BBrilliant STUDY CENTRE - PHYSICS (ONLINE)

CHAPTER -
MODERN PHYSICS

Dual nature of radiation and matter

Quantum theory of light by Max Planck
In 1900, Max Planck proposed that electromagnetic radiation (light) is quantized and exist in elementary
amounts (quanta)
Energy of a photon
According to this proposal, the quantum of light wave of frequency  has the energy

E  h
h  planck’s constant , h = 6.626 x 10-34 Js

hc
OR E    wave length

c  velocity of light
c = 3 x 108 m/s

1240
E(eV) 
(nm)

Properties of photons
1. A source of radiation emits energy in the form of photons and these photons travel in straight line with
the speed of light.
2. Energy of a photon depends upon its frequency and it does not change with charge in medium.
3. With change in medium, the speed and wavelength of the photon change but frequency does not
change.
4. Photons are electrically neutral
5. Under suitable conditions, they can show diffraction.
6. A photon does not exist at rest . Its rest mass is zero.
7. Equivalent mass of photon
E = mc2
h  = mc2

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h h
m 
c 2 c
8. Photon have momentum

h h
P = mc = =
 c
9. Intensity of a light beam  no. of photons present
Number of photons emitted by a source per second
Consider a light bulb of power ‘P’ watt as the source of light energy. It the wavelength of light emitted by
the bulb is  , then energy of each photon emitted by the bulb can be given as

hc
E = h =

N  no. of photons

Nh 
p
t
 No of photons per second

N P p
 
t h hc

We can consider that all the light energy emitted by the source is uniformly distributed in the spherical
region with centre at the source.

Intensity of light
The energy crossing per unit area per unit time perpendicular to the direction of propagation.
 If ‘P’ is the power of source. Intensity at a distance ‘r’ is

P
I W /m2
4r 2

Photon Flex
The number of photons incident on a normal surface per unit area per unit time

E nh 
Intensity I = t A  tA

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 n  I I
 photon flux    
tA
  h  hc

Example :
The sun delivers about 1.4kw/m2 of electromagnetic flux to the earth’s surface. calculate
(a) the total power incident on a roof of dimensions 8 m x 20 m ?
(b) total energy incident on the roof in 1 h ?
Solution
I = 1.4 km /m2 = 1.4 x 103 w/m2

Power
I=
Area
 Power = I x Area
= 1.4 x 103 x 8 x 20
= 224 kw
(b) Total Energy

Energy
I=
time  Area
 Energy = I x time x area
= 1.4 x 1000 x [60 x 60] x [20 x 8]
= 806.4 MJ
Photon density in a light beam
 The number of photons per unit volume
For a uniform cylindrical light beam

Power P
Intensity I = 
Area A
Photon flux, ie, number of photons crossing per unit area per unit time at the cross-sectional area A is

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I
 =
hc
No. of photons per unit area in time ‘t’ second is
= t

 I 
=  t
 hc 
Volume of region crossed by photons in ‘t’ second is
V = (C x t) A
 Total no. of photon is the area A in ‘t’ second

N =  t  A

I
= (tA)
hc

N I (At) I
 Photon density = V  hc (CAt)  (hc)  c


 Photon density  c

FORCE EXERTED BY A LIGHT BEAM ON A SURFACE - Normal Incidence

A) Perfectly absorbing surface

Let P  power of source

momentum of each photon

h
P=


h
Initial momentum = 

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final momentum = 0

h
 change in momentum P =


n P P
No. of photons per second ie, ( )  
t h hc
 Total change in momentum per second

P
ie, F 
C

In terms of intensity
P= I xA

IA
 F C

E I
And radiation pressure = 
A C
B) Perfectly reflecting surface
whole amount of radiation falling on the surface is reflected

h
initial momentum of one photon = (down word)


h
final momentum of the photon = (upword)


2h
change in momentum of one photon = (up word)

Total no. of photons per unit time

n p
ie, ( ) 
t hc

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 Total change in momentum per second

dp  n  2h P 2h 2P
   
dt  t   hc  c

2P
ie, force F 
c

But P = IA

2AI
 F
c

F 2I
Radiation pressure  
A C

C) Partially absorbing and partiality reflecting surface

a  coefficient of absorption
r  coefficient of reflection

O  a  IandO  r  I
But a + r = 1

2h
Change in momentum of one photon when it is reflected =


h
Change in momentum of one photon when it is absorbed =


P
No. of photons incident per second =
hc

 P 
No. of photons reflected per second =  r
 hc 
Force on plate due to reflected photons

 P   2h  2Pr
Fr =  r    c
 hc    

 P 
No. of photons absorbed per second =  a
 hc 

 P 
=   [1  r]
 hc 

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Force on plate due to absorbed photons

 P  h P
Fa =   [1  r]  [1  r]
 hc   C
Total Force
F = Fa + Fr

2P P
= r  (1  r)
c C

P
 F = [l  r]
C
Example
A plate of mass 10 g is in equilibrium in air due to the force excrted by a light beam on the plate.
Calculate power of beam, assume that the plate is perfectly absorbing
Fphoton = Fgravity

P
= mg
C
P = mgc = 10-2 x 10 x 3 x 108 = 3 x 10+7W
Oblique Incidence
A) Perfectly absorbing surface (a =1), r = o

p  power

P
Intensity I =
A cos 

 P = I A cos 
No. of photons incident per unit time

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 n  p (IA cos )

ie,   = 
 t  hc hc
initial momentum of a photon ( at an angle with vertical)

h
=

Final momentum = 0

h
Change in momentum of a photon at an angle with vertical =

Total change is momentum per second

dp h  IA cos  
( )  
  hc 
ie, F =
dt

I A cos 
ie, F 
C

 Force on plate perpendicular to the surface is

IA
F cos  = cos2 
c
Force parallel to the surface
= Fsin 

IA cos  sin 
=
c

Normal force
 Radiation pressure =
Area

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I A Cos2  Icos2 
= 
C A c
B) Perfectly reflecting surface

here a = 0 & r = 1

P
we have I =
A cos 
P = I A cos 
No. of photons incident per second

 n  P  IA cos  
ie     
 t  hc  hc 

2h
Change in momentum of a photon in vertical direction = Cos 

Total change in momentum per second

 n  2h IA cos  2h cos 
a, F =   cos  = x
  
t hc 

2IA cos2 
F  upward direction
c

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F 2Icos2 
Radiation pressure  
A c

C) Partially absorbing or partially reflecting surface

0< r < 1 and 0 < a < 1

a r 1

2h
Change in momentum of a photon when it is reflected = cos  (vertically upward)

Change in momentum of a photon when it is absorbed

h
= (making an angle  with vertical)


 IA cos  
No. of reflected photos N1 =   r
 hc 
Force on plate due to reflected photons

 IA cos   2hcos 
Fr =   r
 hc  

2IA cos2 
Fr = r (vertically downward )
c
No of absorbed photons

 IA cos   IA cos 
N2 =   a  [1  r]
 hc  hc
Force on plate due to absorbed photons

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 IA cos   h
Fa =    (1  r)
 hc  

(IA cos )(1  r)

Fa  (making an angle  with vertical)
c

Component of force perpendicular to surface

F1 = Fr + Fa cos 

2IA cos2  IA cos2 (1  r)

ie, F1 = r
c c

IA cos2 
F1  (1  r)
C

component of force parallel to the surface

IA cos  sin 
F2 = Fasin  = [1  r]
c
Resulting force

F= F12  F22

IA cos 
F =
c
   sin  
IA cos 
=
c
1  r   
 2r cos   1  r 2  2r sin2 

IA cos 
= cos2   r 2 cos2   2r cos2  sin2   r 2 sin2   2r sin2 
c

IA cos 
F = 1  r 2  2r cos2   sin2  
c

IA cos 
F = F 1  r 2  2r cos 2
c

Radiation pressure

F1 Icos2 
=
A

c
1  r 

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Electron Emission
Work function
The minimum energy required by an electron to escape from the metal surface.
Depends on properties of the metal and the nature of the surface.
(1) Thermionic emission : - By suitably heating
(2) Field emission : - By applying very strong electric field
(3) Photo electric emission :- When light of suitable frequency illuminates a metal surface
Experimental study of photo electric effect

(A) Effect of intensity of light on photocurrent

The intensity of light is varied by changing the distance between the light source from the emitter.
The collector A is maintained at a positive potential with resput to emitter ‘C’ so that electrons ejected
from C are attracted towards collector A.
Keeping the frequency of incident light and the accelerating potential fixed, the intensity of light is varied
and the resulting photoelectric current is measured.
Photo current  no. of photoelectrons emitted

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 Photo electric current  intensity of lights

(B) Effect of potential on photoelectric current

Saturation current : - Maximum value of the photoelectric current is called saturation current. Saturation
current corresponds to the case when all the photoelectrons emitted by the emitter plate C reach the
collector plate A.
Stopping potential OR cut off potential
The minimum negative potential ( retarding potential) V0 given to the plate A for which the
photocurrent stops or becomes zero is called stopping potential.
Photoelectric current is zero when the stopping potential is sufficient to repel even the most
energetic photoelectrons, with the maximum kinetic energy.

ie, K max  eV0

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For a given frequency of the incident radiation, the stopping potential is independent of its intensity.
(C) Effect of frequency of incident radiation on stopping Potential

Intensity remains constant and frequency increases. From the graph, it is clear that greater the frequency
of incident light, greater is the maximum kinetic energy of the photo electrons.

 Stopping potential (V0) various linearly with the frequency of incident radiation for a given
photosensitive material.

 There exist a certain minimum cut off frequency 0 for which stopping potential is zero.

 For a frequency  of incident radiation, lower than the cut off frequency 0 , no photoelectric
emission is possible even if the intensity is large.

 The minimum cut-off frequency 0 is called threshold frequency. It is different for different metals
NOTE :
It is found that, if frequency of the incident radiation exceeds the threshold frequency, the photoelectric
emission starts instantaneously without any apparent time lag, even if the incident radiation is very
dim.

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Photoelectric effect and wave theory of light

According to the wave picture of light, the free electrons at the surface of the metal absorb the radiant
energy continuously. The greater the intensity of radiation, the greater are the amplitude of electric &
magnetic fields consequently, the greater the intensity, the greater should be the energy absorbed by
each electron.
 In this picture, the maximum kinetic energy of the photoelectrons on the surface is then expected
to increase with increase in intensity. Also, no matter what the frequency of radiation is , a sufficiently
intense beam of radiation should be able to impart enough energy to the electrons. A threshold
frequency, therefore, should not exist
 According to wave theory it can take hours or more for a single electron to pick up sufficient
energy to overcome the work function and come out of the metal, but photoelectric emission is
instantaneous.
Einstein’s photo electric Equation
According to Einstein, photoelectric emission does not take place by continuous absorption of energy
from radiation. Radiation energy is built up of discrete units - so called quanta of radiation.
Energy of incident photon = work function + maximum KE

h  0  kEmax  Einsteins photo electric equation

Explanation of experimental results

(1) Intensity of light  no. of photons present
 As no. of photons increases, no. of photo electrons also increases
 Intensity of light  photo current
(2) Photo electric emission is possible only when, energy of incident photon is equal to or greater
than 0
(3) As frequency of incident radiation increases, KEmax also increases. As a result stopping
potential also increases.

h  0  eV0

 eV0  h  0

 h  0
V0 =   
 e  e

h 
V0      0
e e
ie.   
y mx  c

 graph b/w frequency and stopping potential is a straight line with +ve slope and -ve Y intercept

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Wave nature of Matter

De Broglie proposed that the wavelength ‘  ’ associated with a particle of momentum ‘P’ is given as :

h h
 
p mv

Where m is the mass of the particle and V its speed.

  De Broglie wave length
The dual aspect of matter is evident in the De Broglie relation
  is smaller for a heavier particle
Example :
De Broglie wave length associated with a ball of mass 0.12 kg moving with a speed of 20 m/s is

h 6.626 1034
   2.76x1034 m
p (0.12  20)
de Broglie wavelength associated with charged particles
(1) For electron (me = 9.1 x 10-31 Kg)

h h
 
2mKE 2mqV

V  accelerating potential

12.27
ie,   A0
V

(2) For protons ( m1 = 1.67 x 10-27 Kg)

h 0.286
  A0
2mqV V

(3) For deutrons [m2 = 2 x 1.67 x 10-27 Kg]

0.202
 A0
V

(4) For  particles [m2 = 4 x 1.67 x 10-27 Kg]

0.101
 A0
V

Heisenberg’s uncertainty principle

It is not possible to measure both position and momentum of an electron [or any other particle] at the
same time exactly
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 x  uncertainty in position
 p  uncertainty in momentum

h
 x.  P = 2

De Broglie explanation
If an electron has definite momentum P, ( ie,  p= 0) by the de broglic relation, it has definite wavelength
 . A wave of definite ( single) wave length extends all over space. By max Born’s [probability interpretation
this means that the electron is not localized in any finite region of space [ ie,  x =  ]

In general, the matter wave associated with the electron is not extended all over space. It is a wave
packet extending over some finite region of space ie,  x is not infinite. Also a wavepacket of finite
extension does not have a single wavelength ie, momentum of the electron will also have an uncertainty
(  p)

Davisson and germer experiment

The wave nature of electron was first experimentally verified by CJ Davisson and L.h. Germer in 1927
and independently by G.P Thomson.

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The energy of the electrons in the primary beam, the angle at which they reach the target, and the
position of the detector could all be varied. Instead of continuous variation of scattered electron intensity
with angle, distinct maxima and minima were observed whose positions depended upon the electron
energy.

Result of the Davisson - Germer experiment, showing how the number of scattered electrons varied
with the angle between incoming beam and crystal surface
De Broglie’s hypothesis suggested that electron waves were being diffracted by the target, such as x-
rays are diffracted by planes of atoms in a crystal.
In a particular case, a beam of 54’eV electrons was directed perpendicularly at the nickel target and a
sharp maximum in the electron distribution accured at an angle of 500 with the original beam.

For constructive interference

Path difference n  2d sin 

The spacing between planes in crystal, d = 0.091nm for  = 650, for n = 1

 = 2d sin 65
= 0.165 nm  This is the wave length of electron beam
By de Broglie’s hypothesis

h 12.27
  A 0  0.166nm
2meV v
This is the predicted wave length of electron which is accelerated through a potential difference of 54V.
This wave length is in agreement with experimental value.

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ATOMS
ATOM MODEL
1) Plum pudding model: J J Thomson (1898)
The +ve charge of the atom is uniformly distributed through out the volume of the atom & -vely charged
electrons are embeded in it like seeds in a water melon.
 failed to explain the origin of spectral lines from the atom and large angle scattering of alpha particles
2) Rutherford’s Planetary Model of atom
According to this model the entire positive charge and most of the mass of the atom is concentrate in
a small volume called the nucleus with electrons revolving around the nucleus just as planets revolves
around the sun.
Rutherford’s  - ray scattering experiment
Zns screen

  particles produces bright flaster on ZnS screen.

A) Scattering angle (O)
Angle of deviation of an  -particle from its original direction is called its scattering angle.

We can show that the number of alpha particles scattered per unit area, N(  ) at scattering angle 

varies inversely as Sin 4 (  2 )

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1
N()
ie, sin (  )
4
2

B) Distance of closest Approach ( size of nucleus)

At the distance, the entire KE of  - particles is converted into electrical potential energy..
PE = KE

1 (Ze)  ze  1
 mv 2
40 r0 2
(Ze)(Ze)
r0 
1
40 ( mv 2 )
2

1
ro 
KE
C) Impact parameter (b)
It is the perpendicular distance of the initial velocity vector of the alpha particle from the central line of
the nucleus.

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 Small pmpact parameter suffers large scattering.

 The relation between impact parameter d and scattering angle  is

2 
1 Ze Cot( 2)
b
40 ( 1 mv 2 )
2

Rutherford’s Model of Hydrogen atom

Electrostatic force of attraction b/w the revolving electron and the nucleus provide required centripetal
force.

mv 2 1 e2
ie, 
r 4 0 r 2

e2
r  orbital radius
40mv 2

Orbital velocity

e2
V2 
4 0mr

e
V
40mr

KE of electron

1 1 e2 e2
KE = mv 2  m 
2 2 40mr 80r

PE of electron

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e 2
PE 
40r

Total Energy

e2
E  KE  PE 
E= 80r  electron is bound to it nucleus.

 Total energy negative means, electron is bound to the nucleus.

Draw backs of Rutherford’s Model
1) It can’t explain the stability of the atom.
2) It can’t explain the atomic spectral lines.
Bohr Model of Hydrogen Atom
Three basic postulates of this model.
1) An electron is an atom could revolve in certain stable orbits without the emission of radiant
energy.
2) Electron revolves around the nucleus only in those orbits for which the angular moment is
h
integral multiple of
2

h
Ln  n  n = 1, 2, 3..............
2

n  principal quantum number..

rn  radius nth orbit

Vn  orbital speed of nth orbit

h
ie, mVnrn  n 
2

3) An electron might make a transition from one of its specified non - radiating orbits to another of lower
energy, by the emission of a photon with energy.

E  Ei  Ef

Bohr Model of Hydrogen Atom

Radius of orbit

mv n2 1 e2

rn 40 rn2

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e2
rn   (1)
4 0 mv n2

But we have

h
mv nrn  n
2

h
v n  n   2
2mrn

Substitute (2) in (1)

e2  4 2m2rn2
rn 
4 0m  n2h2

e2  mrn
I
 0 n2 h 2

  h2 
rn   0 2  n2
 me 

Where a0 = Bohr radius

  h2 
a0 =  0 2  = 0.53 A0
 me 
rn  n2
Velocity of orbit
We have

nh
rn   (3)
2mVn

nh  h2
  0 2 n2
2mVn me

1 0 h  n 2
 n
2Vn e2

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e2 1
Vn  
2 0h n

OR

e2 c
Vn = 
20hc n

1 c
Vn  
137 n

1
Vn 
n
C  velocity of light in vacuum

e2 1

20 hc 137

Total Energy

 e2
En =
80  n

 e2   mc 2
ie, En =
8 0 0h2 n2

 me4
En 
8 0 2 h2 n 2

 13.6 eV
ie, En 
n2

If n = 1  ground state
E1 = -13.6 eV
If n = 2  Ist excited state
E2 = -1.51 eV
If n = 4  3rd excited state
E4 = -0.85 eV
If n = 5  4th excited state
E5 = -0.54 eV

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For hydrogen like atoms

Atoms possess only one electron orbiting around the nucleus
eg : He+ , Li2+, Be3+
Radius of orbit

1
rn = a0 n2  z  atomic number
z
Velocity

 1 c
Vn  z    
 137 n 
Energy

2  13.6 
En = Z  2 
 n 
Ionization Energy
The minimum energy required to free the electron from the ground state of the hydrogen atom is called
ionization energy.
 ionization energy = + 13.6 eV
Atomic Spectra
Each element is associated with a characteristic spectrum of radiation
a) Emission spectrum
When an atomic gas or vapour is excited at low pressure, usually by passing an electric current
through it, the emitted radiation has a spectrum which contains certain specific wavelengths only. A
spectrum of this kind is termed as emission line spectrum. It consists of bright lines on a dark background.
b) Absorption spectrum
When light passes through a gas some are missing in the transmitted light. An arrangement of missing
wavelength is called absorption spectrum.
 It consist of dark lines on a white background
Emission spectrum of hydrogen

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Spectral series
The spacing between lines within certain sets of the hydrogen spectrum decreases in a regular way.
Each of these set is called a spectral series.
Lyman series [ultra violet region]

1 1 1
= R  2  2  n  2,3, 4............
 1 n 
R  Rydbery constant
R = 1.097 x 107 m-1

1 1
  RC  2  2  n  2,3,4..........
1 n 

Balmer Series [Visible region]

1 1 1
= R  2  2  , n = 3, 4, 5 ..........
 2 n 
Paschen series [I R region]

1 1 1
= R  2  2  , n = 4, 5, 6........
 3 n 
Brackett series : [IR region]

1 1 1
= R  2  2  , n = 5, 6, 7 ........
 4 n 
Pfund series : [For infrarad region]

1 1 1

=R  52  n2  , n = 6, 7, 8 ............
 
Explanation of line spectra of hydrogen atom
According to the third postulate of Bohr’s model, when an electron makes a transition from the higher
energy state with quantum number ni to the lower energy state with quantum number nf
then
h  = Eni - Enf

me4 me4
 
802h2ni2 820 h2n2f

me4  1 1 
h  = 82h2  n2  n2 
0  f i 

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BBrilliant STUDY CENTRE - PHYSICS (ONLINE)

hc me4 1 1
= 82h2C  2  2
 0  nf ni 

1 me4  1 1
 2 3  2  2
 80h C  nf ni 

me4
Where  1.03 x 107 m1
802h3 C

This is the value very close to 1.09  107 m-1. This agreement between theorectical and experimental
values of Rydberg constant provided a direct and striking confirmation of Bohr’s model.
Origin of spectral lines.

De Broglie’s Explanation of Bohr’s second postulate of quantization.

Louis de Broglie argued that the electron in its circular orbit must be seen as a particle wave. In
analogy to waves travelling on a string, particle waves too can lead to standing waves under resonant
conditions.
For an electron moving in nth circular orbit of radius rn, the total distance is the circumference of the
orbit, 2rn  n   n  1,2,3..........

where   de Broglie wave length of electron is nth orbit.

h

mVn

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h
 2  rn = n  mV
n

h
 mVn rn = n 
2

h
ie, Ln  n 
2

This de Broglie hypothesis provided an explanation for Bohr’s second postulate for the quantization of
angular momentum of the orbiting electron. The quantized electron orbits and energy states are due to
th wave nature of the electron and only resonant standing waves was persist.

Figure illustrates a standing particle wave on a circular orbit of n = 5

Limitation of Bohr’s Model
1) The Bohr model is applicable to hydrogen like atoms.
2) This model is unable to explain the relative intensities of the frequencies in the spectrum. Experimental
observations depicts that some transitions are more favoured than others
3) Bohr’s model could not explain hyperfine structure of spectral lines
4) It could not account for splitting of spectral live in magnetic and electric fields
Note :
1) H  - line
First line of Balmer series
n = 3 n = 2

1 1 1  9  4  R
   R  22  3 2   36
  

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36
 =
5R

2) H - line

(n = 4)  n = 2

1 1 1 1 1   4  1
R 2  2 R   R 
 2 4   4 16   16 

1 3R 16
   
  16 3R

Series limit of Balmer series

n =   n =2

1  1 1 R
R 2  
 2  4

4

R
Nuclei
Atomic mass unit (u) OR amu
th
 1
One atomic mass unit is defined as   of the mass of carbon (C12) atom.
 12 

1 12
IU = [C ]  1.66  10 27 Kg
12
 Accurate measurement of atomic masses is carried out with a mass spectrometer. Nucleus
contains more than 99.9% of the mass of an atom
Composition of nucleus
 mass of proton mp = 1.672 x 10-27 kg
 mass of neutron mN = 1.674 x 10-27 kg
free neutron is unstable

neutron  e- + proton + 


antineutrino
The composition of a nucleus can be described as

z XA OR zA X

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z  atomic no : no. of protons.

A  mass no : no. of nucleons
A) Isotopes
Atoms which have the same atomic number, but different mass number
eg :

1H1, 1H2 , 1H3

2 He3 , 2He4

B) Isobars
Atoms which have same mass number (A) but different atomic number (Z)

eg : 1H3 , 2He4
C) Isotones
Which contains same number of neutrons

eg : 17 Cl37 and 19 K 39

Size of Nucleus
 Nucleus of mass number A has radius
1 15
R  Ro A 3 Ro  1.2  10 m

Nuclear density

mass UA 3U
 
 = volume 4 R3 A 4R0
3

0
3
 = 2.3 x 10-27 kg/m3
Nuclear density is always remains a constant
Mass Energy Relation
Einstein showed that mass is an another form of energy.
Energy equivalent of mass m is

E  mc 2 C  3x108 m / s

Mass defect
The difference between the mass of nucleus and the sum of the masses of its nucleons is called

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mass defect.

M  [zmp  (A  z)mn ]  m

mp  mass of a proton.
mn  mass of a neutron.
m  mass of nucleus.
Binding Energy of a nucleus
If a certain number of neutrons and protons are brought together to form a nucleus, an energy Eb will
be released in this process. This energy is called binding energy of the nucleus. If we separate a
nucleus into its nucleons, we would have to supply a total energy equal to Eb, to those particles.

 Eb  M  C
2

Binding Energy per nucleon (Ebn)

Average energy per nucleon needed to separate a nucleus into its individual nucleons OR energy
released per nucleon during the formation of nucleus.

Eb
 Ebn 
A

 Ebn is maximum (8.75 MeV) for Fe56

 Ebn is lower for both light nuclei (A<30) and heavy nuclei (A>170)
 Fission
When a nucleus with (A = 240) breaks in to two nuclei, nucleons get more tightly bound. This implies
energy would be released in this process

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Nuclear fusion
When two light nuclei joining to form a heavier nucleus energy is also released. Since stability increases.
Nuclear force
Nuclear forces are the strong forces of attraction which hold together the nucleons.
Properties
1. Nuclear forces are independent of charge.
2. Strangest force in nature
3. Very short range force
4. Nuclear forces show saturation properties, each nucleon interact with its immediate neighbours only.
5. Non - conservative force.
Radioactivity
Henry Bequerel discovered - 1896
Radioactivity was a nuclear phenomenon in which as unstable nucleus undergoes a decay. Three
type of radioactive decay occur in nature
1)  - decay
In which helium nucleus is emitted from unstable nucleus.
Peretrating power is small.
 In  - decay, mass number of product nucleus is decreased by four and atomic number is decreased
by two
A
z X  Z  z Y A  4  2He4

eg : 92 U238  90 Th234  2He4

Spontaneous decay is possible only when :

Total mass of the decay products is less than the mass of the initial nucleus. This difference in
mass appears as kinetic energy of products.
Disintegration energy OR Q - valve

Q = Mx  My  M  C
2

OR
Q = KE of products
Q = KE of Y + KE of alpha {initially the parent nuclei at rest}
By conservation of momentum

MY Vy = M  V

M V
 Vy = My ___ (1)

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1 1
Q= My Vy 2  M V2
2 2

1 M2  V2 1
Q = 2 y M2  2 M V
2
M
y

1 M 
Q= M V2    1
2  My 

 mx  my 
Q = K  M 
 y 

 my 
 K   Q
 M  My 

 A4 
K   Q
4  A  4

A  4
K   Q
 A 

and

A  4
Ky = Q - K  = Q 1 
 A 

A  A  4
Ky = Q  
 A 

4
Ky  Q  
A

2)  - decay
Two types

a)  decay
In which electrons are emitted from unstable nucleus
Nuclear process is

n  p + e  

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In  decay mass number remains constant, but atomic number increases by 1.

b)  decay
In which positions are emitted from unstable nucleous.
Nuclear process is:

p  n  e  
This process is possible only inside the nucleus since, proton has smaller mass than neutron.

During  decay atomic number decreases by 1, and mass number remains same.

c) Gamma Decay 
In which high energy photons are emitted from unstable nucleous.
Like an atom, a nucleous also has discrete energy levels - the ground state and excited states.
Nucleus in an excited state spontaneously decays to its lower energy state (or ground state) by the
emission of photons with energy equal to difference in the two energy levels of the nucleus,. This is
called gamma decay.
A gamma ray is emitted when  or  decay results in a daughter nucles in an excited state. This then
returns to the ground state by a single photon transition or successive transitions involving more than
one photon. A familiar example is the successive emission of gamma rays of energies 1.17 MeV and
1.33 MeV from the deexcitation of 28 Ni60 formed from  decay of 27 Co60

Law of radioactive decay

In any radioactive sample, which undergoes  ,  or  decay, number of nuclei undergoing the decay
per unit time is propotional to the total no. of nuclei in the sample at that time.

dN
ie, N
dt
N  No. of nuclei in the sample
OR

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dN
 N {-ve singn indicates as time increases no. of nuclei decreases}
dt

  decay constant, OR disintegration constant.

N t
dN
N0 N  0 dt
No  initial no. of nucelous

InNNo
N
  t

ln (N) - ln (No) = - t

N
ln   t
No

N
 et
N0

t
N  NO e

 No of nuclei remain undecayed

ie, No. of nuclei in the sample exponentially decays

Half Life (T1/2)

Time required to decay half the number of nucles initially present in the sample

No
When t = T12 , N 
2

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No T 1
ie,  N0 e 2
2

2-1 = e - T 12

2=

ln (2) = T12

ln(2) 0.693
T1  
2  

Mean life OR Average life (  )

Total life time of all radioactive nucleiin the sample

Mean life (  ) =
Total no. of nuclei in t he sample

No  initial no. of nuclei

N  no of nuclei present at time ‘t’
Let ‘dN’ be the no. of nuclei decays in next dt second. Then life time of each ‘dN’ nuclei is (t +d t)  t
 Total age of ‘dN’ nuclei = dN (t)
Total life time of all the atoms in the sample
No

=  t dN
0

No

AV life =  t dN
0

dN
  N
dt

dN =  Ndt   [No et ]dt

By simplifying we get

1



Relation between  & T12

T1  0.693
2

  1.44T1
2

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Activity
The rate of disintegration is activity (A)

 dN 
A    N
 dt 

OR A  No e t

ie, A  A o et where Ao = No

Activity also decreases expontially with time. SI unit of activity is becquerel (Bq)
1Bq = 1 decay / second
Another unit is curie

1curie  1ci  3.7  1010 Bq

Nuclear Fission
Nuclear fission is the phenomenon of splitting of certain heavy nuclei into two nearly equal fragments.

0 n1  U235  92U236  56 Ba144  36kr 89  30 n1  200MeV

92

The fragment product are radioactive nuclei, they emf  particles in succession to achieve stable end
products.
The three neutrons released can produce further fission and the process is going on, resulting in chain
reaction. An uncontrolled chain reaction results in explosion (atom bomb), and a controlled chain reaction
gives controlled release of energy & is achieved in nuclear reactors.

{Only thermal reactions can ca fission in 92 U235 than fast neutrons}

Main parts of a nuclear reactor

1) Nuclear fuel : U235, Pu239
2) Moderator
They are used to slow down neutrons
Eg : Heavy water, graphite, parafine)
3) Control rods : (Cadmium, Boron rods)
Neutrons are absorbed by control rods. Reaction rate is controlled through it.
Multiplication factor (k)

Rate of production of neutrons

K=
Rate of loss of neutrons

If k = 1

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Reactor is said to be critical (steady power generating)

If k > 1
Reactor is said to be super critical
(neutron population will grow exponentially)
If K < 1
Reactor is said to be subcritical
Neutron population will exponentially decay
4) Coolant :
Which is used to remove the heat produced during fission and transfer it from the core of the nuclear
reactor to the surroundings.
Eg : Heavy water, CO2, or liquid sodium.
The coolant passes through the heat exchanger, where it passes heat to water & converts it to steam.
The steam drives a turbine and generates electric power.
Nuclear Fusion
The phenomenon of fusing of two or more lighter nuclei to form a single heavy nucleus. Mass defect in
the process appears as energy.
Thermo nuclear fusion
When fusion is achieved by raising the temperature of the system so that particles have enough
kinetic energy to overcome the coulomb repulsive behaviour, it is called thermonuclear fusion.
Energy production in sun
Fusion reaction in the sun is a multi-step process in which the hydrogen is burned in to helium.
Proton - proton cycle

1H1 1 H1 1 H2  e    0.42MeV  (1)

C  e       1.02 MeV   2 

1H2  1H1  2He3    5.49MeV  (3)

2 He3  2He3  2 He4  1H1  1H1  12.86MeV  (4)

For the fourth reaction to occur, the Ist 3 reaction must occur twice.
2(1) + 2(2) + 2(3) + (4) 

4 1H1  2e  2 He4  2  2   26.7 MeV

ie, Four hydrogen combins to form a helium with a release of 26.7 MeV energy.
Pair production
A photon materialize into as electron and a positron. Rest mass energy of e- or e+ = 0.54 MeV

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BBrilliant STUDY CENTRE - PHYSICS (ONLINE)

 Pair production requires a photon energy of at least 1.02 MeV. Any additional photon energy becomes
kinetic energy of the electron and positron.
Pair annihilaion
e- + e+    
Note :
A radioactive nucleous can decay by two different processes. The half life for the Ist process is T1 and
that for the second process is T2. The effective half life T of the nucleus is given by :

1 1 1
 
T T1 T2

proof:
Decay constant for the first process is given by :

0.693
1   1
T1

for second process

0.693
2    2
T2

The probability that an active nucleus decays by the first process in small time dt is ( 1dt)

Similarly, the probability that it decays by the second process is (  2dt) .

The probability that it decays either by first process or second process is 1dt +  2 dt

If  is the effective decay constant

 dt = 1dt +  2 dt

0.693 0.693 0.693

 
T T1 T2

1 1 1
 
T T1 T2

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X-RAYS
X -ray was discovered by the German physicist W C Roentgen in 1895. X - rays are produced when
highly energetic (fast moving) electrons are made to strike a metal target. A device used to produce X-
rays is generally called an X-ray tube or coolidge tuve.
Electromagnetic radiation with wavelength from 0.1A0 to 100 A0 falls in to the category of X-rays. The
boundaries of this category are not sharp. The shorter wavelength end overlaps gamma rays and the
longer wave length end overlaps ultraviolet rays.

When electrons callide with the atoms of solid, they loose their kinetic energy which is converted into
enrgy in the form of X-rays.
The melting point, specific heat capacity, and atomic number of the target should be high.
When electrons striks with the target, some part of energy is lost and converted into heat. since the
target should not melt or it should absorb the heat, so the melting point and specific heat of the target
should be high.
The nature of emitted X-rays depends on :
1) The current in the filament F
2) Voltage between the filament and the anode
(a) An increase in the filament current increases the number of electrons it emits. Larger number of
electrons means an intense beam of x-rays is produced.

1
(b) An increase in the voltage of the tube increases the kinetic energy of electrons [eV = mv2].
2
When such highly energetic beam of electrons is suddenly stopped by the target, an energetic
beam of x-rays is produced. This way we can control the quality of x-rays, ie , penetrating power
of x-rays.

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BBrilliant STUDY CENTRE - PHYSICS (ONLINE)

(c) Based on penetrating power, X -rays are classified into two types : Hard X-rays, and soft X-rays.
The ones having high energy and hence high penetration power are Hard x-rays and the ones
with law energy and hence low penetration power are soft.
Properties of X-rays
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 positive charge.
3. They cause fluorescence in several materials.
4. They are not deflected by electric and magnetic fields.
5. They affect photographic plates.
6. They show all the properties of the waves except refraction.
Continuous X-rays
Electromagnetic theory predicts that an accelerated electric charge will radiate electromagnetic waves,
and a rapidly moving electrons when suddenly brought to rest is certainly accelerated. X-rays produced
under these circumstances is given the German name bremsstrahlung (braking radiation).
The continues X-rays OR (bremsstrahlung X-rays) produced at a given accelerating potential ‘V’ vary
in wavelength, but none has a wavelength shorter than a certain value  min . This minimum wavelength
corresponds to the maximum energy of the X-rays which in turn is equal to the maximum kinetic
energy qV or eV of striking electrons.

hc
 min 
eV

This wavelength also known as the cut off wavelength or the threshold wavelength.

Characteristic X-rays
The X-ray spectrum typically consists of a broad continuous band containing a series of sharp lines.
The sharp lines super imposed in the continuous spectrum are known as characteristic X-rays, because
they are characteristic of the target material.

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BBrilliant STUDY CENTRE - PHYSICS (ONLINE)

Characteristic X-ray emission occurs when a bombarding electron that collide with a target atom has
suffecient energy to remove and inner shell electron from the atom. the vacancy created in the shell
is filled when an electron from a higher level drops down into it. This transition is accompanied by the
emission of a photon whose energy equals the difference in energy between the two levels.

Let us assume that the incoming electron dislodged an electron from the innermost shell ( K shell.) If
the vecancy is filled by an electron dropping from the next higher -shell ( L shell), then corresponding X-
ray is called K  . If the vacancy is filled by an electron dropping from the m-shell, the k  line is produced.
An L  line is produced as an electron drops from the M shell to the L-shell, and m L line is produced
by a transition from the N-shell to the L-shell.

  min depends only on accelerating voltage applied between the target and filament. It does not depend
upon material of the target, it is same for two different metals.
Mosley’s Law for characteristic spectrum
Although multi electron atoms cannot be analyzed with the Bohr model, Henry moseley in 1914 made

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BBrilliant STUDY CENTRE - PHYSICS (ONLINE)

an effort towards this. Moseley measured the frequencies of characteristic X-rays from a large number
of elements & plotted the square root of the frequency  against the atomic number Z of the element.

 Frequency of X-rays is given by :

 1 1
  CR  2  2   z  b 
 n1 n2 

b  screening constant or shielding constant

For K  line (b =1)

 1 1
k   CR  2   (z  1)
1 4

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