1. M
MO
OD
DE
ER
RN
N P
PH
HY
YS
SI
IC
CS
S
WAVE PARTICLE DUALITY:
Despite their wave nature, electromagnetic radiations, have properties akin to those of particles.
Electromagnetic radiation is an emission with a dual nature, i.e. it has both wave and particle aspects. In particular,
the energy conveyed by an electromagnetic wave is always carried in packets whose magnitude is proportional to
frequency of the wave. These packets of energy are called photons.
Energy of photon is E  h. f , where h is Planck’s constant, and f is frequency of wave.
According to de-Broglie
As wave behaves like material particles, similarly matter also behaves like waves. According to him, a
wavelength of the matter wave associated with a particle is given by
v
h h
p m
   , where m is the mass and v is
velocity of the particle.
If an electron id accelerated through a potential difference of V volt,
then 2
1
v
2
e
m eV
 or
2 V
v
e
e
m


v 2 V
e e
h h
m e m
  
(It is assumed that the voltage V is not more than several tens of Kilovolt)
Brain Teaser:
1. Why is the wave nature of matter not apparent in our daily lives ?
2. A charged and an uncharged particle have the same momentum. Will they have the same
de-Broglie wavelength ?
Illustration 1. Sun gives light at the rate of 1400 Wm-2
of area perpendicular to the direction of light. Assume
λ (sunlight) = 6000 Å. Calculate the
(a) number of photons/sec arriving at 1 m2
area at that part of the earth, and
(b) number of photons emitted from the sun/sec assuming the average radius of Earth’s orbit is
11
1.49×10 m.
Solution: 1400
I  W/m2
; 6000
  Å
(a) E , Energy of the photon
hc
h
  

( 8
3 10
c   m/sec)
Let n be the number of photon received /sec per unit area.
10
21
34 8
(1400 1) (6000 10 )
n 4.22 10
E/Photon 6.63 10 3 10
IA 

  
   
  
(b) Total energy emitted per second = power (watts)
2 10
34 8
Power of Sun(W) (4 ) (6000 10 )
n / sec
E/photon 6.63 10 3 10
I R 

   
 
  
( R average radius of earth’s orbit)
45
1.178 10
 
PHOTOELECTRIC EFFECT:
The photoelectric effect is the phenomenon of emission of electrons by a metallic surface under the action
of light.

2. Observation of the experiments on Photo-Electric Effect:
(i) The emission of photoelectrons is instantaneous.
(ii) the number of photoelectrons emitted per second is proportional to the intensity of the incident light.
(iii) The maximum velocity with which electrons emerge is dependent only on the frequency and not on the
intensity of the incident light.
(iv) There is always a lower limit of frequency called threshold frequency below which no emission takes
place, however high the intensity of the incident radiation may be.
W
Wo
or
rk
k F
Fu
un
nc
ct
ti
io
on
n:
:
The minimum amount of work or energy necessary to take a free electron out of a metal against the
attractive forces of surrounding positive ions inside metals is called the work function of the metal.
0 0
W h
  , where 0
 is the threshold frequency.
An electron can undergo collisions with other electrons, protons or macroscopically with the atom. In this
process it will fritter away its energy. Therefore, electrons with K.E. ranging from 0 to K. E.max will be produced.
E
Ei
in
ns
st
te
ei
in
n’
’s
s P
Ph
ho
ot
to
oe
el
le
ec
ct
tr
ri
ic
c E
Eq
qu
ua
at
ti
io
on
n:
:
According to Einstein, photon energy is utilized for two purposes.
(i) Partly for getting the electron free from the atom and away from the metal surface. This energy is
known as the photoelectric work function of the metal and is represented by 0
W .
(ii) The balance of the photon energy is used up in giving the electron a Kinetic Energy of 2
1
v
2
m .
2
0
1
v
2
h W m
   .
In the case the photon energy is just sufficient to liberate the electron only, the Kinetic Energy of the
electron is zero.
i.e., 0 0
h W
 
where 0
 is the threshold frequency and 0
W is the work function. If the frequency of incident light is less
than 0
 , no photoelectric emission takes place.
Kinetic Energy of photoelectrons is 0 0
( )
KE hf hf h f f
    
0 0
1 1 1 1
12400 V
hc e
   
   
   
   
   
.
Illustration 2. A beam of light has three wavelengths 4144 Å, 4972 Å and 6216 Å with a total intensity of
-3
3.6 × 10 W m-2
equally distributed amongst the three wavelengths. The beam falls normally on
an area 1.0 cm2
of a clean metallic surface of work function 2.3 eV. Assuming that there is no
loss of light by reflection and that each energetically capable photon ejects one electron.
Calculate the number of photo-electrons liberated in 2 seconds.
Solution: Three different wavelengths are incident on metal surface, so first determine which is (are) capable
of ejecting photo-electrons.
For photo-emission, 0
   Given: 0 2.3
W  eV
0 0
/
W hc
 

34 8
0 19
0
6.63 10 3 10
5404
2.3 1.6 10
hc
Å
W


  
  
 
 only wavelengths 4144 Å and 4972 Å will cause photo-emission (6216 Å > 0
 )
Intensity of each incident wavelength 3 3
3.6 10 /3 1.2 10
 
    W/m2
.
[  I is distributed equally among three wavelengths]

3. /sec
/
IA
n
hc


3 4 10
11
34 8
(1.2 10 ) (10 ) 4144 10
/sec( 4144 ) 2.5 10
6.63 10 3 10
n Å
  

   
    
  
3 4 10
11
34 8
(1.2 10 ) (10 ) 4972 10
/sec( 4972 ) 3 10
6.63 10 3 10
n Å
  

   
    
  
 total electrons emitted/sec 11
5.5 10
 
 total electrons emitted in 2 seconds 11
11 10
  .
S
St
to
op
pp
pi
in
ng
g P
Po
ot
te
en
nt
ti
ia
al
l:
:
This is the value of negative potential difference which just stops the electrons with maximum kinetic
energy from reaching the anode. If s
V is the stopping potential, then
2
max
1
V v
2
s
e m
 .
BOHR’S MODEL OF HYDROGEN LIKE ATOMS:
Bohr postulated in his H-atom model.
1. The electron is revolving round the nucleus in stationary orbits.
2. When an electron makes a transition from a higher orbit to a lower stable orbit, the difference in the
energy of the electron is radiated as a photon of energy h .
3. The angular momentum of the electron in the stationary orbits are quantised.
mvr n
  where
2
h


 .
h is Planck’s constant.
n is called the quantum number; 1
n  for the first stable orbit, 2 for the second orbit, etc.
The Bohr model is applicable not only for hydrogen but all hydrogen-like atoms i.e., atoms which have
been ionized to have a single electron revolving round the nucleus.
C
Ci
ir
rc
cu
ul
la
ar
r O
Or
rb
bi
it
ts
s:
:
The atom consists of central nucleus, containing the entire positive charge and almost all the mass of the
atom. The electrons revolve around the nucleus in certain discrete circular orbits. The necessary centripetal force for
circular orbits is provided by the Coulomb attraction between the electron and nucleus. So,
2
2
0
1 ( )( )
4
mv Ze e
r r


where, m  mass of electron
r  radius of circular orbit,
v  speed of electron in circular orbit,
Ze  charge on nucleus,
Z = atomic number,
e  charge on electron 19
1.6 10 C

  
S
St
ta
at
ti
io
on
na
ar
ry
y O
Or
rb
bi
it
ts
s:
:
The allowed orbits for the electrons are those in which the electron does not radiate energy. These orbits are
also called stationary orbits.
Q
Qu
ua
an
nt
tu
um
m c
co
on
nd
di
it
ti
io
on
n (
(B
Bo
oh
hr
r’
’s
s Q
Qu
ua
an
nt
ti
is
sa
at
ti
io
on
n R
Ru
ul
le
e)
):
:
The stationary orbits are those in which angular momentum of the electron is an integral multiple of
( )
2
h


 i.e.,
2
h
mvr n
 
  

 
, n being integer or the principle quantum number

4. R
Ra
ad
di
iu
us
s o
of
f O
Or
rb
bi
it
t:
:
Since, we have
2
2
0
1 ( )( )
4
mv Ze e
r r


…(i)
and
2
nh
mvr 

…(ii)
From (ii),
2
nh
v
mr


Putting in (i), we get
2 2
0
2
n
n h
r
me Z




2
(0.53)
n
n
r Å
Z

So, for H-like atoms
2
n
n
r
Z
 .
V
Ve
el
lo
oc
ci
it
ty
y o
of
f E
El
le
ec
ct
tr
ro
on
n i
in
n n
nt
th
h O
Or
rb
bi
it
t:
:
Since
2
nh
V
mr


and
2 2
0
2
n h
r
me Z




2
0
2
e Z
V
h n
 
  

 

2
0
2
e cZ
V
h c n
  
   
  
 

cZ
V
n
 
  
 
where
2
0
2
e
h c
 

is the Summerfield’s fine structure constant (a pure number) whose value is
1
137
.

1
137
cZ
V
n
 
  
 
i.e. velocity of electron in Bohr’s First Orbit is
137
c
, in Second Orbit is
274
c
and so on.
K
Ki
in
ne
et
ti
ic
c E
En
ne
er
rg
gy
y o
of
f E
El
le
ec
ct
tr
ro
on
n ( )
K
E :
:
Since, we have
2
2
0
1 ( )( )
4
mv Ze e
r r



2
2
0
1
2 8
Ze
mv
r



2
2
0
1
2 8
K
Ze
E mv
r
 

.

5. P
Po
ot
te
en
nt
ti
ia
al
l E
En
ne
er
rg
gy
y (
(U
U)
) o
of
f E
El
le
ec
ct
tr
ro
on
n i
in
n n
nt
th
h O
Or
rb
bi
it
t:
:
0
1 ( )( )
4
Ze e
U
r
 


2
0
4
Ze
U
r
 

T
To
ot
ta
al
l E
En
ne
er
rg
gy
y (
(E
E)
) o
of
f E
El
le
ec
ct
tr
ro
on
n i
in
n n
nt
th
h O
Or
rb
bi
it
t:
:
Total Energy = K.E. + P.E.

2 2
0 0
8 4
Ze Ze
E
r r
 
 

2
0
8
Ze
E
r
 

So, we conclude that
Total Energy = – K.E.
1
2
 (P.E.)
Further, since
2 2
0
2
n h
r
me Z




4 2
2 2 2
0
8
me Z
E
h n
 
  

 

2
2
(13.6)
Z
E eV
n
 
Also
4 2
2 3 2
0
8
me Z
E ch
ch n
 
  

 

2
2
( )
Z
E Rch
n
 
where R  Rydberg’s constant
4
7 1
2 3
0
1.097 10
8
me
m
ch

  

and
Rch  Rydberg’s Energy 18
2.17 10 13.6
J eV

 is the electron energy in first orbit of H atom.
F
Fr
re
eq
qu
ue
en
nc
cy
y o
of
f E
Em
mi
it
tt
te
ed
d R
Ra
ad
di
ia
at
ti
io
on
n:
:
If electron makes a transition (jumps) from final state f
n to the initial state i
n , then frequency of emitted
radiation v is given by
i f
E E hv
 
or 2
2 2
1 1
i f
f i
E E
v Z Rc
h n n
 

  
 
 
 
If c is the speed of light and  the wavelength of emitted or absorbed radiation, then
2
2 2
1 1
f i
c
v Z Rc
n n
 
  
 
 
  
So, wave number ( )
v is given by
2
2 2
1 1 1
f i
v Z R
n n
 
  
 
 
  
This relation holds for radiations emitted by hydrogen-like atoms i.e.
( 1), ( 2), ( 3)
H Z He Z Li Z
 
   and ( 4)
Be Z


If the electron makes a transition from 1
n  to the higher states, it is absorption.

6. H
Hy
yu
ud
dr
ro
og
ge
en
n S
Sp
pe
ec
ct
tr
ru
um
m:
:
Initial
State
Final
State
Wavelength
Formula
First Member-
Second
Member
Series Limit
i
n   ot f
n
Maximum
Wavelength
( 1)
f
n  to
f
n
Lines
Found in
Lyman
i
n = 2, 3,
4, 5, 6, …..
f
n = 1
2 2
1 1 1
1 i
R
n
 
 
 
  
i
n = 2 to f
n = 1
i
n = 3 to f
n = 1
From  to 1
4
R
 
911Å
 
From 2 to 1
4
3R
 
1216Å
 
UV
Region
Balmer
i
n = 3, 4,
5, 6, 7, …..
f
n = 2
2 2
1 1 1
2 i
R
n
 
 
 
  
i
n = 3 to f
n = 2
i
n = 4 to f
n = 2
From  to 2
4
R
 
3646Å
 
From 3 to 2
36
5R
 
6563Å
 
Visible
Region
Paschen
i
n = 4, 5,
6, 7, 8, …..
f
n = 3
2 2
1 1 1
3 i
R
n
 
 
 
  
i
n = 4 to f
n = 3
i
n = 5 to f
n = 3
From  to 3
9
R
 
8204Å
 
From 4 to 3
144
7R
 
18753Å
 
IR Region
Brackett
i
n = 5, 6,
7, 8, 9, …..
f
n = 4
2 2
1 1 1
4 i
R
n
 
 
 
  
i
n = 5 to f
n = 4
i
n = 6 to f
n = 4
From  to 4
16
R
 
14585Å
 
From 5 to 4
400
9R
 
40515Å
 
IR Region
Pfund
i
n = 6, 7,
8, 9, 10, ....
f
n = 5
2 2
1 1 1
5 i
R
n
 
 
 
  
i
n = 6 to f
n = 5
i
n = 7 to f
n = 5
From  to 5
25
R
 
22790Å
 
From 6 to 5
900
11R
 
74583Å
 
Far IR
Region
R
Ry
yd
db
be
er
rg
g C
Co
on
ns
st
ta
an
nt
t:
:
Rydberg Constant is given by
4
2 3
0
8
me
R
ch


where m is mass of electron.
Illustration 3. A single electron orbits around a stationary nucleus of charge Ze
 , where Z is a constant and
e is the magnitude of the electronic charge. It requires 47.2eV to excite the electron from
second Bohr orbit to the third Bohr orbit. Find:
(a) the value of Z ,
(b) the energy required to excite the electron from 3
n  to 4
n  ,
(c) the wavelength of radiation required to remove electron from the first Bohr Orbit to
infinity,
(d) the kinetic energy, potential energy and angular momentum of the electron in the first
Bohr orbit and the radius of first Bohr orbit.
Solution: (a) transition is 1 2
n   2 3
n  ; 47.2
E
  eV
we have 2
2 2
1 2
1 1
13.6
E Z eV
n n
 
  
 
 
 2
2 2
1 1
47.2 13.6
2 3
Z
 
 
 
 
 5
Z 

7. (b) transition is 1 3
n   2 4
n  ; ?
E
 
we have 2
2 2
1 2
1 1
13.6
E Z
n n
 
  
 
 
eV
 2
2 2
1 1
13.6 5 16.53
3 4
E
 
    
 
 
eV.
(c) transition is 1 2
1
n n
    ; E
  ionization energy ?

 2
2 2
1 1
13.6 5 340
1
E
 
    
 

 
eV
34 18
10
19
6.63 10 3 10
36.56 10 36.56
340 1.6 10
hc
Å
E



  
     
  
(d) 2 2
. . [ 13.6 ]/
n
K E E Z n
     
 . .
K E (Ist Bohr orbit) = 2 2
1 [ 13.6 5 ]/1 340
E
       eV
P.E. = 1
2 2 680
n
E E
   eV
Angular momentum 34 34
( / 2 ) 1 (6.63 10 / 2 ) 1.056 10
n h J s
 
        
radius 2 2
( ) [0.53 / ] [0.53 /5] 0.106
r n Z Å I Å
   .
R
Ri
it
tz
z C
Co
om
mb
bi
in
na
at
ti
io
on
n P
Pr
ri
in
nc
ci
ip
pl
le
e:
:
If electron is initially in an excited state with say 3
n  , then it may
transit downward from 3
n  level to 1
n  level directly. Alternatively, it may
first transit from 3 2
n n
   and subsequently from 2 1
n n
   . In the
first case if 31
v be the frequency of the photon emitted
31 3 1
hv E E
  …(i)
In the second case, two different spectral lines (photons) of frequency
32
v and 21
v respectively would be emitted given by
32 3 2
hv E E
  and 21 2 1
hv E E
  …(ii)
(i) can be rewritten as
31 3 2 2 1
( ) ( )
hv E E E E
   
 31 32 21
hv hv hv
 
 31 32 21
v v v
 
n=3
v31
v21
v32
n=2
n=1
Ritz made this discovery empirically (1908) long before Bohr proposed his theory and is known as Ritz
combination principle.
Generalizing, we may write, labeling the photon frequency by appropriate integers, as follows:
sm s m
hv E E
 
 ( ) ( )
sm s n n m
hv E E E E
   
 sm sn nm
hv hv hv
  ( )
m n s
  …(iii)
Since all combinations predicted by (iii)are not actually observed, there has been an imposition of some
rules, the so-called selection rules, to eliminate certain combinations. Bohr’s theory provides, as discussed above, a
proper explanation of the combination principle.
Illustration 4. In a transition to a state of excitation energy 10.19 eV, a hydrogen atom emits a 4890 Å photon.
Determine the binding energy of the initial state.
Solution: The energy of the emitted photon is
3
3
12.40 10 .
2.54
4.89 10
hc eV Å
hv
Å

  
 
eV
The excitation energy ( )
x
E is the energy to excite the atom to a level above the ground state.
Therefore, the energy of the level is

8. 1 13.6 10.19 3.41
n x
E E E eV eV eV
      
The photon arises from the transition between energy states such that 1
u
E E hv
  ; hence
( 3.41 ) 2.54
u
E eV eV
   or 0.87
u
E eV
 
Therefore, the binding energy of an electron in the state is 0.87 eV
Note that the transition corresponds to
1 13.6
4
0.87
u
u
E eV
n
E eV
   and
1
1
13.6
2
3.14
l
E eV
n
E eV
   .
X-RAYS:
X-rays are electromagnetic radiation of very short wavelength (0.1 º
A to 100 º)
A and high energy which
are emitted when fast moving electrons or cathode rays strike a target of high atomic mass.
D
Di
is
sc
co
ov
ve
er
ry
y o
of
f X
X-
-R
Ra
ay
ys
s:
:
X-rays were discovered by Roentgen (1895) who found that a discharge tube, operating at low pressure and
high voltage, emitted a radiation that caused a florescent screen in the neighborhood to glow brightly. Crystals of
barium platinocyanide also showed fluorescence. Results were same, if the discharge tube is wrapped in black paper,
to prevent visible light. This indicated that some unknown radiation (X-rays) were responsible for fluorescence.
Roentgen then confirmed that X-rays are emitted, when cathode rays (electrons) strike the wall of discharge tube.
P
Pr
ro
od
du
uc
ct
ti
io
on
n o
of
f X
X-
-R
Ra
ay
ys
s (
(C
Co
oo
ol
li
id
dg
ge
e’
’s
s T
Tu
ub
be
e)
):
:
X-rays are produced when energetic (fast moving) electrons strike a target such as a metal piece. When
electrons collide with the atoms of solid, they loose their kinetic energy which is converted into radiant energy in the
form of X-rays. The figure shown essential features of a modern X-ray tube developed by Coolidge.
Coolidge’s X-ray tube consists of a glass bulb exhausted to nearly perfect vacuum. The cathode C is the
source of electrons by using a heated filament getting supply from battery B . The anode is made of solid copper bar
A . A high melting metal like platinum or tungsten is embedded at the end of copper rod and serves as target T . A
high d.c. voltage V(50 kV) is maintained between cathode and anode.
X-rays
e
C
F
B
V
e
e
A
The energetic electrons strike the target T and the X-ray are produced. Only about 1-10% of the energy of
the electrons is converted to X-rays and the rest is converted into the heat. The target T as a result becomes very hot
and therefore should have high melting point. The heat generated is dissipated through the copper rod and the anode
is cooled by water flowing through the anode.
The nature of emitted X-rays depends on:
(i) the current in the filament F
(ii) the voltage between the filament and the anode.

9. * An increase in the filament current increase the number of electrons it emits. Larger number of electrons
means an intense beam of X-rays is produced. This way we can control the quantity of X-rays i.e. Intensity
of X-rays.
* An increase in the voltage of the tube increase the kinetic energy of electrons 2
( V 1/ 2mv )
e  . When such
highly energetic beam of electrons are suddenly stopped by the target, an energetic beam of X-rays is
produced. This way we can control the quality of X-rays i.e. penetration power of X-rays .
* Based on penetrating power, X-rays are classified into two types. HARD-rays and SOFT-X-rays. The first
one having high energy and hence high penetration power are HARD-X-rays and another one with low
energy and hence low penetration power are SOFT-X-rays.
Brain Teaser:
3. X-rays are produced when a fast electron hits a proper target. What happens to the electron?
4. Why does the target in an X-ray tube become hot?
P
Pr
ro
op
pe
er
rt
ti
ie
es
s o
of
f X
X-
-R
Ra
ay
ys
s:
:
1. These are highly penetrating rays and can pass through several materials which are opaque to ordinary light.
2. They ionize the gas through which they pass. While passing through a gas, they knock out electrons from
several of the neutral atoms, leaving these atoms with +ve charge.
3. They cause fluorescence in several materials. A plate coated with barium platinocyanide, ZnS (zinc
sulphide) etc becomes luminous when exposed to X-rays.
4. They affect photographic plates especially designed for the purpose.
5. They are not deflected by electric and magnetic fields, showing that they are not charged particles.
6. They show all the properties of the waves except refraction. They show diffraction patterns when passed
through a crystal which behaves like a grating.
A
Ap
pp
pl
li
ic
ca
at
ti
io
on
n o
of
f X
X-
-r
ra
ay
ys
s:
:
X-rays have important and useful applications in surgery, medicine, engineering and studies of crystal
structures.
1. Scientific Applications:
The diffraction of X-rays at crystals opened new dimension to X-rays crystallography. Varous diffraction
patterns are used to determining internal structure of crystals. The spacing and dispositions of atoms of a
crystal can be precisely determined used Bragg’s Law: 2 sin
n d
   .
2. Industrial Applications:
Since X-rays can penetrate through various materials, they are used in industry to detect defects in metallic
structures is Big machines, railway tracks and bridges. X-rays are used to analyse the composition of alloys
and pearls.
3. In Radio Therapy:
X-rays can cause damage to the tissues of body (cells are ionized and molecules are broken). So X-rays
damages the malignant growths like cancer and tumors which are dangerous to life, when it used in proper
and controlled intensities.
4. In Medicine and Surgery:
X-rays are absorbed more in heavy elements than the lighter ones. Since bones (containing calcium and
phosphorus) absorb more X-rays than the surrounding tissues (containing light elements like , ,
H C O ),

10. their shadow is casted on the photographic plate. So the cracks or Fracture in bones can be easily located.
Similarly intestine and digestive system abnormalities are also detected by X-rays.
X
X-
-r
ra
ay
y A
Ab
bs
so
or
rp
pt
ti
io
on
n:
:
The intensity of X-rays at any point may be defined as the energy falling per second per unit area held
perpendicular to the direction of energy flow. The intensity of a X-rays beam decreases during its passage through
the sheet of any material. The decrease in the intensity of X-rays is due to the absorption of X-rays by the material.
Let 0
I be the intensity of incident beam and I be the intensity of beam after penetrating a thickness x of
the material, then 0
x
I I e
 ; where  : coefficient of absorption or absorption coefficient of a material. The
absorption coefficient depends upon wave length of X-rays, density of material and atomic number of material. The
elements of high atomic mass and high density absorb X-rays to a higher degree.
X
X-
-R
Ra
ay
y D
Di
if
ff
fr
ra
ac
ct
ti
io
on
n &
& B
Br
ra
ag
gg
g’
’s
s L
La
aw
w:
:
X-rays are electromagnetic radiation of low wavelength (0.1 º to 100 º)
A A . For studying the diffraction
pattern of X-rays, we require size of slit equal to or order of the wavelength of X-rays. The spacing of atoms or
molecules in crystals is of the order of a few angstroms (Aº). Thus the crystals are used to measure the wavelengths
of X-rays, just like gratings are used to determine wavelengths of visible radiations.
X-rays emanating for X-ray tube pass through two holes in two lead screens and then in falls on the crystal.
After emerging from the crystal (sowing diffraction), X-rays falls on the photographic plate to produce a wavelike
circular patterns. These patterns were called as Laue’s pattern. This experiment also conforms the wave nature of X-
rays.
The simplified version and diagram of X-rays spectrometer is
shown here. It is used to determine wavelength of X-rays, from a tube,
narrowed down to a fine line by means of simple stopper slits, 1
S and
2
S are incident on a crystal at an angle  . The crystal is mounted on a
graduated turn table. The reflected X-rays pass through a detector which
contents of a ionization chamber. The position of ionization chamber
and tube is kept fined. The gas in ionization chamber gets ionized due
to diffracted X-rays and the intensity of reflected X-rays is measured by
electrometer. The angle ‘  ’ of the reflected X-rays is calculated from
graduated turn table for maximum intensity.
x-ray
S1 S2
incident
x-ray 
turn table
crystal
reflected
x-ray
ionisation
chamber

B
Br
ra
ag
gg
g’
’s
s L
La
aw
w a
an
nd
d S
Sp
pe
ec
ct
tr
ro
om
me
et
te
er
r:
:
When beam of x-rays is incident on a crystal. The path difference
between the two rays AOA and BOB is 2 sin
PO P O d
  
   . If the path
difference is m , one gets constructive interference.
2 sin
d m
  
where  = grating angle
d = interatomic distance
m = order of diffraction

A
B
P
O
N
P
A
B

 
d
B
Br
ra
ag
gg
g’
’s
s l
la
aw
w o
of
f x
x-
-r
ra
ay
y d
di
if
ff
fr
ra
ac
ct
ti
io
on
n:
:
The crystal behaves like a grating. The interatomic distance
should be of the same order as  . Different crystals are used for different
wavelength regions. For each wavelength even third orders.
At higher orders the wavelength will overlap as in the optical
diffraction.
1 order
=1
st
m
2 order
=2
nd
m
1 2
3

I

11. X
X-
-R
Ra
ay
y S
Sp
pe
ec
ct
tr
ra
a a
an
nd
d O
Or
ri
ig
gi
in
n o
of
f X
X-
-R
Ra
ay
ys
s:
:
Experimental observation and studies of spectra of X-rays reveal that X-rays are of two type and so are their
respective spectras. Characteristic X-rays and Continuous X-rays.
Characteristic X-rays:
The spectra of this group consists of several radiations with specific sharp wavelengths and frequency
similar to the spectrum (line) of atoms like hydrogen. The wavelengths of this group show characteristic discrete
radiations emitted by the atoms of the target material. The characteristic X-rays spectra helps us to identify the
element of target material.
Origin of Characteristic X-rays:
When the atoms of the target material are bombard with high
energy electrons (or hard X-rays), which posses enough energy to
penetrate into the atom, knockout the electron of inner shell (say K shell,
n = 1). When an electron is missing in the ‘ K ’ shell, an electron from
next upper shell makes a quantum jump to fill the vacancy in ‘ K ’ shell.
In the transition process the electron radiates energy whose frequency
lies in the X-rays region. The frequency of emitted radiation (i.e. of
photon) is given by 2
2 2
1 2
1 1
e
v RZ
n n
 
 
 
 
; where R is constant and e
Z is
effective atomic number. Generally e
Z is taken to be equal to Z  ,
where Z is proton number or atomic number of the element and  is
called the screening constant. Due to the presence of the other electrons.
The charge of the nucleus as seen by the electron will be different is
different shells.
K L
k
e
i
e
i
e
Knocking out e-
of K shell by incident
electron i
e
K L
hv
(x-ray)
emission of X-ray photon (K - series)
* Another vacancy is now created in the ‘ L ’ shell which is again
filled up by another electron jump from one of the upper shell
(M) which results in the emission of another photon, but of
different X-rays frequency. This transition continues till outer
shells are reached. Thus resulting in the emission of series of
spectral line.
* The transitions of electrons from various outer shells to the inner
most ‘ K ’ shell produces a group of X-rays lines called as K-
series. These radiations are most energetic and most penetrating.
K -series is further divided into , ,
K K K
   …. depending upon
the outer shell from which the transition is made.
* The jump of electrons from outer shells to ‘ L ’ shell results in L-
Series X-rays and so on.
K-series
L-series
M-series
n=2
n=3
n=4
n=1
K
K
KK
L
L L
M M
N
N
L
M
K
n=
X-ray Series
If we plot v of K X-rays as a function of Z , the atomic number of material, we obtain a straight line.
The relation of straight line is expressed as v ( )
a Z b
  , where a and b are constant. This relation is called as
Moseley’s Law. It helps to determine the atomic number Z of an atom. Here b is the screening constant.
Note:
For K-series X-rays, some authors take 1
b  , i.e. v ( 1)
a Z
  , although the actual value can be different.

12. Illustration 5. If the K radiation of ( 42)
Mo Z  has a wavelength of 0.71 Å, calculate wavelength of the
corresponding radiation of Cu , i.e., K for ( 29)
Cu Z  assuming 1
  .
Solution: According to Moseley’s Law : ( 1)
v a Z
 
 2
( 1)
Z v
 
or 2
( 1) 1/
Z   

2
0
2
( 1)
( 1)
M Cu
Mo
Cu
Z
Z
 




2
2
0
2
( 1) 41
0.71 1.52
28
( 1)
M
Cu Mo
Cu
Z
Å Å
Z
  
     
 
  
.
Continuous X-rays:
In addition to characteristic X-rays, tubes emit a continuous
spectrum also. The characteristic line spectra is superimposed on a
continuous X-rays spectra of varying intensities. The wavelength of the
continuous X-rays spectra are independent of material. One important
feature of continuous X-rays is that they end abruptly at a certain lower
wavelength for a given voltage.
X-ray photon
hv
Target
atom
K
K
Origin of Continuous X-rays:
If an electron beam of energy e V (electron volts) is incident on the target material; the electrons are
suddenly stopped. If the whole of the energy is converted to continuous radiation, then min
 (corresponding to
energy maximum) /
hc Ve
 where V is the voltage applied.
The classical theory of electromagnetism states that the suddenly accelerated or decelerated electrons emit
radiations of electromagnetic nature called as bremsstrahlung (braking radiation) and wavelength of such radiation is
continuous because the loss in energy is statistical. At the peak, the probability of maximum number of electrons
producing radiation.
The wavelength of X-rays emitted is minimum
corresponding to the electron which hits the target with maximum
speed. This electron is completely stopped and will emit the
photon of highest energy.
As the electrons lose energy by collision, longer
wavelength are produced the shape of the curve is statistical.
Intensity
Wavelength (nm)
0.01 0.1 1.0
min

K
K
K
L
L
M
50 kV
40 kV
30 kV
20 kV
0 0.02 0.04 0.06 0.08 0.10
Wavelength (nm)
Relative
intensity
Illustration 6. When 0.50 Å X-rays strike a material, the photoelectrons from the K shell are observed to move
in a circle of radius 23 mm in a magnetic field of T
2
2 10
 . What is the binding energy of K-
shell electrons?
Solution: The velocity of the photoelectrons is found the F ma
 :

13. 2
v
evB m
R
 or
e
v BR
m

The kinetic energy of the photoelectrons is then
2 2 2
2
1 1
2 2
e B R
K mv
m
 
19 2 2 2 3 2
31
1 (1.6 10 ) (2 10 ) (23 10 )
2 (9 1 10 )
C T m
kg
  

  


15
2.97 10
  J
or 15
16
1
(2.97 10 ) 18.36
1.6 10
keV
K J eV
J


  

The energy of the incident photon is
12.4 .
24.8
0.50
v
hc keV Å
E eV
Å
  

The binding energy is the difference between these two values:
24. 18.6 6.2
BE Ev K keV keV keV
     .
NUCLEAR PHYSICS:
The Nucleus:
It exists at the centre of an atom, containing entire positive charge and almost the whole of the mass. The
electrons revolve around the nucleus to form an atom. The nucleus consists of protons (+ve charge) and neutrons.
(no charge)
 A proton has positive charge equal in magnitude to that of an electron, 19
(1.6 10 )
C

 and a mass equal to
1836 times that of an electron.
 A neutron has no charge and its mass is approximately equal to that of the proton 27
(1.6726 10 kg)

 . (1837
times that of an electron)
 The number of protons in a nucleus of an atom is called as the atomic number (Z) of that atom. The number
of protons and neutrons (together called Nucleons) in the nucleus of an atom is called the mass number (A)
of the atom.
 A particular set of nucleons forming an atom is called a nuclide. It is represented as A
Z X .
 The nuclides having same number of protons (Z), but different number of nucleons (A) are called isotopes.
 The nuclides having the same number of nucleons (A), but different number of protons(Z) are called
isobars.
 The nuclides having the same number of neutrons (A–Z) are called isotones.
Mass Defect & Binding Energy:
The nucleons are bound together in a nucleus and the energy has to be supplied in order to break apart the
constituents into free nucleons. The energy with which nucleons are bound together in a nucleus is called Binding
Energy (B.E.). In order to free nucleons from a bound nucleus, this much of energy ( = B.E.) has to be supplied.
It is observed that the mass of a nucleus is always less than the mass of its constituent (free) nucleons. This
difference in mass is called as mass defect and is denoted as m
 .
If n
m = mass of neutron and p
m = mass of a proton
( , )
M Z A = mass of bound nucleus
Then, . ( ). ( , )
P n
m Z m A Z m M Z A
    

14. This mass-defect is in form of energy and is responsible for binding the nucleons together. From Einstein’s
mass-energy relation,
2
E mc
 ( c speed of light; m mass)
 binding energy 2
.
m c
 
Generally, m
 is measured in amu units. So let us calculate the energy equivalent to 1 amu. It is calculated
in eV (electron volts; 19
1 1.6 10
eV J

  )
27 8 2
6
19
1 1.67 10 (3 10 )
( 1 ) eV 931 10 eV 931MeV
1.6 10
E amu


  
    

 . . (931)MeV
B E m
 
There is another quantity which is very useful in predicting the stability of a nucleus called as Binding
energy per nucleons.
B.E. per nucleons
(931)
MeV
m
A

 .
From the plot of B.E./Nucleons Vs mass number (A), we observe that:
 B.E./Nucleons increases on an average and reaches a
maximum of about 8.7 MeV for 50 80
A   .
 For heavier nuclei, B.E./nucleons decreases slowly as A
increases. For the heaviest natural element 238
U it drops
to about 7.5 MeV.
 From above observation, it is following that nuclei in the
region of atomic masses 50-80 are most stable.
20 60 180 A
mass number
(in MeV)
2
4
6
8
8.5 MeV
A
E
B .
.
Nuclear Forces:
The protons and neutrons are held together by the strong attractive forces inside the nucleus. These forces
are called as nuclear forces.
 Nuclear forces are short-ranged. They exist in small region (of diameter 15
10 1
m fm

 ). The nuclear force
between two nucleons decrease rapidly as the separation between them increases and becomes negligible at
separation more than 10 fm.
 Nuclear force are much stronger than electromagnetic force and gravitational attractive forces.
 Nuclear force are independent of charge. The nuclear force between two proton is same as that between two
neutrons or between a neutron and proton. This is known as charge independent character of nuclear forces.
In a typical nuclear reaction :
1. In nuclear reactions, sum of masses before reaction is greater than the sum of masses after the reaction. The
difference in masses appears in form of energy following the Law of inter-conversion of mass & energy.
The energy released in a nuclear reaction is called as Q value of a reaction and is given as follows.
If difference in mass before and after the reaction is m
 amu
( m
 = mass of reactants minus mass of products)
then Q value (931)MeV
m
 
2. Law of conservation of momentum is also followed.
3. Total number of protons an neutrons should also remain same on both sides of a nuclear reaction.
NUCLEAR FISSION:
The breaking of a heavy nucleus into two or more fragments of comparable masses, with the release of
tremendous energy is called as nuclear fission. The most typical fission reaction occurs when slow moving neutrons
strike 235
92U . The following nuclear reaction takes place.
235 1 141 92 1
92 0 56 36 0
3 200MeV
U n Ba Kr n
    

15.  If more than one of the neutrons produced in the above fission reaction are capable of inducing a fission
reaction (provided 235
U is available), then the number of fissions taking place at successive stages goes
increasing at a very brisk rate and this generates a series of fissions. This is known as chain reaction. If
mass of 235
U sample greater than a certain size called the critical size then it is capable of continuous
fission by itself.
 If the number of fission in a given interval of time goes on increasing continuously, then a condition of
explosion is created. In such cases, the chain reaction is known as uncontrolled chain reaction. This forms
the basis of atomic bomb.
 In a chain reaction, the fast moving neutrons are absorbed by certain substances known as moderators (like
heavy water), then the number of fissions can be controlled and the chain reaction is such cases is known as
controlled chain reaction. This form the base of a nuclear reactor.
Brain Teaser:
5. Does a nucleus have to be bombarded with fast or slow neutrons in order for it to undergo
fission?
NUCLEAR FUSION:
The process in which two or more light nuclei are combined into a single nucleus with the release of
tremendous amount of energy is called as nuclear fusion. Like a fission reaction, the sum of masses before the fusion
(i.e. of light nuclei) is more than the sum of masses after the fusion (i.e. of bigger nucleus) and this difference
appears as the fusion energy. The most typical fusion reaction is the fusion of two deuterium nuclei into helium.
2 2 4
1 1 2 21.6MeV
H H He
  
For the fusion reaction to occur, the light nuclei are brought closer to each other (with a distance of
14
10
m). This is possible only at very high temperature to counter the repulsive force between nuclei. Due to this
reason, the function reaction is very difficult to perform. The inner core of sun is at very high temperature, and is
suitable for fusion, in fact the source of sun’s and other start’s energy is the nuclear fusion reaction.
Brain Teaser:
6. Why has it not been possible so far to control the fusion process and obtain usable energy from
it?
Illustration 7. It is proposed to use the nuclear fusion reaction: H H He
2 2 4
1 1 2
  in a nuclear reactor, of
200 MW rating. If the energy from above reaction is used with a 25% efficiency in the reactor,
how many grams of deuterium will be needed per day? (The masses of H2
1 and 4
2 He are
2.0141 and 4.0026 amu respectively.)
Solution: Let us first calculate the Q value of nuclear function.
2
(931)
Q mc m
    Mev
 (2 2.0141 4.0026) 931
Q     MeV = 23.834 MeV = 6
23.834 10
 eV.
Now efficiency of reactor is 25%
So effective energy used 6 19 13
25/100 23.834 10 1.6 10 9.534 10
J J
 
      
Now 13
9.534 10 J

 energy is released by fusion of 2 deuterium.
 13
(9.534 10 )/ 2

 J/deuterium is released.
Requirement is 200 MW 6
200 10 / 86400
J s
   for 1 day.

16.  no. of deuterium nuclei required
6
25
13
200 10 86400
3.624 10
9.534
10
2

 
  

Number of deuterium nuclei 23
6 10
g
A
  
 25 23
3.624 10 6 10
2
g
   

25
23
2 3.624 10
120.83
6 10
g
 
 

gm/day.
RADIOACTIVITY:
The phenomenon of spontaneous emission of radiation or particles from the nucleus is called radioactivity.
The substances which emit these radiations are called as radioactive substances. It was discovered by Henry
Becquerel for atoms of radium. Later it was discovered that many naturally occurring compounds of heavy elements
like radium, thorium etc also emit radiations.
At present, it is known that all the naturally occurring elements having atomic number greater than 82 are
radioactive. For example some of them are; radium, polonium, thorium, actinium, uranium, radon etc. Later on
Rutherford found that emission of radiation always accompanied by transformation of one element (transmutation)
into another. In actual radioactivity is the result of disintegration of an unstable nucleus. Rutherford studied the
nature of these radiations and found that these mainly consist of , ,
   rays.
 -
- P
Pa
ar
rt
ti
ic
cl
le
es
s 4
2
( )
He :
:
These carry a charge of 2e
 and mass equal to 4 p
m . These are nuclei of helium atoms. The energies of 
– particles vary from 5 MeV to 9 MeV and their velocities vary from 0.01 – 0.1 times of c (velocity of light). They
can be deflected by electric and magnetic fields and have lower penetrating power but high ionizing power.
 –
– P
Pa
ar
rt
ti
ic
cl
le
es
s 0
1
( )
e
 :
:
These are fast moving electrons having charge equal to e
 and mass 9.1
e
m   31
10 kg

. Their velocities
vary from 1% to 99% of the velocity of light (c). They can also be deflected by electric and magnetic field. They
have low ionizing power but high penetrating power. +
particles are positrons.
 –
– R
Ra
ad
di
ia
at
ti
io
on
n 0
0
( )
 :
:
These are electro-magnetic waves of nuclear origin and of very short wavelength. They have no charge and
no mass. They have maximum penetrating power and minimum ionising power. The energy released in a nuclear
reaction is mainly emitted in form of  radiation.
L
La
aw
ws
s o
of
f R
Ra
ad
di
io
oa
ac
ct
ti
iv
ve
e D
De
ec
ca
ay
y:
:
1. RUTHERFORD – SODDY LAWS (Statistical Laws):
 The disintegration of a radioactive substance is random and spontaneous.
 Radioactive decay is purely a nuclear phenomenon and is independent of any physical and chemical
conditions.
 The radioactive decay follows first order kinetics, i.e., the rate of decay is proportional to the number of
undecayed atoms in a radioactive substance at any time t . If dN be the number of atoms (nuclei)
disintegrating in time dt , the rate of decay is given as /
dN dt . From first order of kinetic rate law:
dN
N
dt
  where  is called as decay or disintegration constant.

17. Let 0
N be the number of nuclei at time 0
t  and t
N be the number of nuclei after time t , then
according to integrated first order rate law, we have:
0
t
t
N N e

 0 0
ln 2.303log
t
N N
t
Nt N
  
 The half life 1/ 2
( )
t period of a radioactive substance is defined as the time in which one-half of the
radioactive substance is disintegrated. If 0
N be the number of nuclei at 0
t  , then in a half life T , the
number of nuclei decayed will be 0 / 2
N .
0
t
t
N N e
 …(i)
 0
0
2
T
N
N e
 …(ii)
from (i) and (ii)
/
0
1 1
2 2
t T n
t
N
N
   
 
   
   
n : number of half lives
The mean life ( )
m
T of a radioactive substance is equal to the sum of lift times of all atoms divided by
the number of all atoms 1/
m
T  
Illustration 8. The mean lives of an radio active substance are 1620 and 405 years for   emission and  
emission respectively. Find out the time during which three fourth of a sample will decay if it is
decaying both the  emission and   emission simultaneously.
Solution: When a substance decays by  and  emission simultaneously, the average rate of disintegration
av
 is given by
av  
    
where 
 = disintegration constant for   emission only.

 = disintegration constant for   emission only.
Mean life is given by:
1/
m
T  
 av  
      3
1 1 1 1 1
3.08 10
1620 405
m
T T T

 
     
100
2.303log
25
avt
 
 3 100
(3.08 10 ) 2.303log
25
t

 
 3
1
2.303 log4 449.24
3.08 10
t 
  

years.
2. SODDY FAJAN LAWS (Group-Displacement Laws):
 When a nuclide emits one  – particle 4
2
( )
He , its mass number (A) decreases by 4 units and atomic
number (Z) decreases by 2 units.
4 4
2 2
A A
Z Z
X Y He


   Energy
 When a nuclide emits a  - particle, its mass number remains unchanged but atomic number increases
by one unit.
0
1 1
A A
Z Z
X Y e v
 
    Energy where v is antineutrino.

18. In the nucleus, due to conversion of neutron into proton, antineutrino is produced. It has no charge or mass,
but has momentum. When a proton is converted to a neutron, a neutron and a +ve  -particle is produced,
which is called as positron.  rays are electrons and  are the antielectrons or positrons.
1 1 0
0 1 1
n p e v

   (antineutrino)
1 1 0
1 0 1
p n e

  (positron) + (neutrino)
 When a  particle is produced, both atomic and mass number remain constant.
Antineutrino and neutrino share the energy of electrons and positrons. That is the reason why the energy of
 is continuous and  rays has an energy maximum.
Brain Teaser:
7. When a nucleus undergoes  -decay, is the product atom electrically neutral? in   decay?
A
Ac
ct
ti
iv
vi
it
ty
y o
of
f a
a R
Ra
ad
di
io
oa
ac
ct
ti
iv
ve
e I
Is
so
ot
to
op
pe
e:
:
The activity of a radioactive substance (or radioisotope) means the rate of decay per second or the number
of nuclei disintegrating per second. It is generally denoted by A.

dN
A
dt

If a time 0
t  , the activity of a radioactive substance be 0
A and after time t t
 sec, activity be 0
A then:
0 0
0
t
dN
A N
dt 
 
  
 
 
t t
t t
dN
A N
dt 
 
  
 
 
 0
t
t
A A e
 .
U
Un
ni
it
t o
of
f A
Ac
ct
ti
iv
vi
it
ty
y:
:
The activity is measured in terms of Curie (Ci). 1 curie is the activity of 1 gm of a freshly prepared sample
of radium 226
1/ 2
( 1602
Ra t  yrs.)
1 curie 10
1 3.7 10
Ci
   dps (disintegration per second)
1 dps is also known as 1 Bq (becquerel)  10
1 3.7 10
Ci   Bq
Illustration 9. Radioisotopes of phosphorus 32
P and 35
P are mixed in the ratio of 2:1 of atoms. The activity of
the sample is 2 mCi. Find the activity of the sample after 30 days. 1/2
t of 32
P = 14 days and 1/2
t
of 33
P is 25 days.
Solution: Let 0
A  initial activity of sample.
10
A  initial activity of isotope 1 & 20
A = initial activity of isotope 2.
 0 10 20
A A A
 
Similarly for final activity (Activity after time t )
1 2
t t t
A A A
 
 1 2
1 10 20
t t
t
A A e A e
 
 
Now in the given equation,
0 2
A Ci
   0 10 20 2
A A A
   ….(i)
Initial ratio of atoms of isotopes = 2 : 1
We know from definition of activity,
A N
 
 10 1 10 10 2
20 2 20 20 1
A N N T
A N N T

  

where T represents half life

19.  10
20
A 2 25 50 25
1 14 14 7
A
    …(ii)
On solving equation (i) and (ii), we get: 20 7/16
A  and 10 25/16
A 
1 2
10 20
t
A A e t A e t
 
 

0.693
0.693
30
30
25
14
25 7
16 10
t
A e e
 
 
 
How to solve expressions like this? For example consider the first exponential term:
1.483
0.693 3
14
e


Let 1.483
y e

 ln 1.483
y  

1.483
log
2.303
y    y  antilog
1.483
2.303

 
 
 
So, from above calculations you can derive a general result i.e.,
x
e
 antilog
2.303
x

 
 
 
25 7
0.2265 0.4353 0.5444
16 16
t
A Ci
     .
Illustration 10. A count- rate meter is used to measure the activity of a given sample. At one instant the meter
shows 4750 counts per minute. Five minutes later it shows 2700 counts per minute. Find:
(a) decay constant
(b) the half life of the sample.
Solution: Initial activity 0 /
A dN dt
  at 0
t 
Final activity /
t
A dN dt
  at t t

0
0
t
dN
N
dt 
  &
5
t
t
dN
N
dt 
 
 0
4750
2700 t
N
N

Using 0
2.303log
t
N
t
N
 

4750
(5) 2.303log
2700
 
 1
2.303 4750
log 0.1129min
5 2700

  
 1/ 2
0.693
6.14
0.1129
t   min.