www.nrpschool.com
NEW RAINBOW PUBLIC SCHOOL
PHYSICS
Dual Nature of Matter and Radiation
(Important formulae and Concepts)

Energy of a photon E = h
E
hc

h
c

Momentum of photon P 

Rest mass of photon is zero.

Kinetic mass of photon m 

Einstein's photo electric equation
P
h
c2
h

m
h
c
Ek = h - w

Work function W = h0 =
hc
0
0 - Threshold frequency, 0 - Threshold wavelength

Maximum K.E. ½ mv2
max
= eVs
Where Vs is stopping potential.

de Broglie wavelength associated with moving particle.


h
,
mv
h
2mE k
,

h
2mqv
,
de Broglie wavelength associated with electron

12.27
V
A0
Photoelectric current is directly proportional intensity of incident
radiation. Stopping potential is independent of intensity of incident
light
Photoelectric
current


Prepared by: Abhishek Yadav (NEW RAINBOW PUBLIC SCHOOL)
www.nrpschool.com
Photoelectric
current
Intensity of incident light
3I
2I
I
2I
Potential difference (V)

Stopping potential is directly proportional to frequency of incident
radiation. Maximum kinetic energy of photoelectrons is directly
proportional to frequency of incident radiation as well stopping
potential

Intensity of light never affects stopping potential as well as maximum
kinetic energy of photoelectrons.

When frequency and intensity of incident light are kept field and photo
metal is changed. We observe that stopping potential (Vs) versus
frequency () graphs are parallel straight lines, cutting frequency axis
at different points.
Metal 1
Metal 2
Vs
0
01
02
Frequency 

Work function is defined as the minimum, energy required to free an
electron from its metallic bonding is called work function. It is denoted
by W.
W = h0

Threshold frequency: The minimum frequency of incident light which
is just capable of ejecting electrons from a metal is called the threshold
frequency. It is denoted by 0
Prepared by: Abhishek Yadav (NEW RAINBOW PUBLIC SCHOOL)
www.nrpschool.com

Stopping potential:- The minimum retarding potential applied to
anode of photoelectric tube. Which is just capable of stopping
photoelectric current is called the stopping potential. It is denoted by
V0 (or Vs)
Einstein's photoelectric equation.
Ek = h - W
Or ½ m2 max = h - ho
Vs - Stopping potential
Or eV0 = h - h0
Ek - max. Kinetic energy of electron
Vs 
Other symbols have their.
h h 0
e
e
Usual meaning.
The slope of Ek Versus  graph is h.
The slope of Vs Versus  graphs ( h/e) is
Where symbol have their usual meaning.
QUESTIONS FOR SUPPORTIVE LEARNERS
1.
The wave length of electro magnetic radiation is doubled; How will the
energy of a photon change ?
Ans.
E
hc


1

Clearly when wave length is doubled, the energy of photon is hauled.
2.
What is the stopping potential applied to a photo cell is the maximum.
Kinetic energy of photo electron to 5 eV?
Ans.
Ek = eVo
 5ev = eVo
Or Vo = 5 v
The stopping potential V0 = 5 volt (negative)
3.
Two metal A and B have work function 2 eV and 4 eV respectively
which metal has a lower threshold wavelength for photoelectric effect?
Ans. Work function
W
hc
hc 1
 0  
0
w w
Clearly metal B has lower threshold wavelength.
Prepared by: Abhishek Yadav (NEW RAINBOW PUBLIC SCHOOL)
www.nrpschool.com
4.
Work function of sodium is 2.3 eV. Does sodium show photo electric
emission for light of wavelength 6800A0
Ans.
0 
hc 6.6  10 -34  3  10 8

 5380 A 0
19
w
2.3  1.6  10
Thus 0 < , No photo electric emission will take place.
5.
Two beams one of red light and other of blue light of the same
intensity are incident on a metallic surface to emit photoelectric which
one of the two beams emits electrons of greater kinetic energy?
Ans
W
0 
hc
0
hc
1

w w
The photon of blue light emits electrons of greater kinetic energy than that
of red light, becomes larger wavelength of red that blue.
6.
Ultraviolet light is incident on two photosensitive materials having
work functions W1 and W2 (W1 > W2). In which case will the kinetic
energy of the emitted electron greater? Why?
Ans. K.E. for metal of Work function W2 Will be greater.
As Ek = h - W. small as work function greater K.E.
7.
Ultraviolet radiations of deferent frequencies 1
and
2 are incident on
two photo sensitive, materials having work functions W1 and W2 (W1 >
W2) respectively. The kinetic energy of the emitted photo electrons is
same in both the cases. Which one of the two relations will be of the
higher frequency?
Ans. According to Einstein's photoelectric equation, Ek = h - w
As Ek is same, h1 - w1 = h1 - w2
h1 - w1 = h1 - w2
w1  w2
h
Or 1-2 =
,
As w1 > w2, 1 > 2
So frequency of radiation 1 is higher.
8.
With what purpose famous Davison Germer experiment with electrons
performed?
Prepared by: Abhishek Yadav (NEW RAINBOW PUBLIC SCHOOL)
www.nrpschool.com
Ans. Davison- Germer experiment was performed to verify wave native of
electrons.
9.
Name the experiment which establishes the wave native of particle.
Ans. Davison -Germer experiment.
10. If the potential difference used to accelerate electron is tripled, by what
factor does de Broglie wavelength of electrons beam change.
Ans. De Broglie wavelength becomes
1
3
time.

h
2mev
11. An electron, an alpha particle and a proton have the same kinetic
energy, which one of these particles has (i) the shortest and (ii) the
Ans.
largest, de, Broglie wavelength?
h
1


2mEk
m
For same kinetic energy.
(i)
Out of given particle, the mass of alpha particle is maximum so de
Broglie wavelength associated with alpha particle is shortest.
(ii)
As man of electron is least, so electron has largest de Broglie.
wavelength.
12. An electron and a proton have the same kinetic energy, which one of
the two has the larger wavelength and why?
Ans: An electron has larger wave length as its mass is small.
1
 
m
13. An electron and a proton have the same de Broglie wave length
associated with them. How are their kinetic energies related to each
other?

Ans.
Given e = p
h
2m e E e

h
2m p Fp
h
2mE k

Ee mp

E p me 
1840
i.e. K.e. of electron = 1840  K.e. of proton.
14. A proton and a deuteron have the same velocity, what is the ratio of
their de Broglie wavelengths?
1
 
m for the same velocity
Ans. As
 p m d 2m p


d m p
mp
Prepared by: Abhishek Yadav (NEW RAINBOW PUBLIC SCHOOL)
www.nrpschool.com
15. An electron and alpha particle have the same kinetic energy. How are
the de Broglie. wavelength associated with their related?
16. How
will
the
photoelectric
cement
change
on
decreasing
the
wavelength of incident radiation for a given photo sensitive material?
Ans. It is independent of the frequency of incident light or wavelength of
incident light? e

m
me

e
 86.5


4m p
me

e


1872  4
17. What is the de Broglie wavelength (in A0) associated with an electron
accelerated through potential difference of 100 volt.
Ans.
 
18.
12.27 0
A
v

12.27 0
A  1.22 A 0
100
de-Broglie wavelength associated with an electron accelerated through
a potential difference V is . What will be the de-Broglie wavelength
when the accelerating p.d. is increased to 4V?

1 1
V 
4

,  2    2 
V1 2
1
2
v 2
Short Answer Questions19. For a photosensitive surface, threshold wavelength is 0. Does
photoemission if the wavelength radiation is (i) more than 0 and (ii)
less than 0? Justify your answer.
Ans. Energy of a photon for emission of photoelectrons energy of photon >,
work function or wavelength of incident photon should be less than,
hc
threshold wavelength 0 (i.e. < 0)
E 

20. Two metals  and  when illuminated with appropriate radiations emit
photoelectrons. the work function of  is higher than that . Which
metal has higher valve of threshold frequency and why?
Ans: Work function W = h0
W  0
So threshold frequency of  0 is higher than that of .
Prepared by: Abhishek Yadav (NEW RAINBOW PUBLIC SCHOOL)
www.nrpschool.com
21. An  particle and a proton are accelerated from state of rest through
the same potential difference V. Find the ratio of de-Broglie wavelength
associated with them?
Ans:

 
h
2mqv
h
,
2  4m p 2ev
p 
h
2  4m p ev

1

  :  p  1 : 2 2
p
8
22. Two metals  and  have work function 2 ev and 4eV respectively
which metal will emit electrons when irradiates with light of
wavelength 400 nm and why?
Ans:

hc


6.6  10 -34  3  108
 3.09
400  10 9
This is greater than work function of , but less than work function of
 so metal  will emit electrons.
23. In a plot of photoelectric current versus anode potential, How does.
(i)
The saturation current vary with anode potential for incident
radiations of different frequencies but same intensity.
(ii)
The stopping potential vary for incident radiations of different
intensities but same frequency?
(iii) Photoelectric current vary for different intensities bid same
frequency of incident radiations?
Justify your answer in each case.
Ans. (i)
In photoelectric affect the salvation current does not very with
anode potential with incident radiation of different frequencies.
The reason in that satiations current depends only on intensity of
incident radiation because a single photon can eject a single
electron, however large the frequency of radiation may be.
(ii)
obviously stopping potential is independent of intensity provided
frequency  remains unchanged, that is stopping potential does
not vary with intensity of incident radiations.
(iii) Photoelectric current increases with increase of intensity because
increase in intensity of radiation means increase in umber of
Prepared by: Abhishek Yadav (NEW RAINBOW PUBLIC SCHOOL)
www.nrpschool.com
incident photons As one photon ejects one electron, increase
intensity cause increase in photoelectric current.
24. Show that Bohr's second postulate 'the electron revolves around the
nucleus only in certain fixed orbit without radiating energy can be
explained on the basis of do. Broglie hypothesis of wave nature of
electron.
Ans: The de Broglie wavelength  
Now for electron in orbit
h
mv
2r  n (for nth oribit)
nh
2r 
mv
nh
or mvr 
2
This is Bohr's second postulate. As complete de-Broglie wavelength
may be in certain fixed orbits. So non-radiating electron can be only in
certain fixed orbit.
25. Red light however bright it is, can produce emission of electrons from a
clean zinc surface, but even weak ultraviolet radiation can do so, why?
Ans. The energy of photon of U.V. light is greater than the work function of
Zinc, so ultraviolet can emit photo electrons even intensity is weak.
While the energy of photon of red colour photon is less than work
function of zinc. So the photoelectric emission is in dependent of
intensity.
26. Write the Einstein's photoelectric equation in terms of the stopping
potential and the threshold frequency for a given photosensitive
material. Draw a plot showing the variation of stopping potential
versus frequency of incident radiation.
Ans. Einstein's photoelectric equation in terms
of stopping potential Vs is
eVs = h - 0

or Vs  e  - e
Vs
where 0 is the work function
h
0
0

 = frequency of incident radiation
The graph of stopping potential Vs versus frequency of incident
radiation is shown in Fig.
Prepared by: Abhishek Yadav (NEW RAINBOW PUBLIC SCHOOL)
www.nrpschool.com
27. The wavelength '' of a photon and the de-Broglie wavelength
of
electron how the same value. Show that the energy of photon is
times the kinetic energy of electron, where m, c, h have their usual
2mc
meanings.   h
mv
Ans.
h






E  hc 

Ek   
 1

 mv 2 
 2

1
h2
m 2 2
2
m 
2 2
2hc m 

or

mh2
E
2mc

Ek
h
E
2mc
h
Ek
28. Radiations of frequency 1015 Hz are incident on two photosensitive
surfaces P and Q. Following observations are made.
(i)
For surface P, photoelectric emission occurs but photoelectrons
have zero kinetic energy.
(ii)
For surface Q, photoelectric emission occurs and photoelectrons
have some kinetic energy.
Which of there ha higher works function?
If the incident frequency is slightly reduced, what will happen to
the photoelectric emission in the two cases?
(i)
For surface P, Ek = 0, So, Energy of photon = work function
h = w = h0
= 6.610-34  1015
= 6.610-19 Joule.
(ii)
For surface Q, the photoelectrons have some kinetic energy.
h = w = Ek
Work function of Q is less than that of P
i.e. surface P has higher work function than Q.
As the frequency of incident radiation is slightly reduced energy of
photon will become, less than work function of P, but will be more
Prepared by: Abhishek Yadav (NEW RAINBOW PUBLIC SCHOOL)
www.nrpschool.com
than the work function of Q. Hence surface P will show no
photoelectric emission while Q will show photoelectric emission but
the kinetic energy of photo electrons from surface Q will be lower than
initial value.
29. Radiations of frequency 1015Hz are incident on two photosensitive
surfaces A and B. following observations are recorded:
Surface A : No photoelectric takes place.
Surface B: Photoemission takes place but photoelectrons have zero
energy explain the above observation on the basis of Einstein’s
photoelectric equation. How will the observation with surface B change
when wavelength of light is decreased?
Ek = h - h0
Surface A: As no photo emission takes place; energy of incident photon
is less than the work function.
Surface B: As photoelectric emission takes place with zero kinetic
energy of photoelectrons.
i.e. energy of incident photon is equal to work function when
wavelength of incident light is decreased, the energy of incident photon
become more than the work function, so photoelectrons emitted will
have finite kinetic energy given by.
Ek 
he

-W
30. X-rays of wavelength '' fall on a photosensitive surface, emitting
electrons. Assuming that the work function of the single can be
neglected. Prove that the de-Broglie wavelength of electron emitted will
h
be
2me
h
B 
K.E. of electrons Ek = h + w
2mE k
as W = o, (work function neglected)
h

hc
Ek = h
2m
hc

Ek = 
h
B 
2mc
Prepared by: Abhishek Yadav (NEW RAINBOW PUBLIC SCHOOL)
www.nrpschool.com
31. An electromagnetic wave of wavelength  is incident on a
photosensitive surface of negligible. Work function. If the
photoelectrons emitted from surface have the same de-Broglie
wavelength B, Prove that
 2mc  2
 
 B
 h 
De-Broglie wavelength
B 
h
2mE k
h2
 
2mE k
2
B
or
h2

 hc 
2m 
 
2mc 2
 
B
h
or
 2mc  2
 B
 h 
 
32. Two lines A and B shown in the graph represent the de-Broglie
wavelength () as a function of
1
v
(v is the accelerating potential for
two particles having the same change, which of the two represents the
particle of smaller, mass.
B
The slope of line B is large, so
particle B has smaller mass.

h

2mqv
h
2mq
A

1
V
1
V
33. Ultraviolet light of wavelength 2271 A0 from a 100 w mercum 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 (105 wm-2) red light of
wavelength 6328 A0 produced by a He-Ne laser? (h= 6.6310-34 Js,
C = 3108 m/s)
Ek 
hc
hc

w
Or W   - E k
Prepared by: Abhishek Yadav (NEW RAINBOW PUBLIC SCHOOL)
www.nrpschool.com
6.6  10 34  3  10 8
- 1.6  10 19  1.3
2271  10 10
W  4.2eV
hc
W 
Or
0
0

hc
6.63  10 34  3  10 8

 2977 A 0
w
6.68  10 19
As given wavelength 6328 A0 is greater than 2977 A0 the photocell
will not respond to red light produced by He - Ne lese, How ever intense it
may be.
34. Defined work function of a metal. the threshold frequency of a metal is
f0. When the light of frequency 2f0 is incidents on the metal plate, the
maximum velocity of electrons emitted is v1. When the frequency of
incident radiation is increased to 5f0, the maximum velocity of electrons
emitted is v2 find the ratio of v1 to v2
1
mv 2
2
In first case   2f 0 ,  0  f 0 , v  v1
h   h 0 
1
1
mv 12 
mv 12  hf 0
2
2
In second case   5f 0 ,  0  f 0 , v  v2
h 2f 0  hf 0 
1
1
mv 2
mv 2
2 
2  4hf 0
2
2
(1) from eq n (2) side by side
h 5f 0  hf 0 
Dividing
eq n
- (1)
- (2)
v12
1
v1
1



2
v2
4
v2
2
35. Calculate the de-Broglie. Wavelength of a neutrons kinetic energy 150
eV.

h

2mE k
6.63  10 -34
2  1.67  10 27  750  1.6  10 19
 0.02335 A0
36. Describe Davission and Germer experiment
to demonstrate
he wave
nature of electrons. Draw a labeled diagram of apparatus used
37. State
the laws of photoelectric effect. How have they been explain by
Einstein?
38.Draw a graph showing the variation of photoelectric current with anode
potential of a photocell for (i)
same frequency but different intensities of
Prepared by: Abhishek Yadav (NEW RAINBOW PUBLIC SCHOOL)
www.nrpschool.com
incident radiation (ii) same intensity but different frequencies of incident
radiation.
39. What is photoelectric effect? Explain experimentally the variation of
photoelectric current with (i) intensity
(ii) the p.d. between the plates (iii)
frequency of incident light and hence obtain Einstein‘s photoelectric
equation.
40. Calculate the maximum kinetic energy of electros emitted from a
photosensitive surface of work function3.2eV, for the incident radiation of
wavelength of 300nm
41.The work function of three elements a, B, and C are
5eV,3.8eV and
2.8eV respectively. A radiation of wavelength 4125 A0 is made to be incident
on each of these elements. Show by appropriate calculation in which case
photoelectrons will not be emitted.
42. A source of light is placed at a distance of 0.50m from photocell used
and the cut of potential is found to be V0. If the distance between the light
source and the photocell is made 0.25m. What will be the new cut of
potential?
43.In Davission and germer experiment, state the observation s which led to
(i) show the weave nature of electrons and (ii) confirm de Broglie relation.
44. Deduce de Broglie wavelength of electrons accelerated by a potential of v
volt. Draw a schematic diagram of a localized wave describing the wave
nature of moving electron.
10. Following table gives the values of work function for few photosensitive
metals.
S,No.
Metal
Work
Function
(eV)
1
Na
1.92
2
K
2.15
3
Mo
4.17
If each of these metal is exposed to radiations of wavelength 300nm,
which of them will not emit photoelectrons and why?
Prepared by: Abhishek Yadav (NEW RAINBOW PUBLIC SCHOOL)