orevious Years Questions with Additional
Important Problenms
Ouestion 1. What IS the momentum p of a
o
kraviolet light of wavelength = 332 nm? photon[1]
[ISC 2003]
332x 10m
[1] [ISC 2007]
= 500 nm in SI unit.
Calculate the momentum of a photon
1.6
x
of
10
light
1980
kg-m-s
K=E- Eo
3.75 eV
K=6.25- 2.5
K
K eVo
-
3.75 volt.
Vo= 3.75 volt
3.75 eV = eVo
=
surface has received
10
Stopping potential
A falls
Light of wavelength 5000
Question 6.
E =N
Solution.
N-
on
89
sensitive
==25
100
25 xx 10
[1
energy, how
on a
RADIATION AND MATTER
DUAL NATURE OF
wavelength
function.
anode to stop the
the
[1]
[ISC 2002]
photoelectric
threshold frequency
metal is called the threshold frequency.
hv W+E
instein's photoelectric equation is
Refer to vs|
frequency is shown in figure.
Graph of stopping potential versus
Q.8.
Threshold frequency
Refer to Q. 7.
ISC 2012) Answer. Stopping potential:
graph.
VO
[21 [ISC 1996]
Question 9. Explain stopping potential and threshold
frequency in photoelectric emission. Give an appropriate
where hv is energy of incident photon, W is work function
and Ex is maximum kinetic energy of photoelectron.
a
incident
uphotoelectric
t c emission
emission from a
Work function W=hvo, where vo is the
the work
to release an electron from its metallic surface is called
Answer. Work function: The minimum energy required
12] [ISC 2001]
Question 8. Explain work function and threshold frequency
in photoelectric emission. Give Einstein's equation of
photoelectric emission. What do the symbols stand for?
Kinetic energy of photoelectrons = eVs
current. It is denoted by Vg-
potential applied
How is it related
x 3 x 10
10 x5000x 10
If the
6.63 x 10-S* J-S
6.63
x 10
=2x 102 kg-m-s. surface.
many photons
have fallen on the surface?
P=
Solution. Momentum of photon,
Ouestion 2.
6.6 x10* -s.
of green light of wavelength a
h
6.63 x 10
Question 7. Define stopping potential.
to the kinetic energy of photoelectrons?
hc
109
10- J.
minimum negative
Answer.
1.32 x
photon of energy
11IISC
2009
(1] [ISC 20091
Stopping potential: It is the
10-2 kg-m-s
6.6 x1034
500 x
x
KE-M-s*
Solution. Momentum of photon of wavelength2,
P
=
1.32
=
[1] [iSC 2006]
Question 3. A radio transmitter operates on a wavelength
of 1500 m at a power of 400 kW. Find the energy of the
radio photons in joules.
a
Solution. Energy of each radio photon,
hc
E =
1500
momentum
of
6.6x10* x3x10
Question 4. Find the
=
=
can cause
Threshold frequency: The minimum frequency
10m
which
3.0 eV.
light
3.0x1.6x10
3.0x1.6x101
is speed of light.
Solution. P =*E
P
andc
iere, E is energy of photon
So,
3x108
Monochromatic
x
nm
198
a= 198
Question 5.
whose work
surface of metal,
98 nm is incident on a
potential. [2]
Calculate the stopping
Tunction is 2.5 eV.
Solution.
F-12375 ev =6.25 eV
2.5 ev
E1980
E=Eo + K
Eg
90
Question
VATSAL ISC HANDBOOK of PHYSICS Class XI
Answer. Refer to article 3.
line of form y
[21
=mx C
-
15.
6.6
x
x
hvo
x
frequenc
potenti
energy of
[11 [ISC 200
e work tunctic
1014
calculate
threshold
=
J.
kinetic
=
is the stopping
5VV
maximum
10
1 0 3 * x 5.0
¢
10'"Hz,
3.3
function
xx
Photoelectric
=
=
what
V,
5eV =eVs
(negative).
is 5 eV,
The
Work
16.
=
=
=
so stopping
incident radiation;
potential
The stopping
on
is
potential u
a
functio
and6 eV respectively. Which metal as low
W
W
=
threshold
[ISC 20
photoelectric effect?
wavelength for
=w
threshold
Ao
Work function
400
nm
is
the cathode when
400
eVs
=
Vs = 0.5 V
K=
=
nm
E 3.1 eV
=
4000 Å
eV
E 3.09375 eV
4000
E=12375
A
0.5 eV
of the material of the cathode.
Solution.
We know
E
Eo+K, Eg is work
Eg= E - K
Eg=3.1-0.5
function..
[ISC 2
0.5 Vis required to block the movement of electrons
monochromatic light of wavele
incident on its surface. Find the work func
Question 19. In a photoelectric cell, a retarding potenti
Clearly metal B has lower threshold wavelength.
wavelength
Answer.
photoelectric e
remain unchanged.
have work
metals A and B
Question 18. Two
intensity of
Answer.
increased
eV,
Answer. Ex
5 volt
potential V,
photocoll
r a d i a t i o n in a
The stopping
of
If the intensity
potential vary?
Question 17.
the stopping
how does
does not depend
photoelectron
Question
Solution.
5.0
is 5.0
Question 10. In photoelectric effect, what is meant by the
metal is
[1) [isC 2011] alkali
term 'threshold frequency?
of the metal.
Question 11. What is meant by work function of a metal
[2]
photoelectron
How does the value of work function influence the
kinetic energy of electrons liberated during
emission?
Answer. Work function: The minimum energy required
to free an electron from metallic surface is called the
work function.
Smaller is work function, larger is kinetic energy of emitted
electron.
Question 12. Draw a graph showing the variation of
stopping
slope
Vs
Stopping
potential
h
straight
=.
V
potential with frequency of incident radiation in
relation to photoelectric effect. Deduce an expression for
the
of this graph using Einstein's
equation.
Answer. The graph is
shown in figure.
electric equation
= hv - hvo
m
a
From Einstein's photo-
Ex
eVs = hv - hvo
V , = #v -#vo
is
slope of graph is
Clearly V-v graph
the
Question 13. Draw a graph showing the variation of
stopping potential with frequency of radiation incident
on metal
How can the value of
plate.
Planck's
constant
be determined from the graph?
[2] [ISC 2009]
Answer. The graph is shown in figure of the above
question.
The slope of the graph is
a
By measuring slope of V,-V graph, the value of Planck's
14. For
photosensitive surface work function
constant, h =ex (slope of V,-v graph).
Question
is
W3.3x10=
x
fre
3.3x 10
J taking Planck's constant to be 6.6
Eg 2.6 eV =4.16 x
10
Question
20. Green
Is, find the threshold frequency.
r11
light ejects 101 J.
given
photosensitive
photoelectrons
surface whereas
[ISC 19971 not What
will
Solution. Threshold frequency
Give reason for happen in case
yellow
ligh
of violet
and rec
Answer. The your answer.
3.3x10-19
5x
Vo
frequency of
Hz. light, so
10
violet
violet
h
6.6x 1034
light is more tha
light will eject
of red light
is less
eject electrons.
light
than
electrons.
yellow
light, so redThe
Wnat
i s t h .
to
cause
stion 27. 7he
The
work
function
of
the min
sodium is
inimum
energy for a
iation
5odium surface?
2.28 eV.
radiation
photoelectric
of
=
a
to
AL
tungsten
equal
emission,
542.8 nm.
wavelength of
m
3.648x 1079
10
uhreshold
5.428 x
"= 6.6x10* x3x108
is
5.18 eV.
x10-l0
NATURE OF RADIATION
nm.
AND MATTER|
=
91
ex
energy or
Question 23. Threshold wavelength of a certain met
is
trons emitted is,
=
"6.63x
hc
from
up?
)
10" s,
light
c
emitted from a metal
of wavelength 300 nm is
are
=
3.00x
10°
ms
6.63 x107J=4.14
1.60x 10J/eV
and
eV.
(3] [ISC 2003]
photons of
The energy of the incident
(6.63x 10 Js) x (3.00x 10 ms
=
300x 10m
6.63xx10-19J
0.54 eV.
the metal is
eV.
e V , =4.14 eV-0.54 eV =3.60
Hz to
a
13] [ISC 2014
8x 10 Hz, by how much
surface ga
given photosensitive
stopping potential for
4x 10
Einstein's photoelectric equation.
Question 25. (i) Write
radiation is increasec
(i) lf the frequency of the incident
will the
W
Cii) The work function of
1.60 x10"J/ev
1.6x10x0.54
E eVo (1.6x10C)x0.54 V
is
kinetic energy of the emitted photoelectrons
then the maximum
(i) If V, be the stopping potential,
given by
=
wavelength A is
Solution.
1eV= 1.60x 10).
h
to stop the emission of electrons is
maximum ki
(i) the energy of the incident photons, (i) the
the work
netic energy of the photoelectrons emitted, (ii)
function of the metal. Express all answers in eV. Given
0.54 V. Calculate
incident on it. The minimum negative potential required
surface when ultraviolet
Question 24. Photoelectrons
2.5x 1019J=1.6x10-19.56
2.5x10-19
eV.
-6.6x10 x3x10°206105
792x10)
396x 109 792x10
1
h=6.6x 10" s, c=3.0x 10 mls) 13][ISC 2010
Solutlon. Maximum kinetic energy of the photoelec
posed to ultraviolet light of wavelength 396 nm?
[31 [ISC 2005]
wavelength, ho =2400 Å
2400x10
8.29x10J
=
6.63x104 x3x10 joule
C
Threshold
=
=
eV
8.29 x 10-19
1.6x
1019
hc(g-1)
20
e
quantum of (ii) Stopping potential, V, =ExEx in
photoelectric emission
e eV) = 2.6V.
2.6 V.
from
the
Calculate
the
wavelength of this
radiation.
792
What is the maximum kinetic
13] a
solution.(i)
From
[ISC 2007]1 photoelectrons emitted by this metal if it is
Einstein'
s
photoelectric equation
quantum
hv) for
hv =W+E
(hv)min = W
=
minimum energ
0,.
ener8y Ek
Hnetic
Minim
energy of
ork function, i.e.,
EminW=2.28 ev
x
=2.28x 1.6 x 1019 J
109 J.
3.648
Ine
=
=
mEnergy of a photon of radiation, E = hc
22.
Wavelength, 2
Question
(i)
is 2400 A. When tungsten is illuminated with light of
wavelength 1600 A, find () work function, (i) maximum
kinetic energy of emitted electron and (ii). stopping
potential.
Solution.
Work function, w
o
electron
n) Maximum KE of emitted
E
1600x
x3x105 (2400-1600)
2400x
6.63x1034
= 4.14x 1017J
=
eV 2.6
eV.
2.6eV.
4.14x10-19
1.6x 10
92
=
=
of
(ii)
r
a
b e t w e e n
p
h
o
metal,
called
a
t
o
e
:
i
.
d
n
o
e
de
a n o d
radiation
[3] [ISC 004
and
When
o
h
r
t
are
effect.
s
c
n
i
emitte
of intensity ()
in Ii
[Fig. (a))
ofinten.
is s h o w n in
increase
r
o
Wre emitted.light)
This
t
c
c
e
t
t
a
a
i
c
d
e
l
with
e
l
effect
incident
of
applied
Frequency
Voltage
on
Photoelectric
incident
is
is
called
the
increa
s e of
ase
of voltage
ant [Fig. (b)].. The
current
is
vVo
ph
frequency
than
n
(6)
more
the
inci.
of frequency of
of
c o n s t a n t
with
The graph
increases
value
becomes
increases
radiation.
Photo-current
then
constant
and
photo-current
incident
of
The
applied
maximum
the
independent
saturation current.
is
(c)
Vo frequency(v)
[Pig. (C)J.
ip
frequency
provided
Photo-current
radiation,
threshold
ph
(a)
x
x
1020
=
of
photoelectric enect
2.1 eV.
103x 5 x 10 J
o5x10 Hz
33
W=6.6
=
as
J
J
1.6x10-19
16
33
2.0625 eV
an
experiment
maximum
30. In
of
=
W33x 10-20
leV 1.6 x10
graph
AB
versus
photoelectrons
straight line
a
Question
the
Solution. W=Work
function =hvo
photoelectric emission is 5 x 10'* Hz. Calculate the work
function.
[3] [ISC 2012
Question 29. Threshold frequency of certain metal for
Cii)
Cii)
()
p h e n o m e n o n
is
Answer.
(iin
wavelength
0.1x10l7 =1x1018 J.
h = 6.6310-34
198x 10
i)
VATSAL ISC HANDBOOK of PHYSICS ClassXI
1019
( 8 * 10-4x1015)
Solution. (i) Please refer to equation give
explanation of photoelectric effect (Article 4).
hvo
(ii) We know evo = hv - hvo
Einstein's
e V = hvi -
eV2 = hv2- hvo
e(V2-V1) =h(v2- v1)
16
Monochromatic
light
10(4 x105) volt 16.5 V.
66
1.6x
6.6x 1034
(V2-V) (2-y)
26.
198 nm is incident on a surface of a metallic cathode
Question
whose work function is 2.5 el. How much potential
dierence must be applied between the cathode and the
13) [ISC 2016]
anode of a photocell to just stop the photo-current from
flowing?
= 1980 Å
Solution. Wavelength of incident light =198 nm
12375
=
ev 6.25 ev
=5
1980
(in Ä)
eV
Energy (E) of incident photon is
E
E
E-Eg evs
Eg 2.5 eV
Thus, (6.25 2.5) eV= eV,
3.75 eV= eVs
V, 3.75 volt (Negative)
Question 27. A monochromatic source of light emits light
ofwavelength 198 nm. Calculate
P
i) Energy of each photon.
(ii) Momentum of the photon.
[21 (ISC 2011]
Solution. G) E = nC _ 6.6310*:3 10
198x10
ii)
0.033x 10 25
= 3.3104 kg-m-s.
uestion 28. What is photoelectric effect? With the help
f suitable graph show the
variation of photo-current
sed. Show the variation of photo-current with
Intensity of incident radiation.
in the
kinetic energy EE
of the emi
emit
frequency v of the incident
lig ht
figure ahead.
shown
Find:
(eV)
8
6
10
20
30
v(Hz)-
30
10
1014
Hz.
.iSC 2015]
[3]
photoelectrons emitted by
mWork function of the metal.
=
x
a Threshold trequency of the metal.
v
for the
light ot frequency
Stopping potential
the
=
Ex= hv- hvo
when Ek 0,
can see
v
vVo
intercept over frequency
Hz
is
=
Solution. ) Einstein's photoelectric equation in term of
kinetic energy of photoelectrons is
We
Vo=Threshold frequency
axis.
Uo 10x 10
P=2mE
ie.
V volt,
h
mu
mu
2mEx
P
Ek qV
h
2mqV
2mEk
93
nature
of
experiments.
For electrons q = e = 1.6 x 10 19 C, m =9x 10
A 2 x 10-10m12.27
V
V
h
2mkT
- 150
150 A=1A.
Solution. de-Broglie wavelength of electron
[1] (ISC 2
Question 1. An electron is accelerated under a pote
difference of 150 volts. What is its de-Broglie wavelen
Previous
Questions with Additic
Important Problems
Years
temperature T, Ex = kT
For neutral particles in thermal equilibrium at abso
kg.
For charged particles associated through a potential
The wave associated with material particle is calle
the de-Broglie wave or matter wave. The d
has been confirmed by diffractic
Broglie hypothesis
experiments.
h
If E is kinetic energy of moving material particle, ther
where p is momentum.
particle behaves as waves and the wavelength associatedwith material particle is
(e.g, electrons, protons, a-particles, atoms etc.) may
exhibit wave aspect. Accordingly, a moving material
Louis de-Broglie postulated that the material particles
electrons, positive ions isestablished by
de-Broglie thought dual nature of matter.
de-Broglie Hypothesis: The particle
has dual nature. In analogy with dual nature of light,
DUAL NATURE OF RADIATION AND MATTER
i) Work function of metal = hUg. It is the value of
hvo = 4 ev
intercept over Ex-axis [Y-axis]
For frequency (v) of incident light, Ex is 8 eV, therefore
i) We known E = eVs, where V, is stopping potential.
8 eV eVs
V,=8V
Stopping potential is always negative.
to
of incident light,
(2] [ISC 2019]
Question 31. Define the following with reference
photoelectric effect:
a) Threshold frequency (to)
(b) Stopping potential (V)
of the photoelectric surface.
ution. (a) Minimum frequency
a
from
emission
wnich just causes photoelectric
(So)
photoelectric surface, is called threshold frequency
6) Refer to article 7.
Matter Waves
certain phenomena
exhibits particle aspects in
and absorption of
1. Wave-particle Duality :
Lght
erference, diffraction
eg, photoelectric effect, emission phenomena (e.g.,
other
wave aspects in
i o n while
and polarisation). That is, light
3.
As for same kinetic
energy electron has
of proton; so
Question
h
mass
less than that
wavelength
[11 [isC 2010]
What will be its value if kinetic energy of
where hh is Planck's
mass
P
constant,
h
mu
p is
ouestion exhibit wave nature.
particles
electrons exhibit
has
ofpartic-
[1] [ISC 20
which ma.
momentum
wave
nature.
2mE
h
longest de-Broglie wavelength; ele-
de-Broglie wavelength ^=~p
Expression
for
de-Broglie
e
wave
of
As mass of electron is least, de-Broglie waveler
electron is longest.
for same energy E
Answer.
proton, deuteron, a-particle?
velocity
by G.P Thom
nature is illustrated experimentally
their experiments.
Davisson and Germer through
with
Question 8. Which of the following moving
particles,
Answer. Diffraction is one phenomenon in which moThis
one
where
electron has greater
Plancks their velocity.
velocity.
de-Broglie
is
v
wavelength.
and
m is
de-Broglie wavelength of electrons of kinetic
phenomenon in
State any
Question 7. State
energy E is .
K=E
h
2mK
electrons is made 4E?
Solution. de-Broglie wavelength
Kis the kinetic energy.
Here,
So,
2/2mE 2
h
v2mE
When kinetic energy is 4E, then
de-Broglie
h
2mx 4E
(0,0)
P
[1]
what is the ratio of the de-
and
de-Broglie wavelength
Question 4. (i) What are matter waves?
accelerated by a potential of V volt.
Question
proton
9. Deduce
(ii) Show with the help of labelled graph how their
P
A
same
velocity;
m
ifor
for same
given by
Equation () gives =
P
muv
2mE
h
P 2mEx
2m
P
F-mm m
Since p mu -
where m is mass and v is
velocity of electron. If
kinetic energy of electron, then
h
wavelength associated with electrons of momen
[2] [ISC 2014] associated with accelerated electrons: The de-
wavelength () varies with their linear momentum (p)? Answer.
=
5.
Answer. () The waves associated with material particles
are called matter waves.
(ii)
Question
deuteron have
Broglie wavelengths?
mu
md2p-2
mp mp
Solution. de-Broglie wavelength a =
velocity v
Nd
VATSALISC HAND80OK ofPHYSICS Class Xll
94
h y p o t h e s i s ?
[ISC 2020,
de-Broglie
Question 2. An electron and a proton possess the
Question
6.
What
is
lie hypothesis
to
same kinetic energy, which of the two has greater
de-Broglie wavelength? Write down the formula used.
electrons,
etc.,
de-Broglie wavelength
1
=
All
such a s
particles
l I SC 2005]
Answer.
a-partic
protons,
According
Answer.
moving
aspect.
wave
and the
exhibit
may
wavelength
1ated
wave
behave
16
articles, ateria
atome
amatern
ce
de-Broglie
as
particlets
as
is given
with them
h
mu
Planck's
As for same kinetic energy electron has mass less than that
w h e r e h is
and
of proton; so electron has greater de-Broglie wavelength.
m is m a s s
v
is
constant,
p is
momentum
ofparti
ICes,
their velocity.
Question 3. de-Broglie wavelength of electrons of kinetic Question 7. State any one
phenomenon
in which
moving
[11 [ISC 2015
energy E is A. What will be its value if kinetic
energy Answer.
of particlesDiffraction
exhibit Wa is one phenomenon in which mouwave
[1] (ISC
electrons is made 4E?
Solution. de-Broglie wavelength
n=-
particles,
nature
h
velocity
h
Answer.
2mE
When kinetic energy is 4E, then de-Broglie wavelength
h
through
by
G.P
This
Thome
50n,
their experiments.
moving with
of the following
de-Broglie wavelength;
has longest
sa
electro
2/2mE
a="=de-Broglie wavelength
Vm
for same energy E
As
2m x 4E
nature.
proton, deuteron, a-particle?
K=E
So,
Germer
Which
Question 8.
K is the kinetic energy.
Here
and
wave
experimentaly
illustrated
is
Davisson
v2mK
exhibit
electrons
mass
of electron is least, de-Broglie
wavelength of
electron is longest.
Question 4. (i) What are matter waves?
(i) Show with the help of labelled graph how their
Question 9. Deduce de-Broglie wavelength of electrons
accelerated by a potential of V volt.
wavelength () varies with their linear momentum (p)? Answer. Expression for de-Broglie wavelength
[2]fISC 2014] associated with accelerated electrons: The de-Broglie
Answer. ) The waves associated with material particles
wavelength associated with electrons of momentump is
are called matter waves.
given by
where
Question
5.
deuteron
have
A
proton and
same
velocity;
is
mass
and
v
is
velocity of electron.
kinetic energy of electron, then
(0.0
what is the ratio of the de-
m
| Since p=mu >v=
[1]
de-Broglie wavelength a
If Ek is
the
P
Sroglie wavelengths?
olution.
m
...)
mu
P
=
elocity v
mu
for
m
2
same
P
2m
P 2m
Nd
Tmp
2-
p
Equation (i) gives A=
h
2mE
.(
acceleratin
voltis
fV
Ex=potential
eV
Kinetic enersy
Equation
(ii) gives
x
10
DUAL NATURE OF RADIATION AND
MATTER
electron, then
which
h
a=
2meV
Gubstitutin8 m=9.1
h6.62
of
x
J-s,
10S kg, e
we
angular
momentum of electron is
1.6
=
get
x
is the mass, v
velocity and r is radius
angular momentum of electron L mur.
m
10
C,
of orbit, then
According
Bohr's quantum condition
6.626.62x 10-34
2x9.1x10 x1.6x 101°v
12.27 x10
According
m
expression for
pciated with electron acceleratedde-Broglie
wavelength
to
potential of V
volt.
10. An electron and
photon have same energy
greater associated
wavelength?
00 eV. Which has
[2]
Solution. de-Broglie
wavelength
associated
with
electron
circumference of electron-orbit is integral
multiple of de-Broglie wavelength
associated with electron, i.e.,
2Ttr = n.
2mEe
2me
E Eph
=
Substituting this value in Eq. (i), we get
2tr=n mu mur=n =n
Question 12. Calculate the de-Broglie wavelength of a
neutron of kinetic energy 150 eV.
mass
(say)
E
=100
..ii)
mu
..()
..i)
2
=
eVN
.1)
ofneutron
Solution.
=
1.67x
10
kg
de-Broglie wavelength A =
Dividing Eq. (Gi) by Eq. G) and using Eq. (i), we get
he/ph
E
h/ 2m
Here,
27c
E=
or
2
wavelengths
This is Bohr's quantum condition.
hc
Ephph
Fig. Orbit with
n = 3 complete
..(i)
According to de-Broglie hypothesis
h
Also wavelength of photon of energy Eph is
Given
quantum
condition only those atomic orbits are
allowed as stationary orbits in which
This is required
he
..1)
27T
de-Broglie
to
to
h
mur= n
=227
Ouestion
integral
an
multiple of 27t
=
=
95
2
2
E 150 eV = 150 x 1.6 x10
[2]
2mEk
J
= 2.4 x 10l J.
ph
6.63 x 1034
= m
[2x1.67 x102x2.4x10-l7j
E
=2.342x 10m
ph
As
E= 100 eV, 2mc = 1MeV
E«2mc
Ihat is wavele
elength
e<^ph
associated
with
0.02342 Å.
Question 13. Give two important uses of photoelectric
photon is
as
greater
ompared to electron of same energ
hypothesis lead
11. How did de-Broglie
uestion
orbitst
condition of atomic
Bohr's quantum
quan
to
[2]
condition, "Only
to Bohr's quantum
orbits in
According
stationary
l0se atomic
orbi
are
allowed as
cells. The threshold wavelength for tungsten is 2400 A.
1600 Å.
it S illuminated with light of wavelength
nen
find:
wO
r k function
lunction
(i)
work
(i) maximum K.E. of emitted electrons
96
VATSAL ISC HANDBOOK of PHYSICS Class XI
(ii) stopping potential
(iv) if the metal surface of
tungsten is coated with oxide,
what will happen to the
threshold wavelength?
Explain.
[3]
Solution. (i)
hc
=
W
x
10*
3
x
108
x
0 . 8 2 x 10-10
= 2.4x1 0 - J
associated
with
wavelength
these
hese
electro
e
The de-Broglie
6.63 x 103 x 3 x 10
J
2400 x 10-10
max
6.63 x 1018 J
6.6
is
h
8
6.6 x 10-34
66.310-19
J
8
8.2875
=
x
10-
J
=
(i) Total energy of photon
6.63x 10*
x
1600 x 10-10
12.43125
x
101
x
10-15
three
particle is moving
times
as
a
[NCERT
66.3x3
10-19 J
16
=
2.4
3
J
16
x
as
th
de-Broglie
wavelength
of
the
of
ratio
an electron. The
electron is 1.813 x 10. Calcd ate
particle to that of the
the particle.
the particle's mass and identify
Question 15. A
3x 108
6.63 x3x 10
2 x 9.1x 10
0.1 .
10 m =
wavelength of
de-Broglie
having mass m and veloticy
Solution.
J
a
moving
particle
v:
Maximum K.E. of electrons
(12.43125 8.2875)
4.14375x 10 J
(ii) Since maximum K.E. of electron
=
4.14375 x 101
1.6 x 1019
x
10 J
For
eV
an
h
m
Mass,
electron,
mass
h
me
= 2.58984375 eV = 2.6 eV
Now as
eVo=2.6 eV
So,
Vo=2.6 V
per question
=
3 and
Ve
iv) Work function will decrease so threshold wavelength
1.813
will increase.
Question
14.
X-rays ofwavelength 0.82Å fall
on a
metal
plate. Find the smallest wavelength associated with the
emitted photoelectrons. The work function of the metal
Then,
mass of
the
x10
particle,
m
= me
is zero.
h
6.6 x 10" sec, C = 3 x 10° m/sec,
me =9.1
x
Solution.
10"
m
kg)
The
photoelectron is K
[3]
maximum
hc
=
.
K.E.
of
the
m
emitted
Hence, the
1.673
x
particle
104 kg.
=
(9.1 x
1.673
is neutron
10
1.813 x 10 (3
x107 kg
as mass
of neutron is a
RTANT FORMULAE FOR
DUAL NATURE
Ifis in A, then
E
where
-hv=he
of
energy of photon in
a
hvh
m2
eV is
ranging from
hv
certain maximum limit.
rm2
mmar
= eV
zero
where
V, stopping potential.
=
5. For photoelectric emission to take place, energy
of photon E2 Wor v2 vo or . s ho
c
E =hv-W
6. i) de-Broglie
Equation
wavelength
with
h
mu
h
muhe
1)
2
2mE
mu=ev,
2mqV
Work Function: W=hvo =
associated
moving particle
mv=h(v-vo)
3.
zero to a
(Emax
h
photon
Einstein's Photoelectric
2.
threshold
4. Photoelectrons emitted have kinetic energies
C
=
frequency, ho =
The maximum kinetic energy is
photon p
() Rest mass of a
photon
iv) Kinetic mass of a
threshold
vg
wavelength.
E= 12375 ev
A(in Aev
i) Momentum
97
NUMERICALS
1. Photon:
)Energy of a photon
OF RADIATION AND MATTER
h
i) For electrons, 12.27A
hc
V
EXERCISES
MCQs (MULTIPLE CHOICE QUESTIONS)
Choose the correct alternative from a, b,
c
and d Jor each
questionsgiven below
A proton and a deuteron are accelerated from rest, by
(a) h (Planck's constant)
(b) h/e
(c) elh
(d) hcle
3. Threshold wavelength for a metal having wo
à given potential difference. Which of the following
function W is A, the threshold wavelength for t
IS true with regards to their de-Broglie wavelength
metal with work function 4W is:
p and
(a)
(c) 4A
respectively?
(a)pD
(b) AphpD
(d) None of these.
graph
photoelectric experiment, the slope ofthe
and frequency of
Detween the stopping potential
In a
incident light gives
1] [ISC 2020]
4.
(b) /4
(d) /2
de-Broglie wavelength (2) associated with mov
particle
(a) does not depend upon accelerating potential
(b) Aoc V
98
VATSAL
ISC
HANDBOOK of PHYSICS Class X
(c) A
(d) 2.
p
5.
A
following
same
plate,
are incident on a metal
light rays does
photoelectric effect
not occur. It may
(c) infrared rays
incident
12.
drop
of
q is held in
radius
and
charge
equilibrium between the plates of a
charged parallel plate capacitor when the potential
is V. To keep a drop of radius 2r and
charge 2q in
equilibrium between the same plates, the potential
difference V required is
(a) 4V
(c) 2V
9.
The equation E
(a) both
r
[1]
(b) 8V
(d) V.
OT
photoelecton.
rons
light
(b) intensity of
surface
(c) nature of emitting
(d) all of the above.
[1]
(c) Bohr's theory
(d) quantum theory of light.
oil
enerEy
independentof:
incident light
(a) frequency of
(d) green light.
experiment an
(d) P-particle.
(c) a-particle
11. The maximum
[1]
Photoelectric effect can be explained by:
(a) corpuscular theory of light
(b) wave nature of light
In an
wihith
velocity
(a) Proton
by the incidence of
(a) X-rays
(b) radio wave
8.
waves
(b) Neutron
(d) Kl
Occur
7.
particles
(b) Kl/2
(c) K
6. When ultraviolet
moving
with t
of thehas
asuldled
as longest
longest waves associat
Which
10.
proportional to:
(a) -l/2
d)
(c)
proton is moving with kinetic energy K, the
de-Broglie wavelength associated with it, is directly
=pc is valid
electron and
for:
photon
(b) electron only
(c) photon only
(d) neither electron nor photon.
13. Which of the photon has minimum energy
(b) Ultraviolet light
(a) Red light
(d) X-rays.
(c) Infra-red light
14. Einstein's photoelectric equation is :[1] [lISC 2018
Which of the following graphs in figure represents
variation of de-Broglie wavelength (2) of a particle
having linear momentum (p)?
[11
(a) Emax h a - do
(b) Emax
c) Emax
(d) Emax
hv
+
o
hc
-
*0
15. The energy associated with
light of which of tis
following colours is minimum?
ISC 2019
P
(a) Violet
(c) Green
P
(a)
(b)
(6) Red
(d) Yellow.
ANSWERSs
1.
11.
(a)
(b)
2.
12.
(b)
(c)
3.
13.
(b)
(c)
4.
(d)
5.
14.
(b)
15.
(a)
(b)
6.
(a)
7.
(d)
8.
(a)
9.
(d)
10.
VERY SHORT ANSWER QUESTIONS
reference to photoelectric effect, what is meant
[1] USC 2014]
by threshold wavelength?
what is meant
2. With reference to photoelectric effect,
1.
With
by threshold frequency?
[1
3. For
photoemission should the
radiation greater than or wavelength of in
less than the
idet
reshol"
wavelength.
thre