Dual Nature (Autosaved) 1-30

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PHOTONS
 Some important properties of photons
(a) Photons do not have any rest mass.

(b) Photons possess momentum. The momentum p of a photon is given by
Where h is the Planck constant and λ is the wavelength of light.
(c) The unit of energy used for expressing the energy of a photon is electron volt.1 eV = 1.6
(d) Photons do not possess any charge and therefore, cannot be deflected by either electric or magnetic fields.
(e) Photons travel with a speed of 3 in vacuum. This velocity is denoted by c.
Also c = vλ
Where v is the frequency of photon and λ is the wavelength of light to which the photon belongs.
(f) The mass of a moving photon
(g) All photons of a particular frequency have the same energy (hv), same momentum whatever be the
intensity of radiation.
(h) In a photon-particle collision, conservation of energy and conservation of momentum is valid. But the number of
photons may not be conserved in a collision. Photon may be absorbed or a new photon may be created.
(i) If the intensity of light (radiation) of a given frequency is increased there is an increase in the number of photons
crossing a given area in a given time.
Illustration 1. Monochromatic light of frequency 6 1014 Hz is produced by a laser. The power emitted is
2 10-3 W.
(a) What is the energy of a photon in the light beam ?

(b) How many photons per second on an average, are emitted by the source?
Solution .
(a) E = h v = 6.63 10 -34 6 1014 = 3.98 10 -19 J
(b) Power =
For n number of photons, the energy will be E = n h v
P = .
Illustration 2. The momentum of a photon is 3 kg m s-1, find the wavelength and frequency
associated with the photon.
Concept Classes 3/5 East Punjabi Bagh, 9811741187

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Solution . Here p = 3 10 -29 kg m s-1
We know that
Also v =
Illustration 3. A photon posseses an energy of 1.5 MeV. find its momentum.
Solution . Given E = 1.5 Me V = 1.5 1.6 J
We know that p = .
Illustration 4. Monochromatic light of wavelength 632.8 nm is produced by a helium-neon laser. The power
emitted is 9.42 mW.
(a) Find the energy and momentum of each photon in the light beam.
(b) How many photons per second, on the average, arrive at a target irradiated by this beam?
(Assume the beam to have uniform cross-section which is less than the target area), and
(c) How fast does a hydrogen atom have to travel in order to have the same momentum as that of the
photon? NCERT
Solution. Given, m
Power,
(a) J
Also,
(b) Number of photons emitted per second,
(c) Mass of hydrogen atom ≈ mass of proton

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Illustration 5. The energy flux of sunlight reaching the surface of the earth is 1.388 × . How
many photons (nearly) per square metre are incident on the Earth per second? Assume that the photons in
the sunlight have an average wavelength of 550 nm. NCERT
Solution. Given
Energy flux ∅ = 1.388 ×
Also
Energy of each photon J.
∅
no. of photons per square metre of earth .
Illustration 6. A 100 W sodium lamp radiates energy uniformly in all directions. The lamp is located at the
centre of a large sphere that absorbs all the sodium light which is incident on it. The wavelength of the
sodium light is589 nm. (a) What is the energy per photon associated with the sodium light? (b) At what rate
are the photons delivered to the sphere? NCERT
Solution. Power of sodium lamp, P = 100 W
(a) Energy of photon, J
Rate at which the protons are delivered photons/s.
Illustration 7. Two metals A And B have work functions 4eV and 10eV respectively. Which metal has
higher threshold wavelength ?
Solution.
So metal A with lower work function has higher threshold wavelength.
Illustration 8. Ultraviolet light is incident on two photosensitive materials having work functions and
In which case will the kinetic energy of the emitted electrons be greater ? Why ?
Solution. K.E. of a photoelectron =
Hence the kinetic energy of the emitted electrons will be greater for the sensitive material having smaller
work function

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Illustration 9. Does the ‘stopping potential’ in photoelectric emission depend upon

(i) Intensity of the incident radiation in a photocell ?
(ii) the frequency of the incident radiation ?
Solution. (i) No, the stopping potential does not depend on the intensity of the incident radiation.
(ii) Yes, the stopping potential depends on the frequency of incident radiation. Above the threshold
frequency,
Illustration 10. Electrons are emitted from a photo-sensitive surface when it is illuminated by green light
but electron emission does not take place by yellow light. Will the electrons be emitted when the surface is
illuminated by (i) red light, and (ii) blue light ?
Solution. (i) No, electrons are not emitted by red light,
Because
(ii) Yes, electrons are emitted by blue light, because
Illustration 11. The maximum kinetic energy of a photoelectron is 3 eV. What is its stopping potential ?
Solution. Stopping potential,
ELECTRON EMISSION
The minimum amount of energy required to remove an electron from the metal surface is called the work
function of the metal. Work function is denoted by ∅ or .
Work functions of certain metals
Metal Work function Metal Work function

∅ ∅
Cs 2.14 Al 4.28
K 2.30 Hg 4.49
Na 2.75 Cu 4.65
Ca 3.20 Ag 4.70
Mo 4.17 Ni 5.15
Pb 4.25 Pt 5.65
Types of Electronic Emission

The energy required to remove electrons from the metal surface can be provided by different ways.
(a) Thermionic emission: when electrons are emitted by heating the metal surface, the emission is called thermionic
emission. The emitted electrons are referred as thermal electrons or thermions.

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(b) Field emission: when electrons are emitted by placing the metal surface in a very strong electric field
, the emission is called field emission.
(c) Photoelectric emission: when the electrons are emitted by illuminating

the metal surface by light of a suitable frequency, the emission is called
photoelectric emission.
(d) Secondary emission: when electrons are emitted by striking the metal
surface with a beam of fast moving electrons, then emission is called
secondary emission.
Experimental Study of Photoelectric Effect
Two conducting electrodes, the anode (A) and cathode (C), are enclosed in
an evacuated glass tube. These are connected to an external circuit
consisting of a commutator and a battery Light falls on the surface of
cathode C causes a current to flow in the external circuit which can be
measured by an ammeter ( ). We will now study the impact of the
following three parameters on the photocurrent:
(i) Intensity of the incident radiation,

(ii) Potential V, and
(iii) Frequency of incident radiations.
(i) Intensity of Incident Radiations
While maintaining a constant value of V, as we keep increasing the

intensity of the incident radiations, the photocurrent increases linearly with
increasing intensity. This shows that there is a direct proportionality
relation between the intensity of the incident radiation and the number of
electrons emitted.
(ii) Potential V
If we keep the frequency and intensity of incident radiations constant, then

as the positive potential of the collector A increases the photocurrent
increases initially and achieves a constant value called saturation current.

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When we apply a negative potential on the collector A, then on increasing the (negative) potential the
photocurrent decreases and .the photocurrent drops to zero. The value of negative potential, given to the collector
A, which prevents the flow of photocurrent, is called the stopping potential and sometimes the cut off potential.
An increase in the intensity of light only increases the saturation photocurrent and does not change the value of the
stopping potential
(iii) Frequency of incident Radiations
We now study the relation between the frequency v of the incident

radiation and the stopping potential . We suitably adjust the
same intensity of light radiation at various frequencies and study
the variation of photocurrent with collector plate potential.
The resulting variation is shown in
The graphs shows that

 If the frequency of incident radiations increases keeping the intensity same, the saturation current remains the
same but the stopping potential changes.
 If incident frequency is increased, magnitude of stopping potential
also increases.
 The three graphs lines are for frequencies,
for which the stopping potentials are respectively.
 If we plot a graph between the frequency of incident radiation and
the corresponding stopping potential for different metals we get a
straight line. As shown in Fig.
 The stopping potential varies linearly with the frequency of
incident radiation for a given photosensitive.
 There exists a certain minimum cut-off frequency for which the stopping potential is zero.
We know summarise the experimental features and observations described in this section.
1. For a given photosensitive material and frequency of incident radiation (above the threshold frequency),
the photoelectric current is directly proportional to the intensity of incident light.
2. For a given photosensitive material and frequency of incident radiation, saturation current is found to be
proportional to the intensity of incident radiation whereas the stopping potential is independent of its intensity
3. For a given photosensitive material, there exists a certain minimum cut- off frequency of the incident
radiation, called the threshold frequency, below which no emission of photoelectrons takes place, no matter
how intense the incident light is above the threshold frequency, the stopping potential or equivalently the
maximum kinetic energy of the emitted photoelectrons increases linearly with the frequency of the incident
radiation, but is independent of its intensity

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4. The photoelectric emission is an instantaneous process without any apparent time lag , even
when the incident radiation is made exceedingly dim.
Failure of Wave Theory to Explain Photoelectric Effect

The explanation of photoelectric effect is possible only with the quantum mechanical model of light.
This is so because
1. A wave transfers energy continuously and not in the form of lumps or packets called photons. Hence, using the
wave model, it is impossible to explain the threshold frequency and also the direct proportionality between
stopping potential and frequency of incident light. With the wave model we would expect the stopping potential to
be dependent on the intensity of incident light because of the continuous energy transfer. But this is in
contradiction to the experimental facts.
2. Also, with the wave model we would expect photoemission to occur at all frequencies and not just above the
threshold frequency. This too contradicts the experimental observation that below the threshold frequency, no
photoemission occurs.
3. If light of low intensity is incident on the metal surface, one would expect the electron to take time to accumulate
enough energy to escape the surface. Hence, there would be a time lag between the incident of light and emission
of photoelectrons. But this too contradicts the experimental observation that photoemission occurs within one
nanosecond of incidence of light if the incident light has a frequency greater than the threshold frequency.
Photoelectron emission is, therefore, instantaneous with incidence of light and this cannot be explained by the
wave concept of light.
EINSTEIN’S PHOTOELECTRIC EQUATION : ENERGY QUANTUM OF RADIATION

According to Einstein’s Radiation energy is built up of discrete units. Each quantum of radiant energy has energy hv.
Where h is Plank’s constant and v the frequency of light. In photoelectric effect, an electron absorbs a quantum of
energy (hv) of radiation. If this quantum of energy absorbed exceeds the minimum energy needed for the electron to
escape from the metal surface (work function ∅ ), the electron is emitted with maximum kinetic energy
∅ …….. (i)
More tightly bound electrons will emerge with kinetic energies less than the maximum value. Note that the intensity of
light of a given frequency is determined by the number of photons incident per second. Increasing the intensity will
increase the number of emitted electrons per second. However, the maximum kinetic energy of the emitted
photoelectrons is determined by the energy of each photon.
 According to Eq.(i). depends linearly on v, and is independent of intensity of radiation, in agreement with
observation.
 Since must be non-negative, (Eq.(i) implies that photoelectric emission is possible only if
∅
where ∅ ∅ work function

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threshold frequency
This shows emission is only possible when incident frequency is more than threshold frequency.
∅
Also,
This shows the greater the work function ∅ , the higher the minimum or threshold frequency needed to
emit photoelectrons.
 According to Einstein theory the intensity of radiation is proportional to the number of energy quanta per unit area
per unit time. The greater the number of energy quanta available, the greater is the number of electrons absorbing
the energy quanta and greater therefore, is the number of electrons coming out of the metal (for v > ).
This explains why, for , photoelectric current is proportional to intensity.
 In Einstein’s picture, the basic elementary process involved in photoelectric effect is the absorption of a light
quantum by an electron. This process is instantaneous. Thus, whatever may be the intensity photoelectric emission
is instantaneous. Low intensity does not mean delay in emission, since the basic elementary process is the same.
 ∅ for
As where Stopping potential
∅
( )
This is an important result. It predicts that the versus v curve is a straight line with slope = , independent of
the nature of the material.
PARTICLE NATURE OF LIGHT: THE PHOTON
Photoelectric effect thus gave evidence to the strange fact that light in interaction with matter behaved as it was made
of quanta or packets of energy, each of energy h v.
We can summarise the photon picture of electromagnetic radiation as follows:
1. In interaction of radiation with matter, radiation behaves as if it is made up of particles called photons.
2. Each photon has energy E ( ) and momentum , and speed c, the speed of light.
3. All photons of light of a particular frequency v, or wavelength , have the same energy
E and momentum , whatever the intensity of radiation may be. By increasing
the intensity of light of given wavelength, there is only an increase in the number of photons per second crossing a
given area, with each photon having the same energy. Thus, photon energy is independent of intensity of
radiation.
4. Photons are electrically neutral and are not deflected by electric and magnetic fields.

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5. In a photon-particle collision (such as photon-electron collision), the total energy and total momentum are
conserved. However, the number of photons may not be conserved in a collision. The photon may be absorbed or
a new photon may be created.
Illustration 12. The threshold frequency for a certain metal is 3.3 × Hz. If light of frequency 8.2 ×
Hz is incident on the metal, predict the cut-off voltage for the photoelectric emission. NCERT
Solution. Given
Hz
Also Js and C.
Using
Or V.
Illustration 13. The work function for a certain metal is 4.2 eV. Will this metal give photoelectric emission
for incident radiation of wavelength 330 nm? NCERT
Solution. W = 4.2 eV = 4.2 × 1.6 × J
= 6.72 × J
using
Or Hz
Since
the photoelectric emission for this radiation can not take place.
Illustration 14. Light of frequency 7.21 × is incident on a metal surface. Electrons with a
maximum speed of 6.0 × are ejected from the surface. What is the threshold frequency for
photoemission of electrons ? NCERT
Solution. Here, Hz
Using the relation

10
( )
Hz.
Illustration 15. Light of wavelength 488 nm is produced by an argon laser which is used in the
photoelectric effect. When light from this spectral line is incident on the cathode, the stopping (cut - off)
potential of photoelectrons is 0.38 V. Find the work function of the material from which the cathode is made
NCERT
Solution. Here,
Also Js,
And
Hz.
Using
We get
Illustration 16. Calculate the kinetic energy of photoelectrons (in eV) emitted on shining light of
wavelength 6.2 m on a metal surface. The work function of the metal is 0.1.
Solution. Energy ‘E’ of a photon of wavelength 6.2 m is given by
‘E’
Energy of the most energetic photons emitted will be
K.E. ∅
Illustration 17. Two metals A and B have work functions 2 eV and 4 eV respectively, which metal has a
lower threshold wavelength for photoelectric effect?
Solution. If ∅ is the work function, the threshold frequency , then the corresponding threshold
wavelength λ is given by
∅
∅ or and ∅
Since the threshold frequency is directly proportional to the work function, the threshold frequency is
larger when work function is larger, the threshold frequency is larger. Hence, the threshold frequency is
lower for metal A.
The relation for threshold wavelength is the reciprocal of the relation for frequency, hence, threshold
wavelength is lower for metal B.
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Illustration 18. If the maximum kinetic energy of electron emitted in a photocell is 5 eV, what is the
stopping potential?
Solution. Now, K.E. is stopping potential.
Here K.E.
Or
Remember that if the maximum kinetic energy of emitted electrons is given in electron volts, it is
numerically equal to the stopping potential in volts.
Illustration 19. The work function of caesium is 2.14 eV. Find (a) the threshold frequency for caesium and
(b) the wavelength of the incident light if the photocurrent is bought to zero by a stopping
potential of 0.6 V.
∅
Solution. (a)
(b) ∅ ∅ ∅
Illustration 20. Monochromatic light of frequency 6.0 × Hz is produced by a laser. The power emitted
is 2.0 × . (a) What is the energy of photon in the light beam? (b) How many photons per second, on
an average, are emitted by the source?
Solution.
(a) Each photon has an energy
(b) If N is the number of photons emitted by the source per second, the power P transmitted in the beam
equals N times the energy per photon E, so that P = N E. Then
photons per second.
Illustration 21. The work function of cesium is 2.14 eV, Find (a) the threshold frequency for caesium, and
(b) the wavelength of the incident light if the photocurrent is brought to zero by a stopping potential of 0.60
V.
Solution.

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(a) For the cut-off or threshold frequency, the energy of the incident radiation must be equal to
work function ∅ so that
Thus, for frequencies less than this threshold frequency, no photoelectrons are ejected.
(b) Photocurrent reduces to zero, when maximum kinetic energy of the emitted photoelectrons equals the
potential energy e by the retarding potential . Einstein’s photoelectric equation is
∅ ∅
Or, ∅
Illustration 22. The wavelength of light in the visible region is about 390 nm for violet colour, about 550
nm (average wavelength) for yellow. Green colour and bout 760 nm for red colour.
(a) What are the energies of photons in (eV) at the (i) violet end, (ii) average wavelength, yellow-green
colour, and (iii) red end of the visible spectrum (Take h = 6.63 × J s and 1 eV = 1.6× J.)
(b) From which of the photosensitive materials with work functions listed in using the results of (i), (ii)
and (iii) of (a), can you build a photoelectric device that operates with visible light?
Solution.
(a) Energy of the incident photon,
(1) For violet light,

Incident photon energy,
(2) For yellow-green light, (average wavelength)

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(3) For red light, (higher wavelength end)

(b) For a photoelectric device to operate, we require incident light energy E to be equal to or greater than
the work function ∅ of the material. Thus, the photoelectric device will operate with violet light
(with E = 3.19 eV) photosensitive material Na (With ∅ ), K (with∅ ) and Cs
(with ∅ ). It will also operate with yellow-green light (with E = 2.26 eV) for Cs (with
∅ ) only. However, it will not operate with red light (with E = 1.64 eV) for any of these
photosensitive materials.
Illustration 23. Sketch the graphs, showing the variation of stopping potential with frequency of incident
radiations for two photosensitive materials and having threshold frequencies respectively.
(i) Which of the two metals, A or B has higher work function ?
(ii) What information do you get from the slope of the graphs ?
(iii) What does the value of the intercept of graph ‘A’ on the potential-
axis represent ?
Solution. The graph is shown in fig
(i) Metal A has a higher work function ( ) because .
(ii) As or ( )
Slope of graph
(iii) Intercept of graph on the potential-axis
It depends on the work function of the material.
Illustration 24. Fig shows the variation of stopping potential with the frequency of the incident
radiation for two photosensitive metals and :
(i) Explain which metal has smaller threshold wavelength.

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(ii) Explain, giving reason which metals emits photo electrons having smaller kinetic energy, for the same
wavelength of incident radiation.
(iii) If the distance the light source and metal is doubled how will the stopping potential change ?
Solution. (i) Threshold wavelength,
As
Thus the metal Q has smaller threshold wavelength,
(ii) According to Einstein’s photoelectric equation.
of photoelectron
For the same of incident radiation L.H.S. is a constant. So metals Q with smaller value of will emit
photo-electrons of smaller K.E.
(iii) Stopping potential will remain same because it is independent of intensity and hence of distance
between the light source and the metal surface.
Illustration 25. Fig (a) shows the variation of the stopping potential with the frequency of the
incident radiations for two different photosensitive materials and
(i) What are the values of work functions for and ?
(ii) The values of the stopping potential for and for a frequency of the incident radiations
are and respectively. Show that the slope of the lines equals

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Solution. (i) Work function of metal
Work function of metals
(ii) From fig (b), Slope of lines
Illustration 26. Write Einstein’s photoelectric equation and mention which important features in
photoelectric effect can be explained with the help of this equation.
The maximum kinetic energy of the photoelectrons get doubled when the wavelength of light incident on
the surface changes from to . Derive the expressions for the threshold wavelength and work
function for the metal surface.
Solution. According to Einstein’s photoelectric equation,
Energy of incident photon
Maximum K.E. of photoelectron + Work function of metal
This equation explains the following important features :
(i) depends linearly on frequency v.
(ii) Existence of threshold for any metal surface.
(iii) does not depend on the intensity of incident light.
Derivation for and According to Einstein’s photoelectric equation,
or
For wavelength
.……(1)
For wavelength ,
.……(2)

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From (i) and (ii), we get
or
Work function,
Illustration 27. Light of intensity and frequency is incident on a photosensitive surface and cause
photoelectric emission. What will be the effect on anode current when (i) the intensity of light is gradually
increased, (ii) the frequency of incident radiation is increased, and (iii) the anode potential is increased ? In
each case, all other factors remain the same. Explain, giving justification in each case.
Solution. (i) The anode current increases with the increase in intensity of incident light.
Justification. Larger the intensity of incident radiation, larger is the number of incident photons and hence
larger is the number of electrons ejected from the photosensitive surface.
(ii) The increase in the frequency of incident radiation has no effect on the anode current.
Justification. The incident photon of increased energy cannot eject more than one electron from the metal
surface. It only increases the maximum K.E. of the ejected electron.
(iii) Anode current first increases with the increase in anode potential and then attains a saturation value.
Justification. All the photoelectrons emitted from the metal do not have the same K.E. The increase in
anode potential accelerate more and more electrons towards the anode. A stage is reached when the anode
current attains a saturation value. This happens when all the electrons emitted by the metal get collected by
the anode.
Exercise 1.
1. Blue light can eject electrons from a photosensitive surface while orange light cannot. Will violet and red light
eject electrons from the same surface?
2. The work function of metal A is 4.2 eV. If two photons each of energy 2.5 eV strike an electron of A, will the
emission of electron be possible?
3. Radiation of frequency 1015 Hz is incident in three photosensitive surfaces A, B and C. The following
observations were noticed:
(i) Surface A: Photoemission occurs but the photoelectrons have zero kinetic energy.
(ii) Surface B: No photoemission

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(iii) Surface C: Photoemission occurs and photoelectrons have some K.E. explain
the three observation by Einstein’s photoelectric equation.
4. Give two information, we can get from the Fig.
5. Given below is the graph between frequency (v) of the incident light and maximum
kinetic energy K.E. of emitted photoelectrons. Find the values of (i) threshold
frequency and (ii) work function from the graph.
6. UV light is incident on two photosensitive materials having work unctions W 1 and

W2 (W1 > W2 ). In which case will the kinetic energy of the emitted electrons be greater? Why?
7. What happens when
(i). Intensity of light is increased keeping frequency constant?

(ii) Increasing the frequency of light keeping intensity constant?
8. State how in a photo-cell, the work function of the metal influence the kinetic energy of emitted electrons.
(a) If the intensity of incident radiation is doubled, what changes occur in
(i) the stopping potential and
(ii) the photoelectric current ?
(b) If the frequency of the incident radiation is doubled, what changes occur in the
(i) stopping potential and
(ii) photoelectric current ?
9. If the frequency of the incident radiation on the cathode of a photo-cell is doubled, how will the following
change :
(i) Kinetic energy of the electrons ?
(ii) Photoelectric current ?
(iii) Stopping potential ?
Justify your answer.
(i) Matter waves (de-Broglie Waves)
In 1924, Louis de-Broglie put forward the hypothesis that matter also possesses dual nature on the basis of
following observations:
(a) The whole energy in this universe is in the form of mater and electromagnetic radiation.
(b) The nature loves symmetry, and as the radiation has got dual nature, matter should also possesses dual nature.
(ii) De-Broglie wavelength
The waves which are associated with particles like electrons, protons, neutrons etc. are called matter waves.
These matter waves are also called de-Broglie waves. De-Broglie derived an expression for the wavelength of the

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matter waves .The energy of a photon of frequency is given by,
Similarly for a particle of mass moving with a velocity ,de-Broglie wavelength λ is given by,
This is known as de-Broglie wave equation.
The important applications of de-Broglie waves are given below;
(i) An important application of de-Broglie wave is its use in the construction of electron microscope.
(ii) The necessary condition for the electron in a particular stationary orbit can be calculated with the help of
de-Broglie waves.
From the equation of de-Broglie wavelength, following conclusions can be drawn:

(i) Lighter the particle greater is its de-Broglie wavelength.
(ii) The faster the particle waves, smaller is its de-Broglie wavelength.
(iii) The de-Broglie wavelength of a particle is independent of the charge or nature of the particle.
(iv) The matter waves are not electromagnetic in nature. The electromagnetic waves are produced only by accelerated
charged particles
De – Broglie Wavelength of An Accelerated Electron

Consider an electron accelerated through a potential difference of V volt.
eV = mv2 or v = √
De Broglie wavelength
√
√
Putting h = 6.62 10-34 Js, m = 9.1 10-31 kg and e = 1.6 10-19 C, we get
m
√

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Or or Å
√ √
Illustration 28. Calculate the
(a) Momentum, and

(b) De Broglie wavelength of the electrons accelerated through a potential difference of 56 V. NCERT
Solution. Given V = 56 V.
(a) Momentum of the electron
√ √
(b) Wavelength
√ √
Illustration 29. What is the

(a) Momentum,
(b) Speed, and
(c) De Broglie wavelength of an electron with kinetic energy of 120 eV?
NCERT
Solution. Here, kinetic energy, J
Also, C
And Js.
(a) Momentum √ √
(b) Speed
(c) De Broglie wavelength
Illustration 30. The wavelength of light from the spectral emission line of sodium is 589 nm. Find the
kinetic energy at which NCERT
(a) An electron, and

(b) A neutron, would have the same de Broglie wavelength.

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Solution. Here, nm
Js and
(a) For an electron,

√
( )
J.
(b) For a neutron.

( )
J.
Illustration 31. What is the de Broglie wavelength of
(a) A bullet of mass 0.040 kg travelling at the speed of 1.0 km/s,

(b) A ball of mass 0.060 kg moving at a speed of 1.0 m/s, and
(c) A dust particle of mass 1.0 kg drifting with a speed of 2.2 m/s?
NCERT
Solution.
(a) Here, kg and
m.
(b) Here, and
(c) Here, kg and .
m.
Illustration 32. An electron and a photon each have a wavelength of 1.00 nm. Find
(a) Their momenta,

(b) The energy of the photon, and
(c) The kinetic energy of electron. NCERT
Solution. Given nm m.
Also, kg
And Js.

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(a) Momentum of
It is the same for both electron and photon.
(b) Energy of Photon,
( )
(c) Kinetic energy of electron J
Illustration 33. (a) For what kinetic energy of a neutron will the associated de Broglie wavelength be
1.40 × m?
(b) Also find the de Broglie wavelength of a neutron, in thermal equilibrium with matter, having an average
kinetic energy of kT to 300 K. NCERT
Solution. (a) Here m
Js and kg
( )
(b). Since
√ √
Here and
( )
m
√
Illustration 34. Show that the wavelength of electromagnetic radiation is equal to the de Broglie wavelength
of its quantum (Photon). NCERT
Solution. The de Broglie wavelength of photon is

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But for a photon,
from eqn. (1). wavelength of electromagnetic radiation.

( )
Illustration 35. An electro, α – particle, and a proton have the same kinetic energy. Which of these particles
has the shortest de Broglie wavelength? NCERT
Solution. The wavelength of matter waves is given by
For constant value of E, .

√
Since the mass of α – particles among the given set of particles is maximum, the wavelength associated with
its minimum.
Illustration 36. A particle is moving three times as fast as an electron. The ratio of the de Broglie
wavelength of the particle to that of the electron is 1.813 × . Calculate the particle’s mass and identify
the particle. NCERT
Solution. Let be the mass of particle, be its velocity and be its de Broglie wavelength.
Then given, and
Now …… (1)
And ……… (2)
Dividing (1) and (2),
Or
The mass of the neutron is 1839 times the mass of electron. Therefore, the particle is a neutron.
Illustration 37. What is the de Broglie wavelength of a nitrogen molecule in air at 300 K? Assume that the
molecule is moving with the root mean square speed of molecules at this temperature. (Atomic mass of
nitrogen = 14.0076 u) NCERT
Solution. For nitrogen,

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RMS velocity, √ √
√ .
de Broglie wavelength,
Illustration 38. What is the de Broglie wavelength associated with (a) an electron moving with a speed of
5.4 × m/s, and (b) a ball of mass 150 g travelling at 30.0 m/s?
Solution.
(a) For the electron:
Mass speed . Then, momentum
De Broglie wavelength,
(b) For the ball;
Mass .
The momentum
De Broglie wavelength

24
The de Broglie wavelength of electron is comparable with X-ray wavelengths. However, for the ball
it is about times the size of the proton, quite beyond experimental measurement.
Illustration 39. An electron, an α – particle, and a proton have the same kinetic energy. Which of these
particles has the shortest de Broglie wavelength
Solution.
For a particle, de Broglie wavelength, λ = h/p
Kinetic energy, K =
Then, √
For the same kinetic energy K, the de Broglie wavelength associated with the particle is inversely
proportional to the square root of their masses. A proton ( ) is 1836 times massive than an electron and an
α- particle ( ) four times that of a proton.
Hence, α-particle has the shortest de Broglie wavelength.
Illustration 40. A particle is moving three times as fact as an electron. The ratio of the de Broglie
the particle.
Solution.
De Broglie wavelength of a moving particle, having mass m and velocity v;
Mass,
For an electron, mass
Now, we have and
Then, mass of the particle, ( )( )
Thus, the particle, with this mass could be a proton or a neutron.
Illustration 41. What is the de Broglie wavelength associated with an electron, accelerated through a
potential difference of 100 volts?
Solution. Acceleration potential V = 100 V. The de Broglie wavelength

25
nm
√
The de Broglie wavelength associated with an electron in this case is of the order of X-ray wavelengths.
Illustration 42. A proton and an alpha particle are accelerated through the same potential. Which one of the
two has (i) greater value of de-Broglie wavelength associated with it, and (ii) less kinetic energy ? Justify
your answer.
Solution. (i) de-Broglie wavelength,
For same
√ √ √
Hence proton has a greater value of de-Broglie wavelength.
(ii) Kinetic energy
For same
Hence, proton has less kinetic energy.
Illustration 43.An electron and a proton are accelerated through the same potential. Which one of the two
has (i) greater value of de-broglie wavelength associated with it and (ii) less momentum ? Justify your
answer.
Solution. (i) de-Broglie wavelength,
For same

26
√ √
As so electron has a greater de-Broglie wavelength.
(ii) Momentum √
Or √
As
electron has less momentum.
Illustration 44. Plot a graph showing variation of de-Broglie wavelength versus when is
√
accelerating potential for two particles and carrying same charge but of masses
which one of the two represents a particle of smaller mass and why ?
Solution. de-Broglie wavelength,
√ √ √
versus graph is a straight line with slope

√
For the particle of same charge , slope

√
As the line B with larger slope than that of line A represents the graph for smaller mass
Exercise 2.
1. An electron and a - particle have same kinetic energy. Which one of the two has greater de Broglie
wavelength?
2. An - particle and a proton are accelerated from rest though the same potential difference V. find the ratio of
de Broglie wavelength associated with them.

27
3. de Broglie wavelength associated with an electron accelerated through a potential difference 1 V is .

What will be its wavelength when the accelerating potential is increased to 4 V?
4. A proton and an alpha particle are accelerated through the same potential which one of the two has
(i). greater value of de Broglie wavelength associated with it and
(ii) less kinetic energy justify your answer.
5. A particle of a mass M at rest decays into two particle of masses and having non-zero velocities. What
is the ratio of the de-broglie wavelengths of the zero particles ?
Heisenberg’s Uncertainty Principle:
According to this principle, it is not possible to measure both the position and momentum of an electron for any other
particle) at the same time exactly.
( )
Davisson and Germer Experiment:
Purpose : To established the wave nature of electrons.
Done by :
CJ Davisson and LH Germer in 1927
GP Thomson in 1928
Construction :
1. The experiment arrangement used by Davisson and Germer is schematically shown in Fig.
2. It consists of an electron gun which companies of a tungsten F. coated with barium oxide and heated by a low
voltage power supply (L.T. or battery).

28
3. Electrons emitted by the filament are accelerated to a desired velocity by applying suitable potential voltage
from a high voltage power supply (H.T. or battery). They are made to pass through a cylinder with fine holes
along its axis, producing a fine collimated beam.
4. The beam is made to fall on the surface of a nickel crystal.
5. The electrons are scattered in all directions by the atoms of the crystal. This apparent is enclosed in an
evacuated chamber.
6. The intensity of the electron beam, scattered in a given direction, is measured by the electron detector
(collector). The detector can be moved on a circular scale and is connected to a sensitive galvanometer, which
records the current. The deflection of the galvanometer is proportional to the intensity of the electron beam
entering the collector.
7. The variation of the intensity of scattered electrons is recorded for different angles of scattering θ which is the
angle between the incident and the scattered electron beams.
Observation : The experiment was performed by varying the accelerating voltage from 44 V to 68 V. It was noticed
that a strong peak appeared in the intensity (I) of the scattered electron for an acceleration voltage of 54 V at scattering
angle .
Conclusion : The appearance of the peak in a particular direction is due to the constructive interference of electrons
scattered from different layers of the regularly speed atoms of the crystals.
De Broglie wavelength of electrons
For
√ √ √
From diffraction measurements the wavelength of matter waves nm.
There is an excellent agreement between the theoretical value and the experimentally obtained value of de Broglie
wavelength which confirms wave nature of electrons and the de Broglie relation.
Uses of photocell :
 Uses in the reproduction of sound in motion pictures.

 Uses in the television camera for scanning and telecasting scenes.
 Used in industries for detecting minor flaws or holes in metal sheets.

29
Illustration 45. Calculate the energy and momentum of a photon of wavelength 6000Å
Solution. Energy of a photon
Or
Also the momentum p is given by
Illustration 46. A particle is moving three times as fast as an electron. The ratio of de Broglie wavelength of
the particle to that of the electron is . Calculatethe mass of this particle. Mass of an electron is
Solution. The wavelength of the electron is given by
…….. (i)
Where and are the mass and the velocity of the electron respectively.
Similarly, the wavelength, of the particle is given by
……. (ii)
Where and are the mass and the velocity of the particle respectively.
Dividing eqn. (ii) by eqn. (i), we have
Now
And
Or
This particle must be a proton or a neutron.
Illustration 47. Obtain the de Broglie wavelength of an electron of kinetic energy 100 eV. Mass of an
electron kg, C and .
Solution. The energy E of an electron is

30
The relation between wavelength and kinetic energy E as obtained in the previous question is given by
Laws of Photoelectric Emission

1. For a give metal and frequency of incident radiations, the number of photoelectrons emitted is directly
proportional to the intensity of incident light.
2. For each metal, there exists a certain minimum frequency called threshold frequency , below which there is no
photoelectron emission.
3. When light of frequency greater than the threshold frequency is incident on a metal surface, the maximum
kinetic energy of the photoelectrons emitted, is directly proportional to the frequency of incident light.
4. Photoelectric emission of electron occurs instantaneously. Time lag is s with the incident of light.
5. One photon can eject one electron only. It is not that more one photon are absorbed simultaneously to produce
photoelectric effect.

31
NCERT QUESTION WITH SOLUTION.

Illustration 1. Find the
(a) Maximum frequency, and
(b) Minimum wavelength of X – rays produced by 30 kV electrons.
Solution. Here .
Using
Where and C
Illustration 2. The work function of caesium metal is 2.14 eV. When light of frequency 6 × Hz is
incidence on the metal surface, photoemission of electrons occurs. What is the
(a) Maximum kinetic energy of the emitted electrons.
(b) Stopping potential, and
(c) Maximum speed of the emitted photoelectrons?
Solution. Here J.
(a) Maximum kinetic energy of photoelectrons is
(b) Stopping potential is given by
176
(c)
√ √ .
Illustration 3. The photoelectric cut-off voltage in a certain experiment is 1.5 V. What is the maximum
kinetic energy of photoelectrons emitted?
Solution. Here,
Maximum kinetic energy, J.

Illustration 4. Monochromatic light of wavelength 632.8 nm is produced by a helium-neon laser. The power
emitted is 9.42 mW.
(d) Find the energy and momentum of each photon in the light beam.
(e) How many photons per second, on the average, arrive at a target irradiated by this beam?
(Assume the beam to have uniform cross-section which is less than the target area), and
(f) How fast does a hydrogen atom have to travel in order to have the same momentum as that of the
photon?

32
Solution. Given, m
Power,
(d) J
Also,
(e) Number of photons emitted per second,

(f) Mass of hydrogen atom ≈ mass of proton
Illustration 5. The energy flux of sunlight reaching the surface of the earth is 1.388 × . How
many photons (nearly) per square metre are incident on the Earth per second? Assume that the photons in
the sunlight have an average wavelength of 550 nm.
Solution. Given
Energy flux ∅ = 1.388 ×
Also
Energy of each photon J.
∅
no. of photons per square metre of earth .
Illustration 6. In an experiment on photoelectric effect, the slope of the cut-off voltage versus frequency of
incident light is found to be 4.12 × V s. Calculate the value of Plank’s constant.
Solution. The slope of the graph in this case is
Js.
Illustration 7. A 100 W sodium lamp radiates energy uniformly in all directions. The lamp is located at the
centre of a large sphere that absorbs all the sodium light which is incident on it. The wavelength of the
sodium light is589 nm. (a) What is the energy per photon associated with the sodium light? (b) At what rate
are the photons delivered to the sphere?
Solution. Power of sodium lamp, P = 100 W
(b) Energy of photon, J

(c) Rate at which the protons are delivered photons/s.
Illustration 8. What are the energies of photons at the (a) violet, and (b) red ends of the visible spectrum?
The wavelength of light is about 390nm for violet and about760 nm for red.
Solution. Here m.
m.

33
(a)
(b) J
Illustration 9. The threshold frequency for a certain metal is 3.3 × Hz. If light of frequency 8.2 ×
Hz is incident on the metal, predict the cut-off voltage for the photoelectric emission.
Solution. Given
Hz
Also Js and C.
Using
Or V.
Illustration 10. The work function for a certain metal is 4.2 eV. Will this metal give photoelectric emission
for incident radiation of wavelength 330 nm?
Solution. W = 4.2 eV = 4.2 × 1.6 × J
= 6.72 × J
using
Or Hz
Since
the photoelectric emission for this radiation can not take place.
Illustration 11. Light of frequency 7.21 × is incident on a metal surface. Electrons with a
maximum speed of 6.0 × are ejected from the surface. What is the threshold frequency for
photoemission of electrons ?
Solution. Here, Hz
Using the relation
( )
Hz.
34
Illustration 12. Light of wavelength 488 nm is produced by an argon laser which is used in the
photoelectric effect. When light from this spectral line is incident on the cathode, the stopping (cut - off)
potential of photoelectrons is 0.38 V. Find the work function of the material from which the cathode is made
Solution. Here,
Also Js,
And
Hz.
Using
We get
Illustration 13. Calculate the

(c) Momentum, and
(d) De Broglie wavelength of the electrons accelerated through a potential difference of 56 V.
Solution. Given V = 56 V.
(c) Momentum of the electron
√ √
√
(d) Wavelength
√ √
Illustration 14. What is the

(d) Momentum,
(e) Speed, and
(f) De Broglie wavelength of an electron with kinetic energy of 120 eV?
Solution. Here, kinetic energy, J
Also, C
And Js.
(d) Momentum √ √
√
(e) Speed
(f) De Broglie wavelength
Illustration 15. The wavelength of light from the spectral emission line of sodium is 589 nm. Find the
kinetic energy at which

35
(c) An electron, and

(d) A neutron, would have the same de Broglie wavelength.
Solution. Here, nm
Js and
(c) For an electron,

√
( )
J.
(d) For a neutron.

( )
J.
Illustration 16. What is the de Broglie wavelength of

(d) A bullet of mass 0.040 kg travelling at the speed of 1.0 km/s,
(e) A ball of mass 0.060 kg moving at a speed of 1.0 m/s, and
(f) A dust particle of mass 1.0 kg drifting with a speed of 2.2 m/s?
Solution.
(d) Here, kg and
m.
(e) Here, and
(f) Here, kg and .
m.
Illustration 17. An electron and a photon each have a wavelength of 1.00 nm. Find
(d) Their momenta,
(e) The energy of the photon, and
(f) The kinetic energy of electron.
Solution. Given nm m.
Also, kg
And Js.
(d) Momentum of
It is the same for both electron and photon.
(e) Energy of Photon,
J

36
( )
(f) Kinetic energy of electron J
.
Illustration 18. (a) For what kinetic energy of a neutron will the associated de Broglie wavelength be
1.40 × m?
(b) Also find the de Broglie wavelength of a neutron, in thermal equilibrium with matter, having an average
kinetic energy of kT to 300 K.
Solution. (a) Here m
Js and kg
( )
(b). Since
√ √
Here and
( )
m
√
Illustration 19. Show that the wavelength of electromagnetic radiation is equal to the de Broglie wavelength
of its quantum (Photon).
Solution. The de Broglie wavelength of photon is
But for a photon,
from eqn. (1). wavelength of electromagnetic radiation.

( )
Illustration 20. An electro, α – particle, and a proton have the same kinetic energy. Which of these particles
has the shortest de Broglie wavelength?
Solution. The wavelength of matter waves is given by
For constant value of E, .

√
Since the mass of α – particles among the given set of particles is maximum, the wavelength associated with
its minimum.

37
Illustration 21. A particle is moving three times as fast as an electron. The ratio of the de Broglie
the particle.
Solution. Let be the mass of particle, be its velocity and be its de Broglie wavelength.
Then given, and
Now …… (1)
And ……… (2)
Dividing (1) and (2),
Or
The mass of the neutron is 1839 times the mass of electron. Therefore, the particle is a neutron.
Illustration 22. What is the de Broglie wavelength of a nitrogen molecule in air at 300 K? Assume that the
molecule is moving with the root mean square speed of molecules at this temperature. (Atomic mass of
nitrogen = 14.0076 u)
Solution. For nitrogen,
RMS velocity, √ √
√ .
de Broglie wavelength,
Illustration 23. (a). Estimate the speed with which electrons emitted from a heated cathode of an evacuated
tube impinge on the anode maintained at a potential difference of 500 V with respect to the cathode. Ignore
the small initial speeds of the electrons. The ‘specific charge’ of the electron i.e., its e/m is given to be 1.76
× C .
(b). Use the same formula you employ (a) to obtain electron speed of an anode potential of 10 MV. Do you
see what is wrong ? In what way is the formula to be modified.
Solution. (a). Here

38
Using √ * +
(b). Using the same formula for
√ √
Since v is comparable to the velocity of light, the kinetic energy is not exact relation. The
relativistic expression for kinetic energy [i.e. K.E. = ] must be used, so that
, where .
√
Illustration 24. (a). A monoenergetic electron beam with electron speed of 5.20 is subject to a
magnetic field 1.30 × T normal to the beam velocity. What is the radius of the circle traced by the
beam, given e/m for electron equals 1.76 × .
(b). Is the formula you employ in (a) valid for calculating radius of the path of a 20 MeV electron beam? If
not, in what way is it modiefied?
Solution. (a).
And
For tracing out a circle, qv
(b). The formula employed in part (a) is not valid because with the increase in velocity, mass varies and in
the above formula we have taken m as constant. Instead, is to be considered.
√
Illustration 25. In a Thomson’s set-up determining e/m, the same high tension de supply provides potential
to the anode of the accelerating column, an also to the positive deflecting plates in the region of the crossed
fields. If the supply voltage is double, by what factor should the magnetic field be increased to keep the
electron beam undeflected?
Solution. We know
But
…… (1)
If V is doubled i.e. V’= 2

39
Then …… (2)
from eqns. (1) and (2)
i.e., magnetic field has to be increased to √ B.

Illustration 26. The deflecting plates in a Thomson’s set-up are 5.0 cm long, and 1.5 cm apart. The plates
are maintained at a potential difference of 240 V. Electrons accelerated to an energy of 2.0 keV enter from
one edge of the plates midway in a direction parallel to the plates.
(a) What is the deflection at the other edge of the plates?
(b) At what distance from the undeflected position of the screen does the beam strikes if the screen is
30 cm away from the other edge of the plates?
Solution. Here m
(a) Now, kinetic energy

.
deflection at the other edge of plates is
* +
( )
(b) Now distance of screen D = 30 cm = 0.3 m

Position of spot after deflection, ( )
Where ( )
( ) ( ) .
Illustration 27. In a Thomson’s set-up for determination of e/m, a uniform electric field E = 24.0 kV
set up between two parallel plates of length 6.0 cm each produces a deflection of 10.9 cm on the fluorescent
screen. A magnetic field is then switched on and adjusted to the value T to restore the
beam to its undeflected position. The distance of the screen from the centre of the plates is 40.0 cm.
Determine e/m from the data.
Solution. Here E = 24.0 KV
.
The distance of the spot from the centre of screen is

40
* + ……. (1)
But for no deflection of the beam on the application of magnetic field,
from eqn. (1),
* +
( ) ( )
.
Illustration 28. An electron gun with its anode at a potential of 100 V fires out electrons in a spherical bulb
containing hydrogen gas at low pressure . A magnetic filed of curves the
path of the electrons in a circular orbit of radius 12.0 cm. (The path can be viewed because the gas ions in
the path focus the beam by attracting electrons, and emitting light by electron capture; this method is known
as the ‘fine beam tube’ method.) Determine e/m from the data.
Solution.
Now …….. (i)
Also .
from eqn. (1),
( )
.
Illustration 29. In a Millikan’s oil-drop experiment, a charged oil drop of mass density 880 kg is held
stationary between two parallel plates 6.00 mm apart held at a potential difference of 103 V. When the
electric field is switched off, the drop is observed to fall a distance of 2.00 mm in 35.7 s.
(a) What is the radius of the drop?
(b) Estimate the charge of the drop. How many excess electrons does it carry?
(The upper plate in the experiment is at a higher potential). (Viscosity of air = 1.80 × Ns ,
g = 9.81 m , density of air = 1.29 kg )
Solution. Here,
Distance between plates,
Distance moved by the drop

41
Time taken
Terminal velocity
Also,
Density of air,
Using √ √
(b). If the drop is held stationary in between the plates due to an electric field
[ ]
[ ]
( )
[ ]
The drop contains 5 electrons.

Illustration 30. (Millikan’s experiment not only measured charge of an electron ; it also established a
fundamental fact of nature e. Make sure you understand this point by going through the following exercise):
In a Millikan’s oil – drop set-up, an oil-drop falls with a terminal speed in the absence of any electric
field. When a fixed electric field is switched on, and a radioactive source is kept in the environment, it is
found that the same oil drop, when viewed for long, shows up different terminal speed
(a) What causes the drop to change its terminal speed in the same electric field? (assume size and mass
of the drop remain unchanged).
(b) What key observation on the different terminal speeds of the drop suggest charge quantization?
Solution. (a) The radioactive source ionizes the air around it. A drop, therefore, may acquire additional
electrons or it may lose some electrons producing a variation of electric force on the charged oil drop. This
causes a change in the terminal velocity of the drop.
(b). During the upward motion of drop due to the application of electric field,
Illustration 31. In an experiment of the Millikan’s oil drop set up, an oil drop whose radius is measured by a
separate observation to be 1.0 m, falls down in the absence of an electric field with a certain
terminal velocity. When a horizontal electric field is set up by means of two parallel vertical plates held 10
mm apart at a potential difference of 1500 V, the drop is seen to fall steadily at an angle of with the
vertical. The density of the oil used is 900 kg . Estimate the charge on the drop.
Solution. Here
If the density of medium is zero, then the velocity of the drop in vertical direction is given by the relation

42
……. (i)
Also the horizontal velocity under the action of the horizontal electric field is given by the relation
……. (ii)
[ ]
* +
( )
Illustration 32. (a) An X-ray tube produces a continuous spectrum of radiation with its short wavelength
and at 0.45 Å. What is the maximum energy of a photon in the radiation?
(b). Form your answer to (a), guess what order of accelerating voltage (for electrons) is required in such a
tube.
Solution. (a). Å
(b). Since
The order of acceleration voltage is .

Illustration 33. In an acceleration experiment on high-energy collision of electrons with positrons, a certain
event is interpreted as annihilation of an electron-positron pair of total energy 10.2 BeV into two of
equal energy. What is the wavelength associated with each ? (1BeV = eV)
Solution. Energy carried by the pair of
Energy of each is
Using
Illustration 34. Estimating the following two numbers should be interesting. The first number will tell you
why radio engineers do not need to worry much about photons! The second number tells you why our eye
can never’ count photons, even in barely detectable light.

43
(a) The number of photons’ , per second by a Medium wave transmitter of 10 kW power, emitting radio
waves of wavelength 500 m.
(b) The number of photons entering pupil of our eye per second corresponding to the minimum intensity
of white light that we humans can perceive ( ). Take the area of the pupil to be about
0.4 , and the average frequency of white light to be about Hz.
Solution. (a) Number of photons emitted per second,
Now, Power of transmitter
This number is a very-very large quantity.
(b). Minimum intensity
Area of pupil
Average frequency .
Energy of each Photon, J.
number of photons entering into pupil of the eye per second =
This is quite a small number, but still large enough to be counted.

Comparison of cases (a) and (b) tells us that our eye can not count the number of photons individually.
Illustration 35. Ultraviolet light of wavelength 2271Å from a 100 W mercury source irradiates a photo-cell
made of molybdenum metal. If the stopping potential is – 1.3 V, estimate the work function of the metal.
How would the photo-cell respond to high intensity red light of wavelength 6328 Å
produced by a He-Ne laser?
Solution. Here, Å
Power,
Stopping potential, (in magnitude)
Energy of ultraviolet light photon,
Using the relation
Work function

44
Also, for red light,
Since the energy of red ray photon is less than the work function for the metal, photo cell does not respond
to red light.
Illustration 36. Monochromatic radiation of wavelength 640.2 nm (1nm = m) from a neon lamp
irradiates photosensitive material made of caesium. The stopping voltage is measured to be 0.54 V. The
source is replaced by an iron source and its 427.2 nm line irradiates the same photo-cell. Predict the new
stopping voltage.
Solution.
Using
[ ]
Also, for ,
Using
Illustration 37. A mercury lamp is a convenient source for studying frequency dependence of photoelectric
emission since it gives a number of spectral lines ranging from the UV to the red end of the visible spectrum
In our experiment with rubidium photo-cell, the following lines from a mercury source were used :
The stopping voltages, respectively, were measured to be :
(a) Determine the value of Planck’s constant h.

(b) Estimate the threshold frequency and work function for the material.

45
Solution. From the Einstein photoelectric equation,
……. (1)
For
……. (2)
Apply for
……… (3)
From eqns. (2) and (3), on subtraction,
* + [ ]
( ) ………. (4)
Using and the value of from the table (using SI

units),
[ ]* +
–
……….. (5)
Similarly, taking the set of values:
( )
( )
……….. (6)
Also, taking the set of values :
( )
( )
……….. (7)
From (5), (6) and (7), we conclude that the approximate value of h is .
(b) Now using eqn. (2),

46
Also,
Illustration 38. The work function for the following metals given :
Na : 2.75 eV; K : 2.30 eV ; Mo ; 4.17 ; Ni : 5.51 eV. Which of these metals will not give photoelectric
emission for a radiation of wavelength 3300 Å from a He-Cd laser placed 1m away from the photocell?
What happens if the laser is brought nearer and placed 50 cm away?
Solution. Here, Å
Using the relation
Since the energy E of the incident photon of light is greater than the work functions of all the metals given in
the question, photoelectric emission will occur in all the metals.
The distance of the source does not increase or decrease the energy of the photon of light incident, therefore,
the energy of electrons ejected will not change but the intensity of ejected electrons will increase ( )
and become four times.
Illustration 39. Light of intensity falls on a sodium photo-cell of surface area 2 .

Assuming that the top 5 layers of sodium absorb the incident energy, estimate time required for
photoelectric emission in the wave-picture of radiation. The work function for the metal is given to be about
2 eV. What is the implication of your answer?
Solution. Here,
Taking the approximate radius of an atom as m, the effective area of sodium atom is
.
If there is one free electron per atom, then number of electrons in five layers.
Now Incident power
For absorption of the incident power equally by all electrons, energy absorbed per electron per second is
Time required for photo electric emission to take place
The answer obtain implies that the time of emission of electron is very large and is not in agreement with the
observer time of emission, which is approximately s. Thus wave-picture of radiation is not applicable
for photo-electric emission.

47
Illustration 40. (a). Show that a free electron at rest cannot absorb a photon and thereby acquire kinetic
energy equal to the energy of the photon. Would the conclusion change if the free electron was moving with
a constant velocity?
(b). If the absorption of the photon by a free electron is ruled out as proved in (a) above, how does
photoelectric emission take place at all?
Solution. (a) Let a free electron at rest absorbs a photon of energy hv and gains a momentum p. Then, using
the law of conservation of energy,
Rest mass energy of electron + energy of photon √
Squaring both sides,
………. (1)
Also, according to principle of conservation of linear momentum,
……… (2)
Using eqn. (2) in eqn. (1),
Which is not possible

Conclusions remain the same even if the electron were moving with a constant velocity because the electron
at rest and in uniform motion are in two inertial frames of reference. Therefore, if a process can not occur in
one frame, it also does not occur in the other.
(b). In case of metallic substance, the momentum of the photon is shared by the electron and the lattice as a
whole without sharing the energy. The energy is thus, handed over to the electron only and the photoelectric
emission becomes possible.
Illustration 41. In an experiment on photoelectric emission by on platinum, the energy distribution

of photoelectrons exhibits peaks at a number of discrete energies : 270 keV, 339 keV and 354 keV. The
binding energies of K, L and M shell in platinum are known to be 77 keV, 13 keV and 3.5 keV
approximately. What is the wavelength of the rays with which the data are consistent?
Solution. We know that
Energy of the incident photon = binding energy of electron + K.E. of photoelectron emitted
i.e.,
For K shell, B + E = 270 + 77 = 347 keV
For L shell, B + E = 339 + 13 = 352 keV
For M shell, B + E = 354 + 3.5 = 357.5 keV
Average value of B + E = .

48
Now B + E =
.
Illustration 42. An X-ray pulse is sent through a section of Wilson cloud chamber containing a
supersaturated gas, and tracks of photoelectrons ejected from the gaseous atoms are observed. Two groups
of tracks of lengths 1.40 cm and 2.02 cm are noted. If the range-energy relation for the cloud chamber is
given by R = αE with α = 1 cm/keV, obtain the binding energies of the two levels from which electrons are
emitted. (Wavelength of the X-rays pulse = 4.9 Å)
Solution. Here, the two values of range are 1.40 cm and 2.02 cm.
The two values of energy are (Using R = )
And
Now Å
Energy of incidence photon
Now, using the Einstein’s photoelectric equation,
And .
Illustration 43. Crystal diffraction experiment can be performed using X-rays, or electrons acceleration
through appropriate voltage. Which probe has greater energy? (For quantitative comparison, take the
wavelength of the probe equal to 1Å, which is of the order of inter-atomic spacing in the lattice)
.
Solution. Here Å
Also, .

49
Text Question
Question 4. A ratio transmitter operates at a frequency of 880 kHz and a power of 10 kW. Find the
number of photons emitted by it per second.
Ans 4. n =
= 1.72 Hz.

50
Question 5. What is the frequency of a photon whose energy is 66.3 eV?
Ans 5. E = hv
66.3 1.6 = 6.63 10-34 v
v = 1.6 1016 Hz.
Question 6. The wavelength of an electromagnetic radiation is doubled. What will happen to the
energy of photons?
Ans 6. E = hv = E
Question 7. (a) On what factor does the energy carried by quantum of light depend?
(b) which radiation has higher energy:(i) visible light (ii) X – ray ?
Ans 7. (a) Energy of a quantum of light is directly proportional to its frequency
(Max Planck Theory). (b) X – ray has higher energy since it has more frequency.
Question 8. How many electron volt make one joule?
Ans 8. 1 eV = 1.6 10 -19 joule 1 Joule =
Question 9. What is the momentum of a photon of frequency v?
Ans 9. P =
Question 10. Mention one physical process for the release of electrons from the surface of a metal.
Ans 10. Photoelectric effect.
Question 11. Light from a bulb falls on a wooden table but no photoelectrons are emitted. Why?
Ans 11. In wood, there are no free electrons to be emitted.
Question 12. If the intensity of incident radiation on a metal is doubled, what happens to the
kinetic energy of electrons emitted?
Ans 12. The intensity of incident radiation has no effect on the kinetic energies of the emitted
electrons.
Question 13. What is the effect on the emitted photoelectrons if the wavelength of the incident light
is decreased?
Ans 13. When wavelength of incident light decreases, the frequency and hence, the energies of
the incident photons increase. As a result, the energy of emitted electrons increases.
Exercise 1.

51
Ans 1. Violet light can eject electrons because frequency of violet light (fv)> frequency of
blue light (fB). Red light cannot eject electrons because frequency of red light (fR)<
frequency of orange light (f0).
Ans 2. No because an electron can accept only one photon at an instant. Since energy of one
photon is less than work function, no photoelectron emission will take place.
Ans 3. Einstein’s Photoelectric equation is hv = + K.E.max

For surface A: (K.E)max = 0
hv =
v = Threshold frequency = 1015 Hz
Work function = = 4.14 10-15 1015 = 4.14 eV
The energy of incident photon = 4.14 eV
For surface B: No photoemission occurs.
The work function of surface B is greater than 4.14 eV
For surface C: The work function of surface C is less than 4.14 eV.
Therefore
(K.E.)max = hv - = [ ] eV
Ans 4. Since threshold frequency of caesium is less than that of sodium, therefore, the work function
of caesium is also less than sodium. Also the straight line suggest that v V0
Ans 5. From the graph it is clear that threshold frequency V0 = 1015 Hz and work function is 4 eV.
Ans 6. K.E. = hv – W
For same v, smaller the value of W, grater will be kinetic energy. Therefore, the answer is the material
having work function W2 .
Ans 7. (a) The photoelectric current increases because increasing the intensity of light increases the
number of photons per second per unit area.
(b) The kinetic energy of photoelectron increase because, increasing the frequency increases
the energy of quanta (photon). It has no effect on number of photons/area until intensity of
light is not changed.
Question 21. How does the stopping potential applied to a photocell change, if the distance between the
light source and the cathode of the cell is doubled?
Ans 21. The stopping potential depends on the frequency of incident radiation. As the frequency is
unchanged, the stopping potential remains the same. In fact, by increasing the distance the intensity of light
( 1/r2) decreases. This in turn will decrease the photoelectric current.
Question 22. The frequency (v) of the incident radiation is greater than the threshold frequency (v 0)in a
photocell. How will the stopping potential vary if frequency v is increased, keeping the other factors
constant?
Ans 22. eV0 = hv0. When v is increased, the stopping potential will also increase.

52
Exercise 2.
Ans 1. K.E. =
If K.E. is same than is constant. Since > me . This means that momentum of -
particle is more. Also = . Thus, wavelength of - particle will be less.
Alternatively when K.E. is some, As > me therefore, me

√ √
Ans 2. We know that

√
Therefore, = √
√
Ans 3. Å
√
√ √
i.e., the wavelength becomes half.
Ans 4. (i) for same V we get

√ √
For proton mq = 1 1 = 1
For alpha particle mq = 4 2 = 8
Proton will have greater value of de Broglie wavelength associated with it.
(ii) K.E. = qV
For same V, K.E. q
The charge on proton is smaller than the - particle.Therefore, the kinetic energy of
-particle. Therefore, the kinetic energy of -particle is less.
Question 27. State whether the following statements are true (T) or false (F):
(a) An electron cannot behave like a wave.

(b) Metter has dual nature.
Ans 27. (a) F (b) T
Question 28. What is the conclusion drawn from Davisson and Germer experiment of electron diffraction ?
Ans 28. Electron has wave nature and hence, matter has wave nature.
Question 29. On what principle is electron microscope based?

53
Ans 29. Wave nature of electron.
Question 30. Name the experiment for which the following graph (Fig),
showing the variation of intensity of scattered electrons with the angle of
scattering, was obtained. Also name the important hypothesis that was confirmed
by this experiment.
Ans 30. Davisson and germer electron diffraction experiment which
confirmed the wave nature of electrons.
Question 31. Does the stopping potential in photoelectric emission depend upon
(i) Intensity of the incident radiation in a photocell

(ii) The frequency of incident radiation?
Ans 31. (i) The stopping potential is independent of the intensity of incident radiation.
(ii) The stopping potential depends on the frequency of the incident radiation as follows:
eV0 = hv - hv0
question 32. Explain how Einstein’s photoelectric equation enables us to understand the (i) linear
dependence of the maximum kinetic energy of the emitted electrons on the frequency of the incident
radiation (ii) existence of a threshold frequency for a given photoemitter.
Ans 32. K.E.max = hv - hv0
(i) From the equation it is clear that K.E.max depends on v’ i.e., linear dependence.
(ii) If v < v0’ K.E. max is negative which is not possible, i.e, is below v0 emission will not
take place.
1. If radiation of wavelength 5000 is incident on a surface of work function 1.2eV, find the value
of stopping potential.
∅
[Hint. ∅ ∅
]
The existence of work function show that to cause the emission of a photoelectrons, the incident
photon needs to have some minimum energy, i.e., the work function ∅ is possible only if the
incident photon possesses a certain minimum frequency given by
∅ ….(6)
Eqn. (6) here shows that the threshold frequency of the incident light is ∅. If the frequency of incident
light is less than this threshold frequency, no photoemission occurs.
(3) The threshold frequency of a metal is 2 . What is the work function?
[Hint. ∅
]
The concept (4)can now be written as
or or …(7)

54
Now if the incident light has a wavelength and the ‘the threshold wavelength ‘ is then
( )
Where c is the velocity of light in vacuum.
( ) …(8)
EXERCISE-1
1. Mention one physical process for the release of electrons from the surface of a metal.
2. Light from a bulb falls on a wooden table but no photoelectrons are emitted. Why?
3. If the intensity of incident radiation on a metal is doubled, what happens to the kinetic energy of electrons
emitted?
4. What is the effect on the emitted photoelectrons if the wavelength of the incident light is decreased?

55
5. Why are alkali metals more suitable for the study of photoelectric emission?
6. Why photoelectric effects cannot be explained on the basis of wave theory of light?
7. Does each incident photon essentially eject a photoelectron?
8. Blue light can eject electrons from a photosensitive surface while orange light cannot. Will violet and red
light eject electrons from the same surface?
9. The work function of metal A is 4.2 eV. If two photons each of energy 2.5 eV strike an electron of A, will
the emission of electron be possible?
10. Radiation of frequency Hz is incident in three photosensitive surfaces A, B and C. the following
observations were noticed:
(i) Surface A: Photoemission occurs but the photoelectrons have zero kinetic energy.
(ii) Surface B: No Photoemission .
(iii) Surface C: Photoemission occurs and photoelectrons have some K.E. Explain the three observations
by Einstein’s photoelectric equation.
11. Give two informations, we can get from the Fig.,
12. State whether the following statements are true (T) or false (F)
(a) The smaller the frequency of incident light, the greater is the stopping potential.
(b) The smaller the intensity of incident light, the greater is the stopping potential.
(c) the greater the work function, the smaller is the stopping potential.
(d) The smaller the work function, the greater is threshold frequency.
(e) The greater the energy of photon, the smaller is the stopping potential.
(f) The smaller the frequency of incident light, the greater is the kinetic energy of ejected electrons.
13. Match the following: Assuming ejection of electron of electrons is taking place when metal surface is
illuminated by light.
14. Given below is the graph (Fig., ) between frequency (v) of the incident light and maximum kinetic energy
K.E. of emitted photoelectrons. Find the value of (i) threshold frequency and (ii) work function from the
graph.
15. Two metals A,B have work functions 2 eV and 4 eV respectively. Which metal has a lower threshold
wavelength for photoelectric effect?

56
16. Two metals A and B have work functions 4 eV amd 10 eV respectively , which metal has higher
threshold wavelength?
17. If the stopping potential is 5 V, what is the maximum kinetic energy of photoelectrons ejected?
18. What is the effect on the velocity of the photoelectrons, if the wavelength of the incident light is
decreased?
19. UV light is incident on two photosensitive materials having work functions and ( ). In
which case will the kinetic energy of the emitted electrons be greater? Why?
20. Work function of aluminum is 4.2 eV. If two photons each of energy 2.5 eV are incident on its surface,
will the emission of electrons take place?
21. What happens when
(a) Intensity of light is increased keeping frequency constant ?
(b) Increasing the frequency of light keeping intensity constant?
22. Does the stopping potential in photoelectric emission depend upon (i) intensity of the incident radiation
in a photocell (ii) the frequency of incident radiation ?
23. Plot a graph showing the variation of photoelectric current with anode potential for two light beams
wavelength but different intensity.
24. Explain how Einsteen’s photoelectric equation enables us to understand the (i) linear dependence of the
maximum kinetic energy of the emitted electrons on the frequency of the incident radiations (ii) existence
of a threshold frequency for a given photoemitter.
EXERCISE – 1
ANSWER
1. Photoelectric effect.
2. In wood, there are no free electrons to be emitted.
3. The intensity of incident radiation has no effect on the kinetic energies of the emitted electrons.
4. When wavelength of incident light decreases, the frequency and hence, the energies of the incident
photons increase. As a result, the energy of emitted electrons increases.

57
5. Due to small value of work function.
6. According to wave theory, energy is continuous, and even if we shine a metal surface with radiation of
frequency smaller than threshold frequency then after sufficient time enough energy will be accumulated
by the electron to leave the metal surface. But this is in contradiction to experimental facts. Also according
to wave theory, energy intensity of radiation. So if we choose a radiation of frequency small than the
threshold frequency for shining on metal, then the process of electron emission should start faster for
higher intensity. But in fact no electron emission takes place in this case also. Thus, classical wave theory
cannot explain the experimental facts of photoelectric effect.
7. Not every incident photon causes the emission of photoelectrons.
8. Violet light can eject electrons because frequency of violet light ( ) frequency of blue light . Red
light cannot eject electrons because frequency of red light frequency of orange light .
9. No because an electron can accept only one photon at an instant. Since energy of one photon is less than
work function, no photoelectron emission will take place.
10. Einstein’s photoelectric equation is

For surface A:
Threshold frequency Hz
Work function eV
The energy of incident photon eV
For surface B: No photoemission occurs.
The work function of B is greater than 4.14 eV
For surface C: The work function of surface C is less than 4.14 eV.
Therefore,
[ ]
11. Since threshold frequency of caesium is less than that of sodium, therefore, the work function of caesium
is also less than sodium.
Also the straight line suggest that
12. (a) F (b) F (c) F (d) F (e) F (f) F

13. (a) (ii) (b) (iii) (c) (iv) (d) (i)
14. From the graph it is clear that threshold frequency Hz and work function is 4 eV.
15. We know that Lower the work function, lower is the threshold frequency. Therefore, threshold
frequency of A is lower.
16. . Metal A will have higher threshold wavelength.
17. 5 eV.
18.
19. K.E. = hv – W.
For same v, smaller the value of W, greater will be kinetic energy. Therefore, the answer is the material
having work function .
20. No. because

58
21. (a) The photoelectric current increases because increasing the intensity of light increases the number of
photons per second per unit area.
(b) The kinetic energy of photoelectron increases because, increasing the frequency increases the energy
of quanta (photon). It has no effect on number of photons/area until intensity of light is not changed.
22. (i) The stopping potential is independent of the intensity of incident radiation.
(ii) The stopping potential depends on the frequency of the incident radiation as follows:
23. Refer to Fig.
24. K.
(i) From the equation it is clear that K. depends on v’ i.e., linear dependence.
(ii) If v < is negative which is not possible, i.e., is below , emission will not take place.
SUMMARY OF DUAL NATURE CONCEPT CLASESS (XII)

PHYSICS
 When light and matter interact, energy and momentum are transferred through lumps of energy called
photons.
The energy of a photon of frequency v is E = hv
The momentum of a photon

59
 The phenomenon of emission of electrons from a metallic surface when illuminated by light of
appropriate frequency is called photometric effect.
 Photometric current depends on (a) the intensity of incident light, (b) potential difference between
emitter and collector (c) nature of cathode. It is independent of the frequency of incident light.
 The stopping potential (and maximum kinetic energy) depends on (a) the frequency of incident light (b)
nature of cathode. It is independent of the intensity of incident light.
 For photoelectric effect,
and
 For each metal, there exists a certain minimum frequency called threshold frequency below which there
is no photoelectron emission. The energy corresponding to this frequency is called threshold energy.
 The minimum energy required by an electron escape the surface of the metal is called its work function.
 The experimental observations of photoelectric effect could not be explained by classical wave theory of
light. It could only be explained by particle (Photon) theory of light. It could only be explained by
particle (Photon) theory of light.
 A photocell is a device based on photoelectric effect which converts light energy to electrical energy.
 Radiation in certain situation behaves like wave and in some other like particle. This is called wave-
particle duality.
 Like radiation, matter also has dual nature. The wave associated with matter is called matter wave. The
wavelength is given by.
The wavelength is appreciable for subatomic particles only.
 Davisson and Germer electron diffraction experiment confirmed the wave nature of electrons.

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1

 Some important properties of photons

(a) Photons do not have any rest mass.

Where h is the Planck constant and λ is the wavelength of light.

(f) The mass of a moving photon

(a) What is the energy of a photon in the light beam ?

(a) E = h v = 6.63 10 -34 6 1014 = 3.98 10 -19 J

For n number of photons, the energy will be E = n h v

Concept Classes 3/5 East Punjabi Bagh, 9811741187

Solution . Here p = 3 10 -29 kg m s-1

Illustration 3. A photon posseses an energy of 1.5 MeV. find its momentum.

Solution . Given E = 1.5 Me V = 1.5 1.6 J

(b) Number of photons emitted per second,

(c) Mass of hydrogen atom ≈ mass of proton

Concept Classes 3/5 East Punjabi Bagh, 9811741187

Energy flux ∅ = 1.388 ×

Energy of each photon J.

Solution. Power of sodium lamp, P = 100 W

(a) Energy of photon, J

Rate at which the protons are delivered photons/s.

So metal A with lower work function has higher threshold wavelength.

Concept Classes 3/5 East Punjabi Bagh, 9811741187

Illustration 9. Does the ‘stopping potential’ in photoelectric emission depend upon

(ii) Yes, electrons are emitted by blue light, because

Metal Work function Metal Work function

Types of Electronic Emission

Concept Classes 3/5 East Punjabi Bagh, 9811741187

(c) Photoelectric emission: when the electrons are emitted by illuminating

Experimental Study of Photoelectric Effect

(i) Intensity of the incident radiation,

(i) Intensity of Incident Radiations

While maintaining a constant value of V, as we keep increasing the

If we keep the frequency and intensity of incident radiations constant, then

Concept Classes 3/5 East Punjabi Bagh, 9811741187

(iii) Frequency of incident Radiations

We now study the relation between the frequency v of the incident

The graphs shows that

Concept Classes 3/5 East Punjabi Bagh, 9811741187

Failure of Wave Theory to Explain Photoelectric Effect

EINSTEIN’S PHOTOELECTRIC EQUATION : ENERGY QUANTUM OF RADIATION

Concept Classes 3/5 East Punjabi Bagh, 9811741187

As where Stopping potential

PARTICLE NATURE OF LIGHT: THE PHOTON

Concept Classes 3/5 East Punjabi Bagh, 9811741187

Solution. W = 4.2 eV = 4.2 × 1.6 × J

Using the relation

Concept Classes 3/5 East Punjabi Bagh, 9811741187

Solution. Energy ‘E’ of a photon of wavelength 6.2 m is given by

Energy of the most energetic photons emitted will be

Solution. Now, K.E. is stopping potential.

(a) Each photon has an energy

photons per second.

Concept Classes 3/5 East Punjabi Bagh, 9811741187

(a) Energy of the incident photon,

(1) For violet light,

(2) For yellow-green light, (average wavelength)

Incident photon energy,

(3) For red light, (higher wavelength end)

(i) Which of the two metals, A or B has higher work function ?

(i) Metal A has a higher work function ( ) because .