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MODERN PHYSICS - 1
1

PHOTOELECTRIC EFFECT :
When electromagnetic radiations of suitable wavelength are incident on a metallic surface then electrons are
emitted, this phenomenon is called photo electric effect
...
1

Photoelectron : The electron emitted in photoelectric effect is called photoelectron
...
2

Photoelectric current : If current passes through the circuit in photoelectric effect then the current
is called photoelectric current
...
3

Work function : The minimum energy required to make an electron free from the metal is called
work function
...
It is the minimum for Cesium
...


Work functions of some photosensitive metals
Metal
Cesium
Potassium
Sodium
Lithium

Work function
(ev)
1
...
2
2
...
5

Work function
(eV)
3
...
5
4
...
6

Metal
Calcium
Copper
Silver
Platinum

To produce photo electric effect only metal and light is necessary but for observing it, the circuit is completed
...




intensity
frequency

1

A

2

C

A

V

Rheostat

cell, few volts

Here the plate (1) is called emitter or cathode and other plate (2) is called collector or anode
...
4

Saturation current : When all the photo electrons emitted by cathode reach the anode then current
flowing in the circuit at that instant is known as saturated current, this is the maximum value of
photoelectric current
...
5

Stopping potential : Minimum magnitude of negative potential of anode with respect to cathode for
which current is zero is called stopping potential
...
This voltage
is independent of intensity
...
6

Retarding potential : Negative potential of anode with respect to cathode which is less than
stopping potential is called retarding potential
...
1

Photocurrent 

OBSERVATIONS : (MADE BY EINSTEIN)
A graph between intensity of light and photoelectric current is found
to be a straight line as shown in figure
...
In this
experiment the frequency and retarding potential are kept constant
...
2

A graph between photoelectric current and potential diffrence between cathode and anode is found
as shown in figure
...

Maximum kinetic energy of photoelectron on the cathode = eVs
KEmax = eVs
Whenever photoelectric effect takes place, electrons are ejected
out with kinetic energies ranging from
0 to K
...
e
...


No
...


O

Kinetic energy eVS

2
...
Stopping potential is same, so maximum value of kinetic
energy is not effected
...
4

If light of different frequencies is used then obtained plots are shown in figure
...


2
...



th1







th2 th3





It is clear from graph that there is a minimum frequency of electromagnetic radiation which can
produce photoelectric effect, which is called threshold frequency
...

  th for photo electric effect
Maximum wavelength for photoelectric effect
max = th
...

We have
Kmax = A + B
When
 = th , Kmax = 0
and
B = – Ath
Hence
[Kmax = A( – th)]
and
A = tan  = 6
...

2
...


(from experimental data)

It is also observed that photoelectric effect is an instantaneous process
...


THREE MAJOR FEATURES OF THE PHOTOELECTRIC EFFECT CANNOT BE EXPLAINED
IN TERMS OF THE CLASSICAL WAVE THEORY OF LIGHT
...


c t

Consider a cylindrical volume with area of crosssection A and length c t along the X-axis
...

The energy contained
...
t)A
U
= uav c
...

2 0 0
If we consider light as a wave then the intensity depends upon electric field
...
A
...

But it is observed that photoelectric effect is an instantaneous process
...

then

t=

A

x

3
...
Since the force applied
to the electron is eE, this suggests that the kinetic energy of the photoelectrons should also increased as the light beam is made more intense
...


3
...
However observations shows that there exists for each surface a
characterstic cutoff frequency th, for frequencies less than th, the photoelectric effect does not
occur, no matter how intense is light beam
...
3

The time delay problem : If the energy acquired by a photoelectron is absorbed directly from the
wave incident on the metal plate, the “effective target area” for an electron in the metal is limited and
probably not much more than that of a circle of diameter roughly equal to that of an atom
...
Thus, if the light is feeble
enough, there should be a measurable time lag, between the impinging of the light on the surface
and the ejection of the photoelectron
...
However, no detectable time lag has ever
been measured
...


4

PLANCK’S QUANTUM THEORY :
The light energy from any source is always an integral multiple of a smaller energy value called quantum of
light
...

Here energy is quantized
...

hc
E = h =

and
hc = 12400 eV Å

5
...
The energy
E of a single photon is given by
E = h,
If we apply Einstein’s photon concept to the photoelectric effect, we can write
h = W + Kmax,
(energy conservation)
Equation says that a single photon carries an energy h into the surface where it is absorbed by a single
electron
...
The excess energy (h – W) becomes the electron’s kinetic energy; if the
electron does not lose energy by internal collisions as it escapes from the metal, it will still have this much
kinetic energy after it emerges
...
There is complete agreement of the photon theory with experiment
...
of photons incident per unit time on an area ‘A’ when light of
h
intensity ‘’ is incident normally
...

The second objection (the frequency problem) is met if Kmax equals zero, we have
hth = W,
Which asserts that the photon has just enough energy to eject the photoelectrons and none extra to appear
as kinetic energy
...

The third objection (the time delay problem) follows from the photon theory because the required energy is
supplied in a concentrated bundle
...


Hence Einstein’s equation for photoelectric effect is given by
h = hth + Kmax
Ex
...


Kmax =

hc
hc
– 

th

In an experiment on photo electric emission, following observations were made;
(i)
Wavelength of the incident light = 1
...
5 volt
...

(b) Work function and
(c) Threshold frequency;
(a)
Since
vs = 2
...
26 eV
1980

Kmax = 2
...
76 eV



hth = W = 3
...
6 × 10–19 J

th =

3
...
6  10 19
6
...
1 1014 Hz

Ex
...
5 × 10–3 Wm–2
...
9
eV
...
e
...


Sol
...
1 eV,
,
4000

E2 =

12400
= 2
...
06 eV
6000

12400
= 1
...


Number of photoelectrons emitted per second = No
...
5  10 3 )(10  4 )

1
...
12 × 1012

Ex
...


19

 1
1
1 
= A  E  E  E 


2
3 
 1
1
1 
 1




 3
...
58 2
...


A small potassium foil is placed (perpendicular to the direciton of incidence of light) a distance r (= 0
...
0W
...
8 eV) from the beam to eject an electron? Assume that the
ejected photoelectron collected its energy from a circular area of the foil whose radius equals the radius of a
potassium atom (1
...

If the source radiates uniformly in all directions, the intensity  of the light at a distance r is given by
=

1
...
32 W/m2
...
5 m)2
4r
The target area A is (1
...
3 × 10–20 m2, so that the rate at which energy falls on the target is
given by
P = A = (0
...
3 × 10–20 m2)
2

=

= 1
...

If all this incoming energy is absorbed, the time required to accumulate enough energy for the electron to
escape is

 1
...

1eV


Our selection of a radius for the effective target area was some-what arbitrary, but no matter what reasonable
area we choose, we should still calculate a “soak-up time” within the range of easy measurement
...

1
...
7  10  20 J / s 

Ex
...


A metallic surface is irradiated with monochromatic light of variable wavelength
...
With an unknown wavelength, stopping potential is 3 V
...

Using equation of photoelectric effect
Kmax = E – W
(Kmax = eVs)
12400
12400
–

5000
 = 2262 Å



3 eV =

or
Ex
...


=

12400
– 2
...
35 m and
2 = 0
...

Find the work function of that metal
...
(i)

1
hc
mv 2 
W
2
2
2


...
(i) with Eq
...
64 eV

  
5400
3500
2 
 1

W=

5
...
88 eV

Ans
...
6

A photocell is operating in saturation mode with a photocurrent 4
...
When another monochromatic radiation of wavelength 1650
Å and power 5 mW is incident, it is observed that maximum velocity of photoelectron increases to two times
...
[(No
...


(a)

K1 =

Since

12400
–W
= 7
...
13 – W


...
(iiii)

12400
= 4133 Å
3

Energy of a photon in first case =

or

0 =


...


12400
= 4
...
6 × 10–19 J
Rate of incident photons (number of photons per second)

P1
10 3
= E =
= 1
...
6  10 19
1
Number of electrons ejected

=

4
...
6  10 19

per second

= 3
...
0  1013
1
...


Energy of photon in second case
12400
= 7
...
0  10 3
n2 = E =
= 4
...
7 × 1015
100
= 9
...
4 × 1013) (1
...


Number of electrons emitted per second =

Ex
...


Light described at a place by the equation E = (100 V/m) [sin (5 × 1015 s–1) t + sin (8 × 1015 s–1)t] falls on a
metal surface having work function 2
...
Calculate the maximum kinetic energy of the photoelectrons
...
The one with larger frequency will cause photoelectrons with
largest kinetic energy
...
14 × 10

 8  1015 1 


eV-s) ×  2 s  – 2
...
27 eV – 2
...
27 eV
...
All photons of light of a particular
wavelength  have the same energy E = hc/ and the same magnitude of momentum p = h/
...
Assume absorption and reflection
coefficient of surface be ‘a’ and ‘r’ and assuming no transmission
...

For calculating the force exerted by beam on surface, we consider following cases
...
of photons incident per unit time


=

A  h

hc 
A
=
(upward)
c
force on photons = total change in momentum per unit time
A
=
(upward)
c
A

force on plate due to photons(F) =
(downward)
c

=

pressure =
Case : (II)
when

A

F
=
=
cA
c
A

r = 1, a = 0

intial momentum of the photon
final momentum of photon

h

h
=


=

(downward)
(upward)

h
h
2h
+
=



 energy incident per unit time = A
A
no
...
P

change in momentum

=

=

A
2h
2A

...
of photons incident per unit time

=

A
hc

No
...
r
hc

No
...

hc


Force due to reflected photon (Fr)

A
2h
2 A 

...
9

Sol
...
0 × 10 8 × 10–1

Calculate force exerted by light beam if light is incident on surface at an
angle  as shown in figure
...

Case - I

a = 1,



r=0

initial momentum of photon (in downward direction at an angle  with vertical) =

h


[

]

final momentum of photon = 0
change in momentum (in upward direction at an angle  with vertical) =

h


[

 ]

energy incident per unit time = A cos 
Intensity = power per unit normal area
=

P
A cos 

P = A cos 

A cos 

...
of photons incident per unit time =

A cos 
...

=
hc

c
Force (F) = total change in momentum per unit time

=

A cos 
 on photon and
(direction
c
Pressure = normal force per unit Area

F=



Sol
...

Calculate power of beam
...

Since plate is in air, so gravitational force will act on this
Fgravitational = mg
(downward)
= 10 × 10–3 × 10
= 10–1 N
for equilibrium force exerted by light beam should be equal to Fgravitational
Fphoton = Fgravitational
Let power of light beam be P



Ex
...
of photons incident per unit time

=

h
cos 


h cos 


(upward)
h sin 






h sin 


energy incident per unit time
h

=

A cos   
hc
total change in momentum per unit time

=



A cos   
2h
×
cos 
hc


=


force on the plate =

Pressure
Case III

=

(upward)

2A cos 2 
(downward)
c

2A cos 2 
cA

=

2A cos 2 
c

0 < r < 1,

P =

2 cos 2 
c

a+r=1

change in momentum of photon when it is reflected =

2h
cos 


change in momentum of photon when it is absorbed =

h
(in the opposite direction of incident beam)


(downward)

energy incident per unit time = A cos 
no
...
of reflected photon (nr) =

A cos   
hc

A cos 
...

(1 – r)
hc
force on plate due to absorbed photons Fa = na
...
of absorbed photon (na) =

A cos 
...
10 A perfectly reflecting solid sphere of radius r is kept in the path of a parallel beam of light of large aperture
...

Sol
...
Consider a radius OP of the sphere
Q
making an angle  with OZ
...
Change  to  + d and rotate the radius
o 
about OZ to get another circle on the sphere
...
Consider a
small part A of this ring at P
...
The light is reflected by the sphere along PR
...

(direction along PO )
t
c
The resultant force on the ring as well as on the sphere is along ZO by symmetry
...

t
c

The force acting on the ring is

dF =

2
(2r2 sin d)cos3 
...


7
...

hc

By Einstein’s energy mass relation, E = mc2 the equivalent mass m of the photon is given by,
E
h
h
m 2  2 

...
(ii)
p
mc
Here p is the momentum of photon
...
(iii)
mv p
where p is the momentum of the particle
...
Combining all
these relations Eq
...
(iv)

7
...
(iv), we get a simple formula for calculating de-Broglie
wavelength of an electron
...
(v)
V(in volts )
de-Broglie wavelength of a gas molecule :
Let us consider a gas molecule at absolute temperature T
...
2

3
kT ; k = Boltzman constant
2
h
gas molecule =
3mkT
K
...
=



Ex
...
Find the de-Broglie wavelength associated with
it
...

For an electron, de-Broglie wavelength is given by,

150
=
V
= 1
...


Ex
...

Sol
...


 H2
 He

=

mHe THe
mH2 TH2 =

4 (127  273)

...
J
...
Thomson’s model called the ‘plum pudding’ model is illustrated in figure
...
His idea was taken seriously
...


9
...
The electrons were supposed to move in circular orbits round the
nucleus (like planets round the sun)
...


mv 2 kZe 2
 2
r
r
and total energy = potential energy + kinetic energy
 kZe 2
=
2r
Rutherford’s model of the atom, although strongly supported by evidence for the nucleus, is inconsistent with
classical physics
...
1

9
...


Regarding stability of atom : An electron moving in a circular orbit round a nucleus is accelerating
and according to electromagnetic theory it should therefore, emit radiation continuously and thereby
lose energy
...
If this
happened the radius of the orbit would decrease and the electron would spiral into the nucleus in a
fraction of second
...
In 1913 an effort was made by Neil Bohr to overcome
this paradox
...

As a result, electrons will radiate electromagnetic waves of all frequencies, i
...
, the spectrum of
these waves will be ‘continuous’ in nature
...
Instead they are line spectra
...
Niel Bohr removed the difficulties of Rutherford’s atomic model by the application of Planck’s
quantum theory
...
In stationary orbits electron does not emit radiation, contrary to the predictions of classical electromagnetic theory
...
If the energy of electron
in the higher orbit be E2 and that in the lower orbit be E1, then the frequency  of the radiated waves
is given by
h = E2 – E1

E 2  E1
h
Bohr found that the magnitude of the electron’s angular momentum is quantized, and this magnitude
h
for the electron must be integral multiple of

...
So, according to Bohr’s
postulate,
or

(3)

=

nh
(n = 1, 2, 3
...
The
value of n for each orbit is called principal quantum number for the orbit
...
(ii)

According to Newton’s second law a radially inward centripetal force of magnitude F =

mv 2
is
rn

needed by the electron which is being provided by the electrical attraction between the positive
proton and the negative electron
...
(ii) and (iii), we get

Thus,

rn 


...
(iv)
me 2
e2
vn 
and

...
We’ll denote this minimum radius, called the Bohr
radius as a0
...
529 × 10–10 m = 0
...
(vi)

Eq
...
(vii)

rn  n2

Similarly, substituting values of e, 0 and h with n = 1 in Eq
...
19 × 106 m/s


...
Which is approximately
equal to c/137 where c is the speed of light in vacuum
...
(v), in terms of v1 can be written as,
vn =

v1
n

vn 

or

1
n

Energy levels : Kinetic and potential energies Kn and Un in nth orbit are given by
me 4
1
mvn2 =
2
8 0 n 2h 2
2

Kn =

and

Un = –

1
4 0

e2
rn

=–

me 4
2

4 0 n 2h 2

(assuming infinity as a zero potential energy level)
The total energy En is the sum of the kinetic and potential energies
...
6 eV
...
6 eV

...
6

eV

...
, etc
...

2

E2 = – 3
...
51 eV,
...
1

Hydrogen Like Atoms
The Bohr model of hydrogen can be extended to hydrogen like atoms, i
...
, one electron atoms, the
nuclear charge is +ze, where z is the atomic number, equal to the number of protons in the nucleus
...
Thus, the equations for, rn, vn
and En are altered as under:
rn =

 0 n 2h 2
2

=

n2
a
z 0

where

nmze
a0 = 0
...
19 × 106 m/s
vn =

mz 2 e 4
En = –
where

2
8 0 n 2 h 2

E1 = –13
...
(i)


...
(iii)

10
...

Ground state energy of H atom = –13
...
4 eV
Ground state energy of Li++ Ion = –122
...

n=2
first excited state
n=3
second excited state
n=4
third excited state
n = n0 + 1
n0th excited state

(3)

Ionisation energy (E
...
6 eV
Ionisation energy of He+ Ion = 54
...
4 eV

(4)

Ionisation potential (I
...
) : Potential difference through which a free electron must be accelerated
from rest such that its kinetic energy becomes equal to ionisation energy of the atom is called
ionisation potential of the atom
...
P of H atom = 13
...
P
...
4 V

(5)

Excitation energy : Energy required to move an electron from ground state of the atom to any other
exited state of the atom is called excitation energy of that state
...
6 eV
Energy in first excited state of H-atom = –3
...
2 eV
...

st excitation energy = 10
...

st excitation potential = 10
...

Binding energy or Seperation energy : Energy required to move an electron from any state to n
=  is called binding energy of that state
...

Binding energy of ground state of H-atom = 13
...
13 First excitation potential of a hypothetical hydrogen like atom is 15 volt
...

Sol
...
6 Z2 eV
and
En = 2
n
E0
n = 2, E2 =
4
E0
given
– E0 = 15
4
3E 0
–
= 15
4
E0
for
n = 4,
E4 =
16
E0
third exicitation energy =
– E0
16
15
15   4  15 

=–
E = 

16 0
16 
3

75
=
eV
4
75

third excitation potential is
V
4
10
...
It is
excited to some higher energy state when it acquires some energy from external source
...


A photon corresponding to a particular spectrum line is emitted when an atom makes a transition
from a state in an excited level to a state in a lower excited level or the ground level
...

On Screen :
A photograph of spectral lines of the Lymen, Balmer, Paschen series of atomic hydrogen
...
represents the I, II & III line of Lymen, Balmer, Paschen series
...
6

10
...
6

12
...
78 eV

series limit

1

 (series limit)

91
...
6 eV

I

2

3

656
...
89 eV

II

2

4

486
...
55 eV

III

2

5

434
...
86 eV

series limit

2

 (series limit)

364
...
41 eV

I

3

4

1875
...
66 eV

II

3

5

1281
...
97 eV

III

3

6

1093
...
13 eV

series limit

3

 (series limit)

822

1
...

n =7
n =6
n =5
n =4

Lymen
series

Brackett series

n =1

Paschen
series

–3
...

10
...
28eV
–0
...
54eV
–0
...
51eV

n =3
n =2

Pfund
series

–13
...
If Ei is the
initial energy of the atom before such a transition, Ef is its final energy after the transition, and the
hc
photon’s energy is h =
, then conservation of energy gives,

hc
h =
= Ei – Ef
(energy of emitted photon)

...
The visible line with longest wavelength, or lowest frequency is in the red and is called H, the next line, in the blue-green is called H
and so on
...

This is now called the Balmer series
...
(ii)

n 
2
Here, n = 3, 4, 5
...

R = Rydberg constant = 1
...

Similarly, for n = 4, we obtain the wavelength of H line
...
Using the relation, E =

hc
we can find the photon energies correspond

ing to the wavelength of the Balmar series
...


...
) can be represented by formula similar to Balmer’s formula
...

Lymen Series :

1
n 

Paschen Series :

1
1 
 1
 R  2  2  , n = 4, 5, 6
...


n 
4
1
1
 1
 R 2  2

n
5


 , n = 6, 7, 8

The Lymen series is in the ultraviolet, and the Paschen
...


Pfund Series :

Ex
...

Sol
...
The corresponding
wavelength for H line is,
1
1 
 1
 (1
...
2056 × 107

 = 4
...

3
...
9  10 7
= 6
...


Ex
...
In what region of the electromagnetic spectrum does each series lie?
Sol
...


n 
 (1)


for largest wavelength, n = 2
1
1 1 
 1
...
823 × 107
 max
1 4 

max = 1
...

The shortest wavelength corresponds to n = 
1
1 1 
 1
...
911 × 10 m = 911 Å
Ans
...

Ex
...

From the nth state, the atom may go to (n – 1)th state,
...
So there are
(n – 1) possible transitions starting from the nth state
...
Similarly for other lower states
...
2 + 1
=

n(n  1)
2

Ex
...


(b)
(a)

Find the wavelength of the radiation required to excite the electron in Li++ from the first to the third
Bohr orbit
...
6 eV is the energy of a hydrogen atom in
ground state thus for Li++,
E1 = 9E0 = 9 × (– 13
...
4 eV
The energy in the third orbit is
E
 1 = – 13
...
6 eV = 108
...

Energy required to excite Li++ from the first orbit to the third orbit is given by
E3 – E1 = 8 × 13
...
8 eV
...
4 nm
108
...
Thus, there will be three spectral lines in the spectrum
...
18 Find the kinetic energy potential energy and total energy in first and second orbit of hydrogen atom if
potential energy in first orbit is taken to be zero
...

E1 = – 13
...
60 eV
U1 = 2E1 = –27
...
40 eV

K2 = 3
...
80 eV

Now U1 = 0, i
...
, potential energy has been increased by 27
...
So values of kinetic energy, potential energy and total energy in first orbit are 13
...
60
respectively and for second orbit these values are 3
...
40 eV and 23
...


Ex
...
Two electrons move
close to the nucleus making up a spherical cloud around it and the third moves outside this cloud in a circular
orbit
...

Calculate the ionization energy of lithium in ground state using the above picture
...

In this picture, the third electron moves in the field of a total charge + 3e – 2e = + e
...
The lowest energy is :

E1
13
...
4 eV
4
4
Thus, the ionization energy of the atom in this picture is 3
...

E2 =

Ex
...

(a)
Find the ionization potential of this atom
...

(d)
Find wave number of the photon emitted
for the transition n = 3 to n = 1
...
80 eV
– 1
...
08 eV

n=2

– 5
...
3 eV

n=1

– 15
...


(a)

Ionization potential = 15
...
3
E31 = – 3
...
6) = 12
...
52 volt
...
52 –1
=
Å–1 =
Å
 31
12400
12400
 1
...
3 eV > 6 eV
...
So, Kmin = 6 eV
...
3 eV < 11 eV
...

So, Kmin = (11 – 10
...
7 eV
...
21 A small particle of mass m moves in such a way that the potential energy U = ar2 where a is a constant and
r is the distance of the particle from the origin
...

Sol
...
The necessary centripetal force is provided by the above force
...
(i) and (ii) for r, we get

mvr =

1/ 4

 n 2h 2 

r 
 8am 2 



Ans
...
22 An imaginary particle has a charge equal to that of an electron and mass 100 times the mass of the electron
...
Take the mass of the nucleus to be infinite
...

(a)
Derive and expression for the radius of nth Bohr orbit
...

Sol
...
(i)

The quantization of angular momentum gives,
mp vrn =

nh
2


...
(i) and (ii), we get
r=

n 2h 2  0
zmp e 2

Substituting
mp = 100 m
where m = mass of electron and z = 4
we get,
(b)

rn =

n 2h 2  0
400 me 2

Ans
...
60 eV
and

 z2
En   2
n



m



For the given particle,

E4 =

and

E2 =

(13
...
60) ( 4)2
( 2) 2

× 100 = –1360 eV

× 100 = – 5440 eV

DE = E4 – E2 = 4080 eV


 (in Å) =

12400
= 3
...


Ex
...
It moves in a circular orbit around a nucleus of charge +3e
...
Assuming that the Bohr’s model is applicable to this system, (a) derive an expression for the radius
of the nth Bohr orbit, (b) find the value of n for which the radius of the orbit is approximately the same as that
of the first Bohr orbit for a hydrogen atom and (c) find the wavelength of the radiation emitted when the –
meson jumps from the third orbit to the first orbit
...

(a) We have,

mv 2
Ze 2

r
4 0 r 2
or,

v 2r 

Ze 2
4 0m
nh
2m

The quantization rule is vr =

The radius is r =

( vr )2

=

2

v r
=


...
(ii)

Zme 2
For the given system, Z = 3 and m = 208 me
...

 is the ground state energy of hydrogen atom and hence is equal to – 13
...



From (iii), En = –

1872
n

2

× 13
...
2 eV and E3 =

 25459
...
8 eV
...
4eV
...

9

The wavelength emitted is
=

1240 eV  nm
hc
= 22630
...

E

Ex
...
Consequently, the atoms emit radiations of
only three different wavelength
...

(a)
Determine the initial state of the gas atoms
...

(c)
Find the minimum wavelength of the emitted radiations
...

Sol
...
e
...

Hence nf = 3
...

Hence the emitted wavelength is either equal to, less than or greater than the absorbed wavelength
...

If
ni = 2,
then Ee  Ea
...
6) (Z2)  4  9  = 68


Z=6
12400
E 3  E1

12400
1

(13
...
2

(c)

min =

(d)

Ionization energy = (13
...
6 eV

=

=

= 28
...


Ans
...
33 Å
Ans
...
6
Ex
...
Assuming that Bohr’s postulate regarding the quantisation of angular momentum holds good for this
electron, find
(a)
the allowed values of the radius ‘r’ of the orbit
...

(d)
The total energy of the allowed energy levels
...

(a)
radius of circular path

=

r=

mv
Be


...
(ii)

and v =

nhBe
2m 2

nhBe
1
mv2 =
4m
2

(b)

K=

(c)

e
evr
M = iA =  T  (r2) =
 
2

e
2

=
Now potential energy

11
...
B
=

(d)

Ans
...
Let r1 and r2 be the distance of CM
from nucleus and electron
...

Let r be the distance between the nucleus and the electron
...
So,


r1 =

1
Ze 2
mr22 = 4
0
r2
or

1
 Mr 
Ze 2
 2 =
m 

...
529 Å)

n2 m

Z μ

M

r2

r1
CM

m

 Z 2e 4

 Ze 2
U=
40r

U=

2
4 0 n 2h 2

1 2 1 2 2
1
 =
r 
and
K=
v2
2
2
2
v-speed of electron with respect to nucleus
...
6 eV)

Z2  μ 
 
n2  m 

The expression for En without considering the motion of proton is En = –
while considering the motion of nucleus
...
e
...
26 A positronium ‘atom’ is a system that consists of a positron and an electron that orbit each other
...

Sol
...
We know that
=


Hence

12
...
 Their difference will also be halved
...


Ex
...


What will be the type of collision, if K = 14eV, 20
...
18 eV
(elastic/inelastic/perectly inelastic)
Loss in energy (E) during the collision will be used to excite the atom or electron from one level to another
...

E = {0, 10
...
09 eV,
...
6 eV)
According to Newtonion mechanics
minimum loss = 0
...
E
...
2eV, 12
...

If
K = 20
...
2 eV]
According to quantum mechanics
loss = {0, 10
...
09eV,
...

loss = 10
...
2eV, 12
...
}
loss = 0
elastic collision
loss = 10
...
18 eV
According to classical mechanics E =[0, 12
...
2eV, 12
...
13
...
2 eV inelastic collision
loss = 12
...
28 A He+ ion is at rest and is in ground state
...
Find minimum value of K so that there can be an inelastic collision between these two particle
...

Here the loss during the collision can only be used to excite the atoms or electrons
...
8eV, 48
...
, 54
...
(1)

Z2

m
eV
n2
n
K
Now according to newtonion mechanics
Minimum loss = 0
maximum loss will be for perfectly inelastic collision
...


En = – 13
...
E
...
(5m )
...
( mv 2 ) =
0
5 2
5

4m
He

+

maximum loss = K –

K
4K
=
5
5

 4K 
so loss will be 0,

...
8 eV
5
K > 51 eV
minimum value of K = 51 eV
...
29 A moving hydrogen atom makes a head on collision with a stationary hydrogen atom
...
What is the minimum value of the kinetic
energy of the moving hydrogen atom, such that one of the atoms reaches one of the excitation state
...

Let K be the kinetic energy of the moving hydrogen atom and K’, the kinetic energy of combined mass after
collision
...
(i)

...
2 eV

K
2
Now minimum value of E for hydrogen atom is 10
...

or
E  10
...
(i) and (ii), we get

E =

K
 10
...
4 eV
Therefore, the minimum kinetic energy of moving hydrogen is 20
...


Ex
...

Find the minimum kinetic energy of the neutron for which inelastic (completely or partially) collision may take
place
...
67 × 10–27 kg
...

Suppose the neutron and the hydrogen atom move at speed v1 and v2 after the collision
...
Suppose an energy E is used in this way
...

mv = mv1 + mv2

...
(ii)
2
2
2
2
2
2
From (i),
v = v1 + v2 + 2v1v2 ,
From (ii),

v2 = v12 + v22 +

Thus,

2v1v2 =

2E
m

2E
m

Hence, (v1 – v2)2 – 4v1v2 = v2 –

4E
m

As v1 – v2 must be real, v2 –

4E
0
m

1
mv2 > 2E
...
2 eV
...
2 eV  20
...
31 How many head-on, elastic collisions must a neutron have with deuterium nucleus to reduce its energy from
1 MeV to 0
...

Sol
...

Using C
...
L
...
along direction of motion
2mK 0 =

2mK 1 +

4mK 2

velocity of seperation = velocity of approach

4mK 2

2mK 1

–

2m
m
Solving equaiton (i) and (ii) we get

2mK 0

=

m

K0
9
Loss in kinetic eneryg after first collision
K1 = K0 – K1

K1 =

K1 =

8
K
9 0


...

9 1 9 9

 Total energy loss
K = K1 + K2 +
...
+ n K 0
9
9
9

K =

As,

1
8
1
K0 (1 +
+
...
025) eV

K0 = 106 eV,

Here,

1



n

=

K 0  K
K0

9
Taking log both sides and solving, we get
n=8

=

0
...
32 A neutron with an energy of 4
...
Assuming that upon each collision
neutron is deflected by 45º find the number of collisions which will reduce its energy to 0
...

Sol
...
(i)

...
(iii)

From conservation of energy
K2 = K0– K1
Solving equations (iii) and (iv), we get


...
e
...
Therefore, after n collisions,
K1 =

 1
Kn = K0  
2

n

 1

0
...
6 × 10 )  
2
Taking log and solving, we get
n  24

n

2n 

6

12
...
6  10 6
0
...


Calculation of recoil speed of atom on emission of a photon
momentum of photon = mc =

h


fixed
H-atom in first excited state
hc
=10
...
(i)

According to energy conservation
1
hc
m 2 
= 10
...
2 eV
'

m =

10
...


10
...
2
cm

10
...
The wavelength of x-rays is found between 0
...
These rays are
invisible to eye
...

Its photons have energy around 1000 times more than the visible light
...


13
...
When voltage
is applied across the filament then filament on being heated emits electrons from it
...
Now when electron strikes the target then x-rays
are produced
...
Since, target
should not melt or it can absorb heat so that the melting point, specific heat of target should be high
...

For more energetic electron, accelerating voltage is increased
...
of photons voltage across filament is increased
...
2

Variation of Intensity of x-rays with  is plotted as shown in figure :

1
...

eV
V(involts)

We see that cutoff wavelength min depends only on accelerating voltage applied between target and
filament
...
33 An X-ray tube operates at 20 kV
...
Find the wavelength corresponding to this photon
...

Kinetic energy acquired by the electron is
K = eV = 20 × 103 eV
...
05 × 20 = 103 eV = 103 eV
...
14  10 15 eV  s)  (3  10 8 m / s)
3

10 eV

=

1242 eV  nm
103 eV

 1
...


Charactristic X-rays
The sharp peaks obtained in graph are known as
characteristic x-rays because they are characteristic of
target material
...
= charecteristic wavelength of
material having atomic number Z are called characteristic x-rays and the spectrum obtained is called
characteristic spectrum
...

Characteristic x-ray emission occurs when an energetic
electron collides with target and remove an inner shell
electron from atom, the vacancy created in the shell is filled
when an electron from higher level drops into it
...
If vaccany in K-shell is filled
by an electron from M-shell, K line is produced and so on
similarly
L, L,
...

n=5
n=4

N
K

n=3
K

n=2
K

L L L

M M


V, Z



min 1

2

3

4




V, Z' < Z

V, Z

min

1 ´12 ´2 3 ´3 4

´4



O

N
M

L
x-rays

n=1

K

Ex
...


1

l

2



hc
hc
,
=

E
since energy difference of K is less than K
Ek < Ek
k < k
1 is K and 2 is K

E =

1

Ex
...


MOSELEY’S LAW :
Moseley measured the frequencies of characteristic x-rays for a large number of elements and plotted the
sqaure root of frequency against position number in periodic table
...


1, 1',1'',1'''



Z1

l 1'

l 2'

Z3

l 1"

l 2''

Z4

Z

l

Z2

2, 2',2'',2'''

l 1"'

l 2'''

1

l

2

Wavelength of charactristic wavelengths
...

Moseley’s Law can be derived on the basis of Bohr’s theory of atom, frequency of x-rays is given by

 1
1 
CR  2  2 
...

2 
 1
b  known as screening constant or shielding effect, and (Z – b) is effective nuclear charge
...
36



K
Z1
Z2
1

2



Find in Z1 and Z2 which one is greater
...


 1
1
cR  2  2 
...



 

Ex
...

A second, fainter, characteristic spectrum is also found because of an impurity in the target
...
9 pm (cobalt) and 143
...
What is the impurity?
Sol
...
9 pm
Zx  1
143
...

Solving for the unknown, we find Zx = 30
...

Ex
...


Element

Wavelength of K X-ray

Mo
Co
Sol
...
5 pm

Moseley’s equation is

v  a( Z  b)
Thus,

c
 a( Z1  b)
1


...
(ii)

From (i) and (ii)

 1
1 
  a (Z  Z )
c

1
2
 
2 
 1

 1
c
1 



a = (Z  Z ) 
2 
1
2  1


or,


1
1
(3  10 8 m / s)1/ 2 


12
1/ 2
12
1/ 2 
(178
...
0 × 10 (Hz)
Dividing (i) by (ii),
=

2
Z b
 1
1
Z2  b
178
...
37

or,
or,

Problem 1
...
0 MeV photon
...


Monochromatic light of wavelength 3000 Å is incident nornally on a surface of area 4 cm2
...

Rate at which photons strike the surface

Solution :

E
= 12 MeV/c
...
05 × 1013 photon/s
...
63  10 19 J / photon

Problem 3
...
0 × 10–19 J when light of wavelength 3000
Å falls on a surface
...
0 × 10–19 J ×

1eV

Problem 4
...
5 eV
...

1
...
5 V
...
5 eV =

12
...
Å 12
...
Å

3000 Å
 th
th = 7590 Å
...


Find the de Broglie wavelength of a 0
...


Solution :

 = h/p =

Problem 6
...


Solution :

V=

Problem 7
...


Solution
...
097 × 10–3 Å–1)  2  2 

5 
3

6
...
s
= 6
...

0
...


...


Problem 8
...

10 photons
...


An electron rotates in a circle around a nucleus with positive charge Ze
...
e
r

v=

Problem 10
...
rm

Ans
...

4 0
...

(ii) Find the wavelength of the Ha line (3  2 transition) of positronium
...
4
...

2
2
16 0 n 2h 2

= – 6
...
8 ev
E2 = – 6
...
8 ×

1
22
1
32

eV = – 1
...
76 eV

E (3  2) = E3 – E2 = – 0
...
70) eV
= 0
...


1
...
94

(i) –6
...
7 eV , –0
...


Problem 11
...
It collides head on with a He+ ion in
ground state kept at rest but free to move
...


Solution :

Energy available for excitation =

4K
5

Total energy required for excitation
= 10
...
8 eV
= 51
...

Solution :

4k
= 51
5



k = 63
...
What are the maximum–energy X–rays from
the TV set ?
The electrons in the TV tube have an energy of 20 keV, and if these electrons are brought to rest by
a collision in which one X–ray photon is emitted, the photon energy is 20 keV
...


In the Moseley relation,

Solution :

K transition ?
A is larger for the K transitions than for the K transitions

---