure are changed?

Solution 2:

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Class XI – NCERT – Chemistry Chapter 2
Structure of Atom

(i) Number of electrons present in 1 molecule of methane (CH4)

{1(6) + 4(1)} = 10
23
Number of electrons present in 1 mole i.e., 6.023 × 10 molecules of methane
23 24
= 6.022 × 10 × 10 = 6.022 × 10

23
(ii) (a) Number of atoms of 14C in 1 mole= 6.023 × 10
Since 1 atom of 14C contains (14 – 6) i.e., 8 neutrons, the number of neutrons in 14 g of 14C is (6.023 ×
1023) ×8. Or, 14 g of 14C contains (6.022 × 1023 × 8) neutrons.
Number of neutrons in 7 mg
6.022 1023  8  7 mg

1400mg
= 2.4092 × 1021
–27 14
(b) Mass of one neutron = 1.67493 × 10 kg Mass of total neutrons in 7 g of C
21 –27
= (2.4092 × 10 ) (1.67493 × 10 kg)
–6
= 4.0352 × 10 kg

(iii) (a) 1 mole of NH3 = {1(14) + 3(1)} g of NH3

= 17 g of NH3
= 6.022× 1023 molecules of NH3
Total number of protons present in 1 molecule of NH3
= {1(7) + 3(1)}
= 10
Number of protons in 6.023 × 1023 molecules of NH3
= (6.023 × 1023) (10)
= 6.023 × 1024
⇒ 17 g of NH3 contains (6.023 × 1024) protons.
Number of protons in 34 mg of NH3
6.022 1024  34 mg

1700 mg
= 1.2046 × 1022

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Class XI – NCERT – Chemistry Chapter 2
Structure of Atom
–27
(b) Mass of one proton = 1.67493 × 10 kg
Total mass of protons in 34 mg of NH3
–27
= (1.67493 × 10 kg) (1.2046 × 1022)
–5
= 2.0176 × 10 kg
The number of protons, electrons, and neutrons in an atom is independent of temperature and pressure
conditions. Hence, the obtained values will remain unchanged if the temperature and pressure is changed.

Question 3:
How many neutrons and protons are there in the following nuclei?
13 16 24 56 88
6 C , 8 O , 12 Mg , 26 Fe, 38 Sr

Solution 3:
13
6 C:

Atomic mass = 13
Atomic number = Number of protons = 6
Number of neutrons = (Atomic mass) – (Atomic number)
= 13 – 6 = 7

16
8 O:
Atomic mass = 16 Atomic number = 8 Number of protons = 8
Number of neutrons = (Atomic mass) – (Atomic number)
= 16 – 8 = 8

24
12 Mg :
Atomic mass = 24
Atomic number = Number of protons = 12
Number of neutrons = (Atomic mass) – (Atomic number)
= 24 – 12 = 12

56
26 Fe :
Atomic mass = 56

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Class XI – NCERT – Chemistry Chapter 2
Structure of Atom

Atomic number = Number of protons = 26

Number of neutrons = (Atomic mass) – (Atomic number)
= 56 – 26 = 30

88
38 Sr :
Atomic mass = 88
Atomic number = Number of protons = 38
Number of neutrons = (Atomic mass) – (Atomic number)
= 88 – 38 = 50

Question 4:
Write the complete symbol for the atom with the given atomic number (Z) and Atomic mass (A)
(i) Z = 17, A = 35
(ii) Z = 92, A = 233
(iii) Z = 4, A = 9

Solution 4:
35
(i) 17 Cl
233
(ii) 92 U
(iii) 94 Be

Question 5:
Yellow light emitted from a sodium lamp has a wavelength (λ) of 580 nm. Calculate the frequency (ν)
and wave number ( v ) of the yellow light.

Solution 5:
From the expression,
c

v
We get,
c
v  …….. (i)


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Class XI – NCERT – Chemistry Chapter 2
Structure of Atom

Where,
ν = frequency of yellow light
c = velocity of light in vacuum = 3 × 108 m/s
–9
λ = wavelength of yellow light = 580 nm = 580 × 10 m Substituting the values in expression (i):
3 108
v  5.17 1014 S 1
580 109
Thus, frequency of yellow light emitted from the sodium lamp
–1
= 5.17 × 1014 s
1
Wave number of yellow light, v 

1
 9
 1.72 106 m1
580 10

Question 6:
Find energy of each of the photons which
(i) correspond to light of frequency 3× 1015 Hz.
(ii) have wavelength of 0.50 Å

Solution 6:
(i) Energy (E) of a photon is given by the expression,
E = hv
Where,
–34
h = Planck’s constant = 6.626×10 Js
15
ν = frequency of light = 3×10 Hz
–34
Substituting the values in the given expression of E: E = (6.626×10 ) (3×1015)
–18
E = 1.988×10 J
(ii) Energy (E) of a photon having wavelength (λ) is given by the expression,
hc
E=

–34
h = Planck’s constant = 6.626×10 Js
c = velocity of light in vacuum = 3×108 m/s Substituting the values in the given expression of E:

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Class XI – NCERT – Chemistry Chapter 2
Structure of Atom

E
 6.626 10 310  J
34 8

0.50 1010
E  3.98 1015 J

Question 7:
–10
Calculate the wavelength, frequency and wave number of a light wave whose period is 2.0 × 10 s.

Solution 7:
1
Frequency (ν) of light 
Period
1
 10
 5.0 109 s 1
2.0 10 s
c
Wavelength (λ) of light 
v
Where,
c = velocity of light in vacuum = 3×108 m/s
Substituting the value in the given expression of λ
3 108
  6.0 102 m
5.0 109
1 1
Wave number  v  of light   1.66 101 m1  16.66m
 6.0 102

Question 8:
What is the number of photons of light with a wavelength of 4000 pm that provide 1 J of energy?

Solution 8:
Energy (E) of a photon = hν
Energy (En) of ‘n’ photons = nhν
E
n n
hc
Where,
–12
λ = wavelength of light = 4000 pm = 4000 ×10 m

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Class XI – NCERT – Chemistry Chapter 2
Structure of Atom

c = velocity of light in vacuum = 3 × 108 m/s

–34
h = Planck’s constant = 6.626 × 10 Js
Substituting the values in the given expression of n:
1   4000 1012 
n  2.012 1016
 6.626 10 310 
34 8

Hence, the number of photons with a wavelength of 4000 pm and energy of 1 J are 2.012×1016.

Question 9:
–7
A photon of wavelength 4 × 10 m strikes on metal surface, the work function of the metal being 2.13
eV. Calculate
(i) the energy of the photon (eV),
(ii) the kinetic energy of the emission, and
–19
(iii) the velocity of the photoelectron (1 eV= 1.6020 × 10 J).

Solution 9:
hc
(i) Energy (E) of a photon = hv 

Where,
–34
h = Planck’s constant = 6.626 × 10 Js
c = velocity of light in vacuum = 3 × 108 m/s
–7
λ = wavelength of photon = 4 × 10 m
Substituting the values in the given expression of E:

E
 6.626 10 310   4.969510
34 8
19
J
7
4 10
–19
Hence, the energy of the photon is 4.97×10 J.
(i) The kinetic energy of emission Ek is given by
 hv  hv0
  E  W  eV
 4.9695 1019 
 19 
eV  2.13eV
 1.6020 10 

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Class XI – NCERT – Chemistry Chapter 2
Structure of Atom

= (3.1020 – 2.13) eV
= 0.9720 eV
Hence, the kinetic energy of emission is 0.97 eV.
(ii) The velocity of a photoelectron (ν) can be calculated by the expression,
1 2
mv  hv  hv0
2
2  hv  hv0 
 v
m
Where,  hv  hv0  is the kinetic energy of emission in Joules and ‘m’ is the mass of the photoelectron.
Substituting the values in the given expression of v:
2   0.9720 1.6020 1019 
v J
9.10939 1031 kg
 0.3418 1012 m2 s2
–1
v = 5.84 × 105 ms
–1
Hence, the velocity of the photoelectron is 5.84 × 105 ms .

Question 10:
Electromagnetic radiation of wavelength 242 nm is just sufficient to ionise the sodium atom. Calculate
–1
the ionisation energy of sodium in kJ mol .

Solution 10:
N Ahc
Energy of sodium (E) 



6.0231023 mol1 6.626 1034 Js3108 ms1 
242 109 m
= 4.947×105 J mol–1
= 494.7×103 J mol–1
–1
= 494 kJ mol

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Class XI – NCERT – Chemistry Chapter 2
Structure of Atom

Question 11:
A 25 watt bulb emits monochromatic yellow light of wavelength of 0.57 µm. Calculate the rate of
emission of quanta per second.

Solution 11:
–1
Power of bulb, P = 25 Watt = 25 Js
hc
Energy of one photon, E = hν =

Substituting the values in the given expression of E:

E
 6.626 10 310   34.87 10
34 8
20
J
 0.57 10  6

–20
E = 34.87×10 J
Rate of emission of quanta per second
25
 20
 7.169 1019 s 1
34.87 10

Question 12:
Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of
wavelength 6800 Å. Calculate threshold frequency  v0  and work function  w0  of the metal.

Solution 12:
–10
Threshold wavelength of radian  0   6800 Å = 6800 × 10 m
Threshold frequency  v0  of the metal
c 3 108 ms 1
   4.411014 s 1
0 7
6.8 10 m
Thus, the threshold frequency  v0  of the metal is 4.41×1014 s-1.
Hence, work function  w0  of the metal = hv0
–34
= (6.626×10 Js) (4.41×1014 s–1)
–19
= 2.922×10 J

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Class XI – NCERT – Chemistry Chapter 2
Structure of Atom

Question 13:
What is the wavelength of light emitted when the electron in a hydrogen atom undergoes transition from
an energy level with n = 4 to an energy level with n = 2?

Solution 13:
The ni = 4 to nf = 2 transition will give rise to a spectral line of the Balmer series. The energy involved in
the transition is given by the relation,
1 1
E  2.18 1018  2  2 
 n1 n f 
Substituting the values in the given expression of E:
1 1
E  2.18 1018  2  2 
4 2 
1  4 
 2.18 1018 
 16 
 3 
 2.18 1018  
 16 
–19
E = – (4.0875 × 10 J)
The negative sign indicates the energy of emission.
hc
Wavelength of light emitted    
E
hc
Since E 

Substituting the values in the given expression of λ:


 6.626 10  3 10 
34 8

4.0875 1019
  4.8631107 m
  486.3 109 m
 486 nm

Question 14:

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Class XI – NCERT – Chemistry Chapter 2
Structure of Atom

How much energy is required to ionise a H atom if the electron occupies n = 5 orbit? Compare your
answer with the ionization enthalpy of H atom (energy required to remove the electron from n =1 orbit).

Solution 14:
The expression of energy is given by,
  2.18 1018  Z 2
En 
n2
Where,
Z = atomic number of the atom
n = principal quantum number
For ionization from n1 = 5 to n2   ,
E  E  E5
  2.18 1018 J  12     2.18 1018 J  12 
     J
      5 
2 2
   
 2.1810 18
J
–20
Hence, the energy required for ionization from n = 5 to n =  is 8.72 × 10 J. Energy required for n1 =
1 to n =  ,
E  E  E5
  2.18 1018 J  12     2.18 1018 J  12 
     J
  
    
2 2
1
   
  2.18 1018 J  1  0
= 2.18 1018 J
Hence, less energy is required to ionize an electron in the 5 th orbital of hydrogen atom as compared to
that in the ground state.

Question 15:
What is the maximum number of emission lines when the excited electron of an H atom in n = 6 drops to
the ground state?

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Class XI – NCERT – Chemistry Chapter 2
Structure of Atom

Solution 15:
When the excited electron of an H atom in n = 6 drops to the ground state, the following transitions are
possible:

Hence, a total number of (5 + 4 + 3 + 2 + 1) 15 lines will be obtained in the emission spectrum.

The number of spectral lines produced when an electron in the nth level drops down to the ground state is
n  n  1
given by. Given, 
2
n=6
6  6 1
Number of spectral lines   15
2

Question 16:
–18 –1
(i) The energy associated with the first orbit in the hydrogen atom is –2.18 × 10 J atom . What is the
energy associated with the fifth orbit?
(ii) Calculate the radius of Bohr’s fifth orbit for hydrogen atom.

Solution 16:
(i) Energy associated with the fifth orbit of hydrogen atom is calculated as:
  2.18 1018  2.18 1018
E5  
 5
2
25

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Class XI – NCERT – Chemistry Chapter 2
Structure of Atom
–20
E5 = –8.72×10 J
(ii) Radius of Bohr’s nth orbit for hydrogen atom is given by,
rn = (0.0529 nm) n2
For,
n=5
r5 = (0.0529 nm) (5)2
r5 = 1.3225 nm

Question 17:
Calculate the wave number for the longest wavelength transition in the Balmer series of atomic
hydrogen.

Solution 17:
For the Balmer series, ni = 2. Thus, the expression of wave number  v  is given by,
 1 
 1.097 107 m1 
1
v = 2  2
  2 n f 
 
Wave number v  is inversely proportional to wavelength of transition. Hence, for the longest
wavelength transition,  v  has to be the smallest.
For  v  to be minimum, nf should be minimum. For the Balmer series, a transition from ni = 2 to nf = 3 is
allowed. Hence, taking nf = 3, we get:
–1
v = 1.5236 × 106 m
 1 1
v = 1.097 107 m1   2  2 
  2 3 
 
1 1
v = 1.097 107 m1    
4 9
94
v = 1.097 107 m1   
 36 
 5
v = 1.097 107 m1   
 36 

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Class XI – NCERT – Chemistry Chapter 2
Structure of Atom

v  1.5236 106 m1

Question 18:
What is the energy in joules, required to shift the electron of the hydrogen atom from the first Bohr orbit
to the fifth Bohr orbit and what is the wavelength of the light emitted when the electron returns to the
ground state? The ground state electron energy is –2.18 1011 ergs .

Solution 18:
Energy (E) of the nth Bohr orbit of an atom is given by,
  2.18 1018  Z 2
En 
n2
Where,
Z = atomic number of the atom
–11
Ground state energy = –2.18×10 ergs
–11 –7
= –2.18×10 ×10 J
–18
= –2.18×10 J
Energy required to shift the electron from n = 1 to n = 5 is given as:
E = E5  E1
  2.18 1018  1
2

   2.18 1018 
 5
2

 1
  2.18 1018  1  
 25 
 24 
  2.18 1018     2.0928 1018
 25 
hc
Weight of emitted light 
E


 6.626 10 310 
34 8

 2.0928 10  18

 9.498 108 m

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Class XI – NCERT – Chemistry Chapter 2
Structure of Atom

Question 19:
–18
The electron energy in hydrogen atom is given by En = (–2.18 × 10 )/n2 J. Calculate the energy
required to remove an electron completely from the n = 2 orbit. What is the longest wavelength of light in
cm that can be used to cause this transition?

Solution 19:
2.18 1018
En  J
n2
Given,
E  E  E2
 2.18 1018   2.18 1018 
     J
      2 
2 2
  
 2.18 1018 
  0 J
 4 
Energy required for ionization from n = 2 is given by,
–18
= 0.545 × 10 J
–19
∆E = 5.45 × 10 J
hc

E
Here, λ is the longest wavelength causing the transition.


6.626 10 310   3.647 10
34 8
7
m
19
5.45 10
–10
= 3647 × 10 m
= 3647 Å

Question 20:
7 –1
Calculate the wavelength of an electron moving with a velocity of 2.05 × 10 ms .

Solution 20:

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Class XI – NCERT – Chemistry Chapter 2
Structure of Atom

According to de Broglie’s equation,

h

mv
Where,
λ = wavelength of moving particle
m = mass of particle, v = velocity of particle, h = Planck’s constant
Substituting the values in the expression of λ:
6.626 1034 Js

9.10939 1031 kg  2.05 107 ms1 
  3.548 1011 m
–1
Hence, the wavelength of the electron moving with a velocity of 2.05×107 ms is 3.548 1011 m

Question 21:
–31 –25
The mass of an electron is 9.1 × 10 kg. If its K.E. is 3.0 × 10 J, calculate its wavelength.

Solution 21:
From de Broglie’s equation,
h

mv
Given,
Kinetic energy (K.E.) of the electron = 3.0 × 10–25 J
1
Since K.E.  mv 2
2
2 K .E.
Velocity  v  
m
2  3.0 1025 J 

9.10939 1031 kg

 6.5866 104
V = 811.579ms-1
Substituting the value in the expression of λ:

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Class XI – NCERT – Chemistry Chapter 2
Structure of Atom

6.626 1034 Js

9.10939 1031 kg 811.579ms1 
  8.9625 107 m
Hence, the wavelength of the electron is 8.9625×10–7 m.

Question 22:
Which of the following are isoelectronic species i.e., those having the same number of electrons?
Na+, K+, Mg2+, Ca2+, S2-, Ar

Solution 22:
Isoelectronic species have the same number of electrons.
Number of electrons in sodium (Na) = 11
Number of electrons in (Na+) = 10
A positive charge denotes the loss of an electron.
Similarly,
Number of electrons in K+ = 18
Number of electrons in Mg2+ = 10
Number of electrons in Ca2+ = 18
A negative charge denotes the gain of an electron by a species.
Number of electrons in sulphur (S) = 16
∴ Number of electrons in S2- = 18
Number of electrons in argon (Ar) = 18
Hence, the following are isoelectronic species:
(1) Na+ and Mg2+ (10 electrons each)
(2) K+, Ca2+, S2- and Ar (18 electrons each).

Question 23:

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Class XI – NCERT – Chemistry Chapter 2
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(i) Write the electronic configurations of the following ions: (a) H– (b) Na+ (c) O2–(d) F–
(ii) What are the atomic numbers of elements whose outermost electrons are represented by (a) 3s1 (b)
2p3 and (c) 3p5?
(iii) Which atoms are indicated by the following configurations? (a) [He] 2s1 (b) [Ne] 3s2 3p3 (c) [Ar] 4s2
3d1.

Solution 23:
(i) (a) H– ion
The electronic configuration of H atom is 1s1.
A negative charge on the species indicates the gain of an electron by it.
∴ Electronic configuration of H– = 1s2
(b) Na+ ion
The electronic configuration of Na atom is 1s2 2s2 2p6 3s1.
A positive charge on the species indicates the loss of an electron by it.
∴ Electronic configuration of Na+ = 1s2 2s2 2p6 3s0 or 1s2 2s2 2p6
(c) O2– ion
The electronic configuration of 0 atom is 1s2 2s2 2p4.
A dinegative charge on the species indicates that two electrons are gained by it.
∴ Electronic configuration of O2- ion = 1s2 2s2 p6
(d) F– ion
The electronic configuration of F atom is 1s2 2s2 2p5.
A negative charge on the species indicates the gain of an electron by it.
∴ Electron configuration of F– ion = 1s2 2s2 2p6

(ii) (a) 3s1

Completing the electron configuration of the element as 1s2 2s2 2p6 3s1.
∴ Number of electrons present in the atom of the element
= 2 + 2 + 6 + 1 = 11
∴ Atomic number of the element = 11
(b) 2p3
Completing the electron configuration of the element as 1s2 2s2 2p3.

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∴ Number of electrons present in the atom of the element = 2 + 2 + 3 = 7

∴ Atomic number of the element = 7
(c) 3p5
Completing the electron configuration of the element as 1s2 2s2 2p5.
∴ Number of electrons present in the atom of the element = 2 + 2 + 5 = 9
∴ Atomic number of the element = 9

(iii) (a) [He] 2s1

The electronic configuration of the element is [He] 2s1 = 1s2 2s1.
∴ Atomic number of the element = 3
Hence, the element with the electronic configuration [He] 2s1 is lithium (Li).
(b) [Ne] 3s2 3p3
The electronic configuration of the element is [Ne] 3s2 3p3 = 1s2 2s2 2p6 3s2 3p3.
∴ Atomic number of the element = 15
Hence, the element with the electronic configuration [Ne] 3s2 3p3 is phosphorus (P).
(c) [Ar] 4s2 3d1
The electronic configuration of the element is [Ar] 4s2 3d1 = 1s2 2s2 2p6 3s2 3p6 4s2 3d1.
∴ Atomic number of the element = 21
Hence, the element with the electronic configuration [Ar] 4s2 3d1 is scandium (Sc).

Question 24:
What is the lowest value of n that allows g orbitals to exist?

Solution 24:
For g-orbitals, l = 4.
As for any value ‘n’ of principal quantum number, the Azimuthal quantum number (l) can have a value
from zero to (n – 1).
∴ For l = 4, minimum value of n = 5.

Question 25:
An electron is in one of the 3d orbitals. Give the possible values of n, l and m, for this electron.

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Solution 25:
For the 3d orbital:
Principal quantum number (n) = 3
Azimuthal quantum number (l) = 2
Magnetic quantum number (ml) = –2, –1, 0, 1, 2

Question 26:
An atom of an element contains 29 electrons and 35 neutrons. Deduce
(i) the number of protons and
(ii) the electronic configuration of the element.

Solution 26:
(i) For an atom to be neutral, the number of protons is equal to the number of electrons.
∴ Number of protons in the atom of the given element = 29
(ii) The electronic configuration of the atom is
1s2 2s2 2p6 3s2 3p6 4s2 3d10.

Question 27:
Give the number of electrons in the species H 2 , H2 and O2

Solution 27:
H 2 :
Number of electrons present in hydrogen molecule (H2) = 1 + 1 = 2
∴ Number of electrons in H 2 = 2 – 1 = 1 H2:
H2 :
Number of electrons in H2 = 1 + 1 = 2
O2 :
Number of electrons present in oxygen molecule (O2) = 8 + 8 = 16
∴ Number of electrons in O2 = 16 – 1 = 15

Question 28:

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Class XI – NCERT – Chemistry Chapter 2
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(i) An atomic orbital has n = 3. What are the possible values of l and ml?
(ii) List the quantum numbers (ml and l) of electrons for 3d orbital.
(iii) Which of the following orbitals are possible?
1p, 2s, 2p and 3f

Solution 28:
(i) n = 3 (Given)
For a given value of n, l can have values from 0 to (n – 1).
∴ For n = 3
l = 0, 1, 2
For a given value of l, ml can have (2l + 1) values.
For l = 0, m = 0
l = 1, m = –1, 0, 1
l = 2, m = –2, –1, 0, 1, 2
∴ For n = 3
l = 0, 1, 2
m0 = 0
m1 = –1, 0, 1
m2 = –2, –1, 0, 1, 2
(ii) For 3d orbital, l = 2.
For a given value of l, ml can have (2l + 1) values i.e., 5 values.
∴ For l = 2
m2 = –2, –1, 0, 1, 2
(iii) Among the given orbitals only 2s and 2p are possible. 1p and 3f cannot exist. For p-orbital, l = 1.
For a given value of n, l can have values from zero to (n – 1).
∴ For l is equal to 1, the minimum value of n is 2.
Similarly,
For f-orbital, l = 4.
For l = 4, the minimum value of n is 5. Hence, 1p and 3f do not exist.

Question 29:
Using s, p, d notations, describe the orbital with the following quantum numbers.
(a) n = 1, l = 0;

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Class XI – NCERT – Chemistry Chapter 2
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(b) n = 3; l =1
(c) n = 4; l = 2;
(d) n = 4; l =3.

Solution 29:
(a) n = 1, l = 0 (Given) The orbital is 1s.
(b) For n = 3 and l = 1 The orbital is 3p.
(c) For n = 4 and l = 2 The orbital is 4d.
(d) For n = 4 and l = 3 The orbital is 4f.

Question 30:
Explain, giving reasons, which of the following sets of quantum numbers are not
1
A n=0 l=0 ml = 0 m 
2 s

1
B n=1 l=0 ml = 0 ms  
2
C n=1 l=1 ml = 0 1
ms  
2
D n=2 l=1 ml = 0 1
ms  
2
E n=3 l=3 ml = – 3 ms  
1
2
F n=3 l=1 ml = 0 1
ms  
2

Solution 30:
(a) The given set of quantum numbers is not possible because the value of the principal quantum number
(n) cannot be zero.
(b) The given set of quantum numbers is possible.
(c) The given set of quantum numbers is not possible.
For a given value of n, ‘l’ can have values from zero to (n – 1). For n = 1, l = 0 and not 1.
(d) The given set of quantum numbers is possible.
(e) The given set of quantum numbers is not possible. For n = 3,
l = 0 to (3 – 1)
l = 0 to 2 i.e., 0, 1, 2
(f) The given set of quantum numbers is possible.

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Question 31:
How many electrons in an atom may have the following quantum numbers?
1
(a) n = 4, and ms  
2
(b) n = 3, l = 0

Solution 31:
(a) Total number of electrons in an atom for a value of n = 2n2
∴ For n = 4,
Total number of electrons = 2 (4)2
= 32
The given element has a fully filled orbital as 1s2 2s2 2p6 3s2 3p6 4s2 3d10.
Hence, all the electrons are paired.
1
∴ Number of electrons (having n = 4 and ms   ) = 16
2
(b) n = 3, l = 0 indicates that the electrons are present in the 3s orbital. Therefore, the number of electrons
having n = 3 and l = 0 is 2.

Question 32:
Show that the circumference of the Bohr orbit for the hydrogen atom is an integral multiple of the de
Broglie wavelength associated with the electron revolving around the orbit.

Solution 32:
Since a hydrogen atom has only one electron, according to Bohr’s postulate, the angular momentum of
that electron is given by:
h
mvr  n ..................1
2
Where,
n = 1, 2, 3, …
According to de Broglie’s equation:

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Class XI – NCERT – Chemistry Chapter 2
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h

mv
h
or mv  ..................  2 

Substituting the value of ‘mv’ from expression (2) in expression (1):
hr h
n ........... 3
 2
Since ‘2πr’ represents the circumference of the Bohr orbit (r), it is proved by equation (3) that the
circumference of the Bohr orbit of the hydrogen atom is an integral multiple of de Broglie’s wavelength
associated with the electron revolving around the orbit.

Question 33:
What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition n = 4
to n = 2 of He+ spectrum?

Solution 33:
For He+ ion, the wave number  v  associated with the Balmer transition, n = 4 to n = 2 is given by:

1 1 1
v  RZ 2  2  2 
  n1 n2 
Where,
n1  2
n2  4
Z = atomic number of helium
1 21 1
v   R  2   
  4 16 
 4 1 
 4R  
 16 
1
3R
v 
 4
4
 
3R

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Class XI – NCERT – Chemistry Chapter 2
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According to the question, the desired transition for hydrogen will have the same wavelength as that of
He+.
2 1 1  3R
R 1  2  2  
 n1 n2  4
 1 1  3
 2  2  … (1)
 1
n n2  4
By hit and trail method, the equality given by equation (1) is true only when
n1 = 1and n2 = 2.
The transition for n2 = 2 to n = 1 in hydrogen spectrum would have the same wavelength as Balmer
transition n = 4 to n = 2 of He+ spectrum.

Question 34:
Calculate the energy required for the process Heg  
 He2g  e

The ionization energy for the H atom in the ground state is 2.18×10–18 J atom-1

Solution 34:
Energy associated with hydrogen-like species is given by,
 Z2 
En  2.18 1018  2  J
n 
For ground state of hydrogen atom,
E  E  E1
  2 
18 1
 0   2.18 10  2   J
  1  
 
E  2.18 1028 J
For the given process, Heg  
 He2g  e
An electron is removed from n = 1 to n = ∞.

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Class XI – NCERT – Chemistry Chapter 2
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E  E  E1
  2 
18  2 
 0   2.18 10  2   J
  1  
 
18
E  8.72 10 J
The energy required for the process 8.72 1018 J

Question 35:
If the diameter of a carbon atom is 0.15 nm, calculate the number of carbon atoms which can be placed
side by side in a straight line across length of scale of length 20 cm long.

Solution 35:
1 m = 100 cm
1 cm = 10–2 m
Length of the scale = 20 cm
= 20×10–2 m
Diameter of a carbon atom = 0.15 nm
= 0.15×10–9 m
One carbon atom occupies 0.15×10–9 m.
Number of carbon atoms that can be placed in a straight line
20 102 m

0.15 109 m
 133.33107
 1.33109

Question 36:
2×108 atoms of carbon are arranged side by side. Calculate the radius of carbon atom if the length of this
arrangement is 2.4 cm.

Solution 36:
Length of the given arrangement = 2.4 cm

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Class XI – NCERT – Chemistry Chapter 2
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Number of carbon atoms present = 2×108

∴ Diameter of carbon atom
2.4 102 m

2 108 m
 1.2 1016 m
Diameter
Radius of carbon atom 
2
1.2 1010 m

2
 6.0 1011 m

Question 37:
o
The diameter of zinc atom is 2.6 A . Calculate (a) radius of zinc atom in pm and (b) number of atoms
present in a length of 1.6 cm if the zinc atoms are arranged side by side lengthwise.

Solution 37:
Diameter
(a) Radius of zinc atom 
2
o
2.6 A

2
 1.3 1010 m
 130 1012 m
 130 pm
(b) Length of the arrangement = 1.6 cm
= 1.6×10–2 m
Diameter of zinc atom = 2.6×10–10 m
Number of zinc atoms present in the arrangement

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1.6 102 m

2.6 1010 m
 0.6153 108 m
 6.153 107

Question 38:
A certain particle carries 2.5×10–16C of static electric charge. Calculate the number of electrons present
in it.

Solution 38:
Charge on one electron = 1.6022×10–19 C
⇒ 1.6022×10–19C charge is carried by 1 electron.
Number of electrons carrying a charge of 2.5×10–16 C
 2.5 1016 C 
 1 19 
 1.6022 10 
= 1.560 × 103 C
=1560 C

Question 39:
In Milikan’s experiment, static electric charge on the oil drops has been obtained by shining X-rays. If
the static electric charge on the oil drop is –1.282×10–18C, calculate the number of electrons present on
it.

Solution 39:
Charge on the oil drop = 1.282×10–18 C
Charge on one electron = 1.6022×10–19C
∴ Number of electrons present on the oil drop

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Class XI – NCERT – Chemistry Chapter 2
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1.282 1018 C

1.6022 1019 C
 0.8001101
 8.0

Question 40:
In Rutherford’s experiment, generally the thin foil of heavy atoms, like gold, platinum etc. have been
used to be bombarded by the α-particles. If the thin foil of light atoms like aluminium etc. is used, what
difference would be observed from the above results?

Solution 40:
A thin foil of lighter atoms will not give the same results as given with the foil of heavier atoms.
Lighter atoms would be able to carry very little positive charge. Hence, they will not cause enough
deflection of α-particles (positively charged).

Question 41:
35
79
Symbols 35 Br and 79Br can be written, whereas symbols 79 Br and 35 Br
are not acceptable. Answer briefly.

Solution 41:
The general convention of representing an element along with its atomic mass (A) and atomic number (Z)
A
is Z X
79 81
Hence 35 Br is acceptable but 35 Br is not acceptable.
79 35
Br can be written but Br cannot be written because the atomic number of an element is constant, but
the atomic mass of an element depends upon the relative abundance of its isotopes. Hence, it is necessary
to mention the atomic mass of an element.

Question 42:
An element with mass number 81 contains 31.7% more neutrons as compared to protons. Assign the
atomic symbol.

Solution 42:

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Let the number of protons in the element be x.

∴ Number of neutrons in the element
= x + 31.7% of x
= x + 0.317x
= 1.317x
According to the question,
Mass number of the element = 81
(Number of protons + number of neutrons) = 81
 x  1.317 x  81
2.317 x  81
81
x
2.317
x  35
Hence, the number of protons in the element i.e., x is 35.
Since the atomic number of an atom is defined as the number of protons present in its nucleus, the atomic
number of the given element is 35.
81
The atomic symbol of the element is 35 Br .

Question 43:
An ion with mass number 37 possesses one unit of negative charge. If the ion contains 11.1% more
neutrons than the electrons, find the symbol of the ion.

Solution 43:
Let the number of electrons in the ion carrying a negative charge be x.
Then,
Number of neutrons present
= x + 11.1% of x
= x + 0.111x
= 1.111x
Number of electrons in the neutral atom = (x – 1)
(When an ion carries a negative charge, it carries an extra electron)
∴ Number of protons in the neutral atom = x – 1
Given,

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Mass number of the ion = 37

∴ (x – 1) + 1.111x = 37
2.111x = 38
x = 18
Number of electrons = 18; Number of protons = 18 – 1 = 17
Atomic number of the ion = 17; Atom correspondence to ion = cl
1
∴ The symbol of the ion is 37
17 Cl

Question 44:
An ion with mass number 56 contains 3 units of positive charge and 30.4% more neutrons than electrons.
Assign the symbol to this ion.

Solution 44:
Let the number of electrons present in ion A3 be x
Number of neutrons in it = x + 30.4% of x = 1.304 x
Since the ion is tripositive,
⇒ Number of electrons in neutral atom = x + 3
∴ Number of protons in neutral atom = x + 3
Given,
Mass number of the ion = 56
(x + 3) + (1.304x) = 56
2.304x = 53
53
x
2.304
x = 23
Number of protons = x + 3 = 23 + 3 = 26
3
The symbol of the ion 56 26 Fe

Question 45:
Arrange the following type of radiations in increasing order of frequency:
(a) radiation from microwave oven
(b) amber light from traffic signal
(c) radiation from FM radio

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(d) cosmic rays from outer space and

(e) X-rays.

Solution 45:
The increasing order of frequency is as follows:
Radiation from FM radio < amber light < radiation from microwave oven < X- rays < cosmic rays
The increasing order of wavelength is as follows:
Cosmic rays < X-rays < radiation from microwave ovens < amber light < radiation of FM radio.

Question 46:
Nitrogen laser produces a radiation at a wavelength of 337.1 nm. If the number of photons emitted is 5.6
× 1024, calculate the power of this laser.

Solution 46:
Power of laser = Energy with which it emits photons
Nhc
Power  E 

Where,
N = number of photons emitted
h = Planck’s constant
c = velocity of radiation
λ = wavelength of radiation
Substituting the values in the given expression of Energy (E):

E
5.6 10  6.626 10 Js 3 10
24 34 8
ms 1 
337.110 m 9

= 0.3302×107 J
= 3.33×106 J
Hence, the power of the laser is 3.33×106 J.

Question 47:
Neon gas is generally used in the sign boards. If it emits strongly at 616 nm, calculate
(a) the frequency of emission,

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Class XI – NCERT – Chemistry Chapter 2
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(b) distance traveled by this radiation in 30 s

(c) energy of quantum and
(d) number of quanta present if it produces 2 J of energy.

Solution 47:
Wavelength of radiation emitted = 616 nm = 616×10–9 m (Given)
(a) Frequency of emission  v 
c
v

Where,
c = velocity of radiation
λ = wavelength of radiation
Substituting the values in the given expression of:  v 
3.0 108 m/s
v
616 109 m
= 4.87×108 ×109 ×10–3 s–1
ν = 4.87×1014 s–1
Frequency of emission (ν) = 4.87×1014 s–1
(b) Velocity of radiation, (c) = 3.0×108 ms–1
Distance travelled by this radiation in 30 s
= (3.0×108 ms–1) (30 s)
= 9.0×109 m
(c) Energy of quantum (E) = hν (6.626×10–34 Js) (4.87×1014 s–1)
Energy of quantum (E) = 32.27×10–20 J
–20
(d) Energy of one photon (quantum) = 32.27×10 J
–20
Therefore, 32.27×10 J of energy is present in 1 quantum. Number of quanta in 2 J of energy
2J

32.27 1020 J
= 6.19 × 1018

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Class XI – NCERT – Chemistry Chapter 2
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= 6.2 × 1018

Question 48:
In astronomical observations, signals observed from the distant stars are generally weak. If the photon
–18
detector receives a total of 3.15×10 J from the radiations of 600 nm, calculate the number of photons
received by the detector.

Solution 48:
hc
From the expression of energy of one photon (E), E 

Where,
λ = wavelength of radiation
h = Planck’s constant
c = velocity of radiation
Substituting the values in the given expression of E:

E 
 6.626 10 Js 310 ms 
34 8 1

 600 10 m 9

E = 3.313×10–19 J
Energy of one photon = 3.313×10–19 J
Number of photons received with 3.15×10–18 J energy
3.15 1018 J

3.313 1019 J
= 9.5
≈ 10

Question 49:
Lifetimes of the molecules in the excited states are often measured by using pulsed radiation source of
duration nearly in the nano second range. If the radiation source has the duration of 2 ns and the number
of photons emitted during the pulse source is 2.5 × 10 15, calculate the energy of the source.

Solution 49:

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Frequency of radiation (ν),

1
v
2.0 109 s
ν = 5.0×108 s–1
Energy (E) of source = Nhν
Where,
N = number of photons emitted
h = Planck’s constant
ν = frequency of radiation
Substituting the values in the given expression of (E):
E = (2.5×1015) (6.626×10–34 Js) (5.0×108 s–1)
E = 8.282×10–10 J
Hence, the energy of the source (E) is 8.282×10–10 J.

Question 50:
The longest wavelength doublet absorption transition is observed at 589 and 589.6 nm. Calculate the
frequency of each transition and energy difference between two excited states.

Solution 50:
(i) Given,
Wavelength associated with first transition, λ1 =589 nm =589 × 10–9 m
Wavelength associated with second transition, λ2 = 589.6 nm =589.6 × 10–9 m
c 3 108 ms1
Frequency of first wavelength is v1 = 
1 589 10–9 m
= 5.093 × 1014 s–1
c 3 108 ms1
And, frequency of second wavelength is v2 = 
2 589.6 10–9 m
= 5.088 × 1014s-1
(ii) Energy difference between two excited states is given as,
ΔE = hv1 – hv2
Or, ΔE = h(v1 – v2)
= 6.626 × 10-34 Js × (5.093 × 1014 s-1 – 5.088 × 1014 s-1)

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= 6.626 × 10-34 Js × 0.005 × 1014 s-1

= 3.31 × 10-22 J

Question 51:
The work function for caesium atom is 1.9 eV. Calculate (a) the threshold wavelength and (b) the
threshold frequency of the radiation. If the caesium element is irradiated with a wavelength 500 nm,
calculate the kinetic energy and the velocity of the ejected photoelectron.

Solution 51:
It is given that the work function (W0 ) for caesium atom is 1.9 eV.
hc
(a) From the expression, W0  , we get:
0
hc
0 
W0
Where,
0 = threshold wavelength
h = Planck’s constant
c = velocity of radiation
Substituting the values in the given expression of ( 0 ):

0 
 6.626 10 14
Js 3.0 108 ms1 
1.9 1.602 1019 J
0  6.53107 m
Hence, the threshold wavelength 0 is 653 nm.
(b) From the expression, W0  hv0 , we get:
W0
v0 
h
Where,
v0 = threshold frequency
h = Planck’s constant
Substituting the values in the given expression of v0

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Class XI – NCERT – Chemistry Chapter 2
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1.9 1.602 1019 J

v0 
6.626 1034 Js
(1 eV = 1.602×10–19 J)
v0 = 4.593×1014 s-1
Hence, the threshold frequency of radiation ( v0 ) is 4.593×1014 s–1.
(c) According to the question:
Wavelength used in irradiation (λ) = 500 nm
Kinetic energy = h (ν - v0 )
1 1 
= hc   
  0 
   
  6.626 1034 Js 3.0 108 ms 1   0 
 0 
  653  500109 m 
 1.9878Jm  18 2 
  653 50010 m 


1.9878 10 15310 
26 9

 653500
= 9.3149×10-20 J
Kinetic energy of the ejected photoelectron = 9.3149×10–20 J
1
Since K.E  mv2  9.3149 1020 J
2
2  9.3149 1020 J 
v
9.10939 1011 m2 s 2
v  2.04511011 m2 s2
v = 4.52×105 ms–1
Hence, the velocity of the ejected photoelectron (v) is 4.52×105 ms–1.

Question 52:

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Class XI – NCERT – Chemistry Chapter 2
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Following results are observed when sodium metal is irradiated with different wavelengths. Calculate (a)
threshold wavelength and, (b) Planck’s constant.
λ (nm) 500 450 400


v  10–5 cm s –1  2.55 4.35 5.35

Solution 52:
 
(a) Assuming the threshold wavelength to be 0 nm  0 109 m , the kinetic energy of the radiation is
given as:
1
h  v  v0   mv2
2
Three different equalities can be formed by the given value as:
1 1  1
hc     mv 2
  0  2
  1
  m  2.55 10 10 ms 
1 1 5 2 1
hc  
 500 10 0 10 m  2
9 9

hc  1 1 1
   m  2.55 103 ms 1  1
2

10 m  500 0  2
9

Similarly,
hc  1 1 1
   2
2
    m 3.45 103 ms 1
10 m  450 0  2
9

hc  1 1 1
   m  5.35 103 ms 1   3
2

10 m  400 0  2
9

Dividing equation (3) by equation (1):

 0  400 
 400   5.35 103 ms 1 2
 0 

 0  500   2.55 103 ms 1 2
 500 
 0 

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50  2000  5.35  28.6225

2

  
40  2000  2.55  6.5025
50  2000
 4.40177
40  2000
17.60700  50  8803.537  2000
6805.537
0 
12.607
0  539.8nm
0  540nm
Threshold wavelength  0  = 540 nm
Note: part (b) of the question is not done due to the incorrect values of velocity given in the question.
50  2000  5.35  28.6225
2

  
40  2000  2.55  6.5025
50  2000
 4.40177
40  2000
17.60700  50  8803.537  2000
6805.537
0 
12.607
0  539.8nm
0  540nm

Question 53:
The ejection of the photoelectron from the silver metal in the photoelectric effect experiment can be
stopped by applying the voltage of 0.35 V when the radiation 256.7 nm is used. Calculate the work
function for silver metal.

Solution 53:
From the principle of conservation of energy, the energy of an incident photon (E) is equal to the sum of
the work function (W0 ) of radiation and its kinetic energy (K.E.) i.e., E = W0 + K.E.

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Class XI – NCERT – Chemistry Chapter 2
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W0 = E – K.E.
hc
Energy of incident photon (E) = Where,

c = velocity of radiation
h = Planck’s constant
λ = wavelength of radiation
Substituting the values in the given expression of E:

E
 6.626 10 34
Js  3.0 108 ms 1 
256.7 109 m
 7.744 1019 J
7.744 1019
1.602 1019 eV
E = 4.83 eV
The potential applied to silver metal changes to kinetic energy (K.E) of the photoelectron. Hence,
K.E = 0.35 V
K.E = 0.35 eV
Work function, W0 = E – K.E.
= 4.83 eV – 0.35 eV
= 4.48 eV
50  2000  5.35  28.6225
2

  
40  2000  2.55  6.5025
50  2000
 4.40177
40  2000
17.60700  50  8803.537  2000
6805.537
0 
12.607
0  539.8nm
0  540nm

Question 54:
If the photon of the wavelength 150 pm strikes an atom and one of its inner bound electrons is ejected out

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Class XI – NCERT – Chemistry Chapter 2
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with a velocity of 1.5×107 ms–1, calculate the energy with which it is bound to the nucleus.

Solution 54:
Energy of incident photon (E) is given by,
hc
E



 6.626 10 Js 3.0 10
34 8
ms 1 
150 10 m  12

 1.3252 1015 J
 13.252 1016 J
Energy of the electron ejected (K.E)
1
 me v 2
2
  9.10939  1031 kg 1.5  107 ms 1 
1 2

2
= 10.2480 × 1017 J
= 1.025 1016 J
Hence, the energy with which the electron is bound to the nucleus can be obtained as:
= E – K.E
= 13.252×10–16 J – 1.02×10–16 J
= 12.227×10–16 J
12.227 1016
 eV
1.602 1019
 7.6 103 eV
50  2000  5.35  28.6225
2

  
40  2000  2.55  6.5025
50  2000
 4.40177
40  2000
17.60700  50  8803.537  2000

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Class XI – NCERT – Chemistry Chapter 2
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6805.537
0 
12.607
0  539.8nm
0  540nm

Question 55:
Emission transitions in the Paschen series end at orbit n = 3 and start from orbit n and can be represented
as v = 3.29×1015 (Hz) [1/32 - 1/n 2 ]
Calculate the value of n if the transition is observed at 1285 nm. Find the region of the spectrum.

Solution 55:
Wavelength of transition = 1285 nm
= 1285×10–9 m (Given)
1 1
v = 3.29×1015  2  2 
3 n 
(Given)
c
Since v 

3.0 10 ms 1
8

1285 109 m
v = 2.33×1014 s1
Substituting the value of ν in the given expression,
1 1 
3.29 1015   2   2.331014
9 n 
1 1 2.33 1014
 
9 n2 3.29 1015
1 1
 0.7082 101  2
9 n
1
 2  1.1101  0.7082 101
n

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Class XI – NCERT – Chemistry Chapter 2
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1
2
 4.029  102
n
1
n
4.029  102
n = 4.98
n≈5
Hence, for the transition to be observed at 1285 nm, n = 5. The spectrum lies in the infra-red region.

Question 56:
Calculate the wavelength for the emission transition if it starts from the orbit having radius 1.3225 nm
and ends at 211.6 pm. Name the series to which this transition belongs and the region of the spectrum.

Solution 56:
The radius of the nth orbit of hydrogen-like particles is given by,
0.529n2 o
r A
Z
52.9n2
r pm
Z
For radius ( r1 ) = 1.3225 nm
= 1.32225×10–9 m
= 1322.25×10–12 m
= 1322.25 pm
rZ
n12  1
52.9
1322.25 Z
n12 
52.9
Similarly,
211.6Z
n22 
52.9
2
n1 1322.5

n22 211.6

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Class XI – NCERT – Chemistry Chapter 2
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n12
 6.25
n22
n1
 2.5
n2
n1 25 5
 
n2 10 2
 n1 = 5 and n 2 =2
Thus, the transition is from the 5th orbit to the 2nd orbit. It belongs to the Balmer series. Wave number
v  for the transition is given by,
1 1
1.097 107 m1  2  2 
2 5 
 21 
 1.097 107 m1  
 100 
 2.303 106m–1
1
Wavelength (λ) associated with the emission transition is given by,  
v
1

2.303 106 m1
= 0.434×10–6 m
λ = 434 nm

Question 57:
Dual behavior of matter proposed by de Broglie led to the discovery of electron microscope often used
for the highly magnified images of biological molecules and other type of material. If the velocity of the
electron in this microscope is 1.6×106 ms–1, calculate de Broglie wavelength associated with this
electron.

Solution 57:
From de Broglie’s equation,
h

mv

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Class XI – NCERT – Chemistry Chapter 2
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6.626 1034 Js

9.10939 1031 kg 1.6 106 ms1 
= 4.55×10–10 m
λ = 455 pm
de Broglie’s wavelength associated with the electron is 455 pm.

Question 58:
Similar to electron diffraction, neutron diffraction microscope is also used for the determination of the
structure of molecules. If the wavelength used here is 800 pm, calculate the characteristic velocity
associated with the neutron.

Solution 58:
From de Broglie’s equation,
h

mv
h
v
m
Where,
v = velocity of particle (neutron)
h = Planck’s constant
m = mass of particle (neutron)
λ = wavelength
Substituting the values in the expression of velocity (v),
(6.626 1034 ) Kg m 2 s 1

1.675 1027 kg 8 1010 m 
6.626 103 1
 ms
1.675  8
 4.94  102 ms –1
v  494 ms –1
Velocity associated with the neutron = 494 ms–1

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Class XI – NCERT – Chemistry Chapter 2
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Question 59:
If the velocity of the electron in Bohr’s first orbit is 2.19×10 6 ms–1, calculate the de Broglie wavelength
associated with it.

Solution 59:
According to de Broglie’s equation,
h

mv
Where,
λ = wavelength associated with the electron
h = Planck’s constant
m = mass of electron
v = velocity of electron
Substituting the values in the expression of λ:
6.626 1034 Js

9.10939 1031 kg  2.19 106 ms1 
100
  3.32 1010 m  3.32 1010 m 
100
  332 1012 m
 = 332 pm
Wavelength associated with the electron = 332 pm

Question 60:
The velocity associated with a proton moving in a potential difference of 1000 V is 4.37×105 ms–1. If the
hockey ball of mass 0.1 kg is moving with this velocity, calculate the wavelength associated with this
velocity.

Solution 60:
According to de Broglie’s expression,
h

mv
Substituting the values in the expression,

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6.626 1034 Js

 0.1kg   4.37 105 ms 1 
  1.516 1038 m

Question 61:
If the position of the electron is measured within an accuracy of +0.002 nm, calculate the uncertainty in
the momentum of the electron. Suppose the momentum of the electron is h/4π m × 0.05 nm, is there any
problem in defining this value.

Solution 61:
From Heisenberg’s uncertainty principle,
h 1 h
x  p   p  .
4 x 4
Where,
∆x = uncertainty in position of the electron
∆p = uncertainty in momentum of the electron
Substituting the values in the expression of ∆p:
1 6.626 1034 Js
p  
0.002 nm 4   3.14 
1 6.626 1034 Js
 
2 1012 m 4  3.14
–1
= 2.637×10–23 Js m
∆p = 2.637 × 10–23 kg ms–1 (1 J = 1 kg ms2s–1)
Uncertainty in the momentum of the electron = 2.637×10 –23 kg ms–1.
h
Actual momentum 
4 m  0.05nm
6.626 1034 Js

4  3.14  5.0 1011 m
= 1.055 × 10–24 kg ms–1
Since the magnitude of the actual momentum is smaller than the uncertainty, the value cannot be defined.

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Class XI – NCERT – Chemistry Chapter 2
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Question 62:
The quantum numbers of six electrons are given below. Arrange them in order of increasing energies. If
any of these combination(s) has/have the same energy lists:
1. n = 4, l = 2, ml = –2 , ms = –1/2
2. n = 3, l = 2, ml= 1 , ms = +1/2
3. n = 4, l = 1, ml = 0 , ms = +1/2
4. n = 3, l = 2, ml = –2 , ms = –1/2
5. n = 3, l = 1, ml = –1 , ms= +1/2
6. n = 4, l = 1, ml = 0 , ms = +1/2

Solution 62:
For n = 4 and l = 2, the orbital occupied is 4d.
For n = 3 and l = 2, the orbital occupied is 3d.
For n = 4 and l = 1, the orbital occupied is 4p.
Hence, the six electrons i.e., 1, 2, 3, 4, 5, and 6 are present in the 4d, 3d, 4p, 3d, 3p, and 4p orbitals
respectively.
Therefore, the increasing order of energies is 5(3p) < 2(3d) = 4(3d) < 3(4p) = 6(4p) < 1 (4d).

Question 63:
The bromine atom possesses 35 electrons. It contains 6 electrons in 2p orbital, 6 electrons in 3p orbital
and 5 electrons in 4p orbital. Which of these electron experiences the lowest effective nuclear charge?

Solution 63:
Nuclear charge experienced by an electron (present in a multi-electron atom) is dependant upon the
distance between the nucleus and the orbital, in which the electron is present. As the distance increases,
the effective nuclear charge also decreases.
Among p-orbitals, 4p orbitals are farthest from the nucleus of bromine atom with (+35) charge. Hence,
the electrons in the 4p orbital will experience the lowest effective nuclear charge. These electrons are
shielded by electrons present in the 2p and 3p orbitals along with the s-orbitals. Therefore, they will
experience the lowest nuclear charge.

Question 64:

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Class XI – NCERT – Chemistry Chapter 2
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Among the following pairs of orbitals which orbital will experience the larger effective nuclear charge?
(i) 2s and 3s, (ii) 4d and 4f, (iii) 3d and 3p

Solution 64:
Nuclear charge is defined as the net positive charge experienced by an electron in the orbital of a multi-
electron atom. The closer the orbital, the greater is the nuclear charge experienced by the electron(s) in it.
(i) The electron(s) present in the 2s orbital will experience greater nuclear charge (being closer to the
nucleus) than the electron(s) in the 3s orbital.
(ii) 4d will experience greater nuclear charge than 4f since 4d is closer to the nucleus.
(iii) 3p will experience greater nuclear charge since it is closer to the nucleus than 3f.

Question 65:
The unpaired electrons in Al and Si are present in 3p orbital. Which electrons will experience more
effective nuclear charge from the nucleus?

Solution 65:
Nuclear charge is defined as the net positive charge experienced by an electron in a multi-electron atom.
The higher the atomic number, the higher is the nuclear charge. Silicon has 14 protons while aluminium
has 13 protons. Hence, silicon has a larger nuclear charge of (+14) than aluminium, which has a nuclear
charge of (+13). Thus, the electrons in the 3p orbital of silicon will experience a more effective nuclear
charge than aluminium.

Question 66:
Indicate the number of unpaired electrons in: (a) P, (b) Si, (c) Cr, (d) Fe and (e) Kr.

Solution 66:
(a) Phosphorus (P): Atomic number = 15
The electronic configuration of P is: 1s2 2s2 2p6 3s2 3p3
The orbital picture of P can be represented as:

From the orbital picture, phosphorus has three unpaired electrons.

(b) Silicon (Si): Atomic number = 14
The electronic configuration of Si is: 1s2 2s2 2p6 3s2 3p2

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Class XI – NCERT – Chemistry Chapter 2
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The orbital picture of Si can be represented as:

From the orbital picture, silicon has two unpaired electrons.

(c) Chromium (Cr): Atomic number = 24
The electronic configuration of Cr is: 1s2 2s2 2p6 3s2 3p6 4s1 3d5
The orbital picture of chromium is:

From the orbital picture, chromium has six unpaired electrons.

(d) Iron (Fe): Atomic number = 26
The electronic configuration is: 1s2 2s2 2p6 3s2 3p6 4s2 3d6
The orbital picture of chromium is:

From the orbital picture, iron has four unpaired electrons.

(e) Krypton (Kr): Atomic number = 36
The electronic configuration is: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6
The orbital picture of krypton is:

Since all orbitals are fully occupied, there are no unpaired electrons in krypton.

Question 67:
(a) How many sub-shells are associated with n = 4?
(b) How many electrons will be present in the sub-shells having ms value of –1/2 for n = 4?

Solution 67:
(a) n = 4 (Given)
For a given value of ‘n’, ‘l’ can have values from zero to (n – 1).
 l = 0, 1, 2, 3
Thus, four sub-shells are associated with n = 4, which are s, p, d and f.
(b) Number of orbitals in the nth shell = n2
For n = 4

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Number of orbitals = 16
 1
If each orbital is taken fully, then it will have 1 electron with ms value of    .
 2
 1
 Number of electrons with ms value of    is 16.
 2

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