Dual Nature of Matter and Radiation
(Important formulae and Concepts)

Energy of a photon E = h
E
hc

h
P

P
Momentum of photon
c

Rest mass of photon is zero.


h

h
m
m 2
Kinetic mass of photon
c
c
Einstein's photo electric equation
h
Ek = h - w

hc
Work function W = h0 = 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


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
fied 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

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 eVs = h - h0
electron
Ek - max. kinetic energy of
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?
hc
hc
1
W
 0 

0
w w
Ans. Work function
clearly metal B has lower threshold wavelength.
4.
Work function of sodium is 2.3 eV. Does sodium show
Ans.
photo electric emission
for light
of wavelength 6800A0
-34
8
hc 6.6  10  3  10
0 

 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?
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.
11. An electron, an alpha particle and a proton have the
same kinetic energy, which one of these particles has (i)
Ans.
the shortest and (ii) the largest, de, Broglie wavelength?
h
1


2mEk
m
h
2mev
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
1
 
small.
m
13. An electron and a proton have the same de Broglie wave
length associated with them. How are their kinetic
Ans.
energies related to each
other?
h

2mE k
Given e = p
h
2m e E e

h
2m p Fp

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?
2m p
1
m
p
 
 d 
m for the same velocity
d m p
mp
Ans. As
15. An electron and alpha particle have the same kinetic
energy. How are the de Broglie. wavelength associated
4m p
e their

mrelated?
with


 e  1872  4

me
me

Ans. 
e
 86.5

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?
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
Ans.
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
E 
hc

For
emission of
photoelectrons energy of photon >, work function or
wavelength of incident photon should be less than,
threshold wavelength 0 (i.e. < 0)
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 .
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 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.
h

Ans: The de Broglie wavelength
mv
Now for electron in orbit2r  n (for nth oribit)
2r 
nh
mv
or mvr 
nh
2
This is Bohr's second postulate. As complete de-Broglie
wavelength may be in certain fixed orbits. So nonradiating 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 
0
h
-
e
e
Vs
where 0 is the work function
0

 = frequency of incident radiation
The graph of stopping potential Vs versus frequency of
incident radiation is shown in Fig.
27. The wavelength ''
of a photon and the de-Broglie
h

wavelength
of electron how the same value.
mv
2mc
Show that the energy of photon is
times the
h
kinetic energy of electron, where m, c, h have their
usual meanings.
Ans.






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 surface 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 occur and
photoelectrons have some kinetic energy.
Which of there ha a higher work 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 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 Enstein'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 be h
2me
B 
h
2mE k
neglected)

h
2m
B 
h
2mc
hc

K.E. of electrons Ek = h + w
as W = o, (work function
Ek = h
hc
Ek = 
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
 2mc  2
 B
 h 
h2

 hc 
2m 
 
2mc 2
 
B
h
or
 
32. Two lines A and B shown in the graph represent the deBrogle wavelength () as a function of
1
v
(v is the
accelerating potential for two particles having the same
B
change, which of the two represents the particle
of
smaller, mass.
The slope of line B is large, so
particle B has smaller mass.
h
h
1


2mqv
2mq
A

V
1
V
33. Ultraviolet light of wavelength 2271 A0 from a 100 w
mercumm source irradiates a photo cell made of
mulybdenum 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 
Or
Or
W 
hc
hc


w
- Ek
6.6  10 34  3  108
- 1.6  10 19  1.3
10
2271  10
W  4.2eV
hc
W
0
0 
hc
6.63  10 34  3  108

 2977 A 0
19
w
6.68  10
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 22 
mv 22  4hf 0
2
2
n
(1) from eq (2) side by side
h 5f 0  hf 0 
Dividing eq n
v12
1
v
1

 1 
2
v2
4
v2
2
- (1)
- (2)
35. Calculate the de-Broglie. Wavelength of a neutrons
kinetic energy
150 eV.
h


6.63  10-34
2  1.67  10 27  750  1.6  10 19
2mE k
 0.02335 A0
36.Describe
Davission
demonstrate
and
Germer
experiment
to
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)
frequency
same
but different intensities of 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
emittedfrom 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 usedand 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 Briglie 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?
EDITED BY
Mr. S.K. BHAT(VICE PRINCIPAL)
Mr. A.K.SHARMA(P.G.T)