Education

1. QUANTUM PHYSICS

2. Compton Effect
When high frequency monochromatic radiation like gamma rays or x - rays are incident on a substance
and get scattered, then the scattered radiation has two components, one of which has the same
wavelength as of incident radiation and the other component has a wavelength greater than that of
incident wavelength. This phenomenon is called as Compton effect
Before
collision
After
collision
E = hn
Eʹ = hn ʹ
After
collision
Kinetic energy of e- = E - Eʹ
Before
collision 𝞴ʹ > 𝞴
The component having the same wavelength
as incident radiation is said to be unmodified
component and the component having greater
wavelength is called as modified component
Compton effect can be explained by using Planck’s
quantum theory of radiation. Compton effect is due to
elastic collision between the incident photons and
electrons present in scattering substance.

3. If a photon of energy hn strikes a substance, a part of photon energy is transferred to the electron in the
substance and the electron gains kinetic energy and starts to recoil.
The remaining photon energy comes out as scattered radiation, so scattered radiation has energy less
than that of incident radiation. Hence the scattered radiation has a frequency less than that of incident
radiation and wavelength greater than incident radiation.
The wavelength of scattered photon can be determined by using energy and momentum conservation
principles.
ɵ
ɸ
Incident Photon
Recoil Electron
Electron
at Rest
E = hn
P =
𝒉
𝞴
=
𝒉ʋ
𝒄
E = m0c2
P = zero
E =𝒉ʋ′
P =
𝒉
𝞴′ =
𝒉ʋ′
𝒄
E = mc2
P = mv
Consider a photon with energy hn striking a substance, the
photon collides elastically with an electron having a rest mass mo
present in the substance. The electron gains kinetic energy and
moves with velocity v and relativistic mass m. The scattered
photon has a frequency n ’ and energy hn ’

4. Now we shall consider the energy and momentum values of the photon and electron before and after collision.
Before collision
• Energy of photon and electron are hn and m0c2
• Momentum of photon and electron are
𝐡ʋ
𝐜
and zero
After collision
• Energy of photon and electron are hʋ′
and mc2
• Momentum of photon and electron are
𝐡ʋ′
𝒄
and mv
Using the principle of conservation of energy and momentum
Energy before collision = Energy after collision
h n + moc2 = h n ’ + mc2
h(n – n ’) + moc2 = mc2
Squaring on both sides
h2(n – n ’)2 + mo
2c4 + 2h(n – n ’)moc2 = m2c4
h2(n2+ n ’2 - 2 n n ’) + mo
2c4 +2h(n – n ’)moc2 = m2c4
h2 n 2+ h2 n ’2 - 2 h2 n n ’ + mo
2c4 + 2h(n – n ’)moc2 = m2c4
(1)
[h(n – n ’) + moc2]2 = m2c4

5. Momentum before collision = Momentum after collision
𝐡ʋ′
𝒄
𝐡ʋ′
𝒄
sin ɵ
𝐡ʋ′
𝒄
cos ɵ
ɵ
𝐡ʋ
𝐜
+ o =
𝐡ʋ′
𝐜
cos ɵ + mv cos ɸ
mv cos ɸ =
𝐡ʋ
𝐜
-
𝐡ʋ′
𝐜
cos ɵ
mv cos ɸ =
𝐡
𝐜
(n – n I cos ɵ)
mvc cos ɸ = h (n - n I cos ɵ)
Squaring on both sides
m2v2c2 cos2 ɸ = h2 [n 2 + n I 2 cos2 ɵ − 2 n n I cos ɵ]
[mvc cos ɸ ]2= [h (n - n I cos ɵ)]2
m2v2c2 cos2 ɸ = h2 n 2 + h2 n I 2 cos2 ɵ − 2h2 n n I cos ɵ (2)
ɸ
mv cos ɸ
mv
sin
ɸ
mv
Momentum, along incident photon
direction after collision
Momentum, along incident photon
direction before collision =

6. Momentum, perpendicular to direction of
incident photon after collision
Momentum, perpendicular to direction of
incident photon before collision =
0 + 0 =
𝐡ʋ′
𝐜
sin ɵ − mv sin ɸ
mv sin ɸ =
𝐡ʋ′
𝐜
sin ɵ
mvc sin ɸ = 𝐡ʋ′sin ɵ
Squaring on both sides
m2v2c2 sin2 ɸ = 𝐡2 ʋ ′𝟐
sin2 ɵ (3)
Add equations (2) and (3)
m2v2c2 cos2 ɸ + m2v2c2 sin2 ɸ = h2 n 2 + h2 n I 2 cos2 ɵ - 2h2 n n I cos ɵ + 𝐡2n ′𝟐 sin2 ɵ
m2v2c2 (cos2 ɸ + sin2 ɸ) = h2 n 2 + h2 n I 2 (cos2 ɵ + sin2 ɵ)- 2h2 n n I cos ɵ
m2v2c2 = h2 n 2 + h2 n I 2 - 2h2 n n I cos ɵ (4)

7. Subtracting equation (4) from equation (1)
m2c4 − m2v2c2 = h2 n 2 + h2 n ’2 − 2 h2 n n ’ + mo
2c4 + 2h(n – n ’)moc2 − [ h2 n 2 + h2 n I 2 − 2h2 n n I cos ɵ]
m2c2 (c2 − v2) = h2 n 2 + h2 n ’2 − 2 h2 n n ’ + mo
2c4 +2h (n – n ’)moc2 − h2 n 2 − h2 n I 2 + 2h2 n n I cos ɵ
m2c2 (c2 − v2) = − 2 h2 n n ’ + mo
2c4 +2h (n – n ’)moc2 + 2h2 n n I cos ɵ
m2c2 (c2 − v2) = − 2 h2 n n ’ ( 1 − cos ɵ) + mo
2c4 + 2h (n – n ’)moc2
From theory of relativity the rest mass m0 is related to the mass m when the particle is moving is given by
𝐦 =
𝐦𝟎
𝟏 −
𝐯𝟐
𝐜𝟐
Squaring
𝒎𝟐
=
𝐦𝟎
𝟐
(𝟏 −
𝐯𝟐
𝐜𝟐)
𝒎𝟐
(𝟏 −
𝐯𝟐
𝐜𝟐) = 𝐦𝟎
𝟐
𝒎𝟐(
𝐜𝟐
−𝐯𝟐
𝐜𝟐 ) = 𝐦𝟎
𝟐
𝒎𝟐(c2 – v2) = 𝒎𝟎
𝟐 𝒄𝟐
(5)

8. 𝒎𝟐
(
𝐜𝟐
−𝐯𝟐
𝐜𝟐 ) = 𝐦𝟎
𝟐
𝒎𝟐
(c2 – v2) = 𝒎𝟎
𝟐 𝒄𝟐
Substituting the value of 𝒎𝟐(c2 – v2) in equation (5)
𝒎𝟎
𝟐 𝒄𝟐 x c2 = − 2 h2 n n ’ ( 1 − cos ɵ) + mo
2c4 + 2h (n – n ’)moc2
𝒎𝟎
𝟐 𝒄𝟒 = − 2 h2 n n ’ ( 1 − cos ɵ) + mo
2c4 + 2h (n – n ’)moc2
2 h2 n n ’ ( 1 − cos ɵ) = 2h (n – n ’)moc2
h n n ’ ( 1 − cos ɵ) = (n – n ’)moc2
(n – n ’)
n n ’ =
h
moc2 ( 1 − cos ɵ)
n
n n ’
−
n ’
n n ’
=
h
moc2 ( 1 − cos ɵ)

9. 1
n ’
−
1
n
=
h
moc2 ( 1 − cos ɵ)
Multiply by c
c
n ’
−
c
n
=
hc
moc2 ( 1 − cos ɵ)
𝞴 ’ – 𝞴 =
h
moc
( 1 − cos ɵ)
As,
c
n ’
= 𝞴 ’ and
c
n
= 𝞴
Where 𝞴 ’ is the wavelength of the scattered photon and 𝞴 is the wavelength of the incident photon.
ɵ is the angle through which the incident photon is scattered.
1. When ɵ = zero, 𝞴 ’ – 𝞴 = zero, i.e., there is no scattering along the direction of incidence.
𝞴 ’ – 𝞴 =
h
moc
( 1 − cos 90°)
𝞴 ’ – 𝞴 =
h
moc
𝞴 ’ – 𝞴 =
6.63 x 10−34
9 x 10−31x 3 x108
𝞴 ’ – 𝞴 = 0.02426 x10-10 m= 0.02426 Å
This difference in wavelength is a constant and is called as Compton wavelength
2. When, ɵ =
π
𝟐

10. 3. When, ɵ = π
𝞴 ’ – 𝞴 =
h
moc
( 1 − cos 180°)
𝞴 ’ – 𝞴 =
2h
moc
𝞴 ’ – 𝞴 =
2 x 6.63 x 10−34
9 x 10−31x 3 x 108
𝞴 ’ – 𝞴 = 0.04852 x 10-10 m = 0.04852 Å
𝞴 ’ = 𝞴 +
h
moc
( 1 − cos ɵ)
In Compton scattering as ɵ varies from 0° to 180 ° , the wavelength of the scattered photon
varies from 𝞴 to 𝞴 +
𝟐 h
moc
, provided the wavelength of the incident photon is sufficiently small.
cos 180° = -1

11. A high frequency radiation of wavelength 3Å undergoes Compton’s scattering. Find the wavelength
of the photon scattered along direction which makes an angle 60o with incident angle direction.
Problem and solution
𝞴 ’ – 𝞴 =
h
moc
( 1 − cos ɵ)
𝞴 ’ – 𝞴 =
6.63 x 10−34
9 x 10−31x 3 x 108
( 1 − cos 60°)
𝞴 ’ – 𝞴 =
6.63 x 10−34
9 x 10−31x 3 x 108
( 1 −
𝟏
𝟐
)
𝞴 ’ – 𝞴 = 0.0121 x 10-10
𝞴 ’ = 𝞴 + 0.0121 x 10-10
𝞴 ’ = 3 x 10-10 + 0.0121 x 10-10
𝞴 ’ = 3.0121 x 10-10 m
𝞴 ’ = 3.0121 Å

12. Matter waves or de Broglie waves
When tiny particles (like electron) are moving they have a wave nature associated with them and
the wave associated with moving particle is called as matter wave or de Broglie wave.
The de Broglie wavelength is given by
𝞴 =
𝒉
𝒎𝒗
Where mv is the momentum(P) of the moving particle
Derivation of de Broglie wavelength for matter waves
According to Einstein's mass energy relation
E = mc2
Also, E = hn
mc2 = hn
mc2 =
𝐡𝐜
𝝺
mc =
𝐡
𝝺
𝞴 =
𝒉
𝒎𝒗
Where c = v and v is the velocity of the moving particle

13. For the particles which are accelerated by applying a voltage V, the de Broglie wavelength is
𝝺 =
𝐡
𝟐𝐦𝐞𝐕
For the particles which are moving with kinetic energy E, the de Broglie wavelength is
𝝺 =
𝐡
𝟐𝐦𝐄
Physical significance of wave function (𝞇)
1. The wave function (𝞇) is the wave displacement of the de Broglie wave.
2. It is a variable quantity which characterises de Broglie wave.
3. The wave function relates the wave nature and the particle nature statistically.
4. It is a complex quantity and we cannot measure it.
5. It describes the state of de Broglie wave.
6. The square of the modulus of wave function (I𝞇I2) gives the probability of finding the particle
in a given volume and is called as probability density (P)
7. The probability of the finding particle in the given volume dxdydz is given by,
8. If the particle is definitely present in the volume, then P = 1 and this condition is known as
Normalisation condition
P = I𝞇I2 dxdydz
P = I𝞇I2 dxdydz = 1

14. If the particle is not present in the given volume then,
P = I𝞇I2 dxdydz = 0
Also, I𝞇I2 = I𝞇𝞇∗I
𝞇∗ is the complex conjugate of 𝞇
Consider a particle of mass m moving with a wave velocity u. According to de Broglie hypothesis the
moving particle has a wave nature associated with it and a wave equation is required to describe the
moving particle. Let x, y, z be Cartesian coordinates of the particle and 𝞇 be the wave function which is the
wave displacement of the de Broglie wave.
Schrodinger's time independent wave equation
The wave equation in partial differential form can be given as,
𝐝𝟐𝞇
𝐝𝐱𝟐 +
𝐝𝟐𝞇
𝐝𝐲𝟐 +
𝐝𝟐𝞇
𝐝𝒛𝟐 =
𝟏
𝒖𝟐
𝐝𝟐𝞇
𝐝𝒕𝟐
𝐝𝟐𝞇
𝐝𝐱𝟐 +
𝐝𝟐𝞇
𝐝𝐲𝟐 +
𝐝𝟐𝞇
𝐝𝒛𝟐 −
𝟏
𝒖𝟐
𝐝𝟐𝞇
𝐝𝒕𝟐 = 0 (1)
The solution of the above equation will be of the form
𝞇 = 𝞇𝟎𝐞−𝐢𝛚𝐭

15. 𝒅𝟁
𝒅𝒕
= -i 𝟂 𝞇𝟎𝐞−𝐢𝛚𝐭
𝒅𝟐
𝟁
𝒅𝒕𝟐 = − (− i2𝟂2)𝞇𝟎𝐞−𝐢𝛚𝐭
𝒅𝟐
𝟁
𝒅𝒕𝟐 = − 𝟂2 𝞇 𝐀𝐬, 𝞇 = 𝞇𝟎𝐞−𝐢𝛚𝐭
Substituting the above value of
𝒅𝟐
𝟁
𝒅𝒕𝟐 in equation (1)
𝐝𝟐𝞇
𝐝𝐱𝟐 +
𝐝𝟐𝞇
𝐝𝐲𝟐 +
𝐝𝟐𝞇
𝐝𝒛𝟐 −
𝟏
𝒖𝟐 (− 𝟂2 𝞇) = 0
As, 𝟂 = 2πn, 𝟂2 = 4π2n2
𝐝𝟐𝞇
𝐝𝐱𝟐 +
𝐝𝟐𝞇
𝐝𝐲𝟐 +
𝐝𝟐𝞇
𝐝𝒛𝟐 +
4π2n2𝞇
𝒖𝟐 = 0
As, u = n l
𝐝𝟐𝞇
𝐝𝐱𝟐 +
𝐝𝟐𝞇
𝐝𝐲𝟐 +
𝐝𝟐𝞇
𝐝𝒛𝟐 +
4π2n2𝞇
n2 l
𝟐 = 0

16. 𝐝𝟐𝞇
𝐝𝐱𝟐 +
𝐝𝟐𝞇
𝐝𝐲𝟐 +
𝐝𝟐𝞇
𝐝𝒛𝟐 +
4π2𝞇
l
𝟐 = 0
As, 𝞴 =
𝒉
𝒎𝒗
𝐝𝟐𝞇
𝐝𝐱𝟐 +
𝒅𝟐𝞇
𝒅𝐲𝟐 +
𝒅𝟐𝞇
𝒅𝒛𝟐 +
4π2𝐦𝟐
𝐯𝟐
𝞇
h
𝟐 = 0 (2)
For a moving particle, Total energy = Potential energy + Kinetic energy
E =
𝟏
𝟐
𝒎𝒗𝟐 + V
E – V =
𝟏
𝟐
𝒎𝒗𝟐
2(E – V) = 𝒎𝒗𝟐
Multiplying by m
2m(E – V) = 𝒎𝟐𝒗𝟐
Substituting 𝐟𝐨𝐫 𝒎𝟐𝒗𝟐 in equation (2)
𝒅𝟐𝝍
𝒅𝒙𝟐 +
𝒅𝟐𝝍
𝒅𝒚𝟐 +
𝒅𝟐𝝍
𝒅𝒛𝟐 +
𝟖𝝅𝟐
𝒎(𝑬−𝑽)𝝍
𝒉𝟐 = 𝟎
This equation is Schrodinger's time independent wave equation

17. 𝒅𝟐
𝒅𝒙𝟐 +
𝒅𝟐
𝒅𝒚𝟐 +
𝒅𝟐
𝒅𝒛𝟐 = ▽ 2
Generally
▽
2
iscalled as Laplace operator.
So Schrodinger's equation can be written as
▽ 2
ψ +
𝟖𝝅𝟐
𝒎
𝒉𝟐 𝑬 − 𝑽 𝝍 = 𝟎
In quantum mechanics
ħ =
𝒉
𝟐𝝅
h = ħ 𝟐𝝅
h 2 = 4 𝝅𝟐 ħ 2
▽ 2
ψ +
𝟖𝝅𝟐
𝒎 𝑬−𝑽 𝝍
4 𝝅𝟐
ħ 2 = 𝟎
Substituting for h2 in the above equation
▽ 2
ψ +
𝟐 𝒎 𝑬−𝑽 𝝍
ħ 2 = 𝟎
If the particle is a free particle its potential
energy is zero(V=0) and so the Schrodinger's
time independent equation becomes
▽ 2
ψ +
𝟐 𝒎𝑬 𝝍
ħ 2 = 𝟎
This is another form of Schrodinger's time
independent equation and here the potential
energy and force acting on the particle are
independent of time.

18. Schrodinger's time dependent wave equation
Consider a particle of mass m moving with a velocity and having a total energy E and potential energy
V. The Schrodinger's time dependent wave equation is given as
▽ 2
ψ +
𝟐 𝒎 𝑬−𝑽 𝝍
ħ 2 = 𝟎
The wave function ψ is given by
𝝍 = 𝝍0 𝒆−𝒊𝝎𝒕
𝐝𝛙
𝐝𝐭
= - i ω 𝝍
As, ω = 2πn
𝒅𝝍
𝒅𝒕
= - i 2πn 𝝍
As, E = hn, n =
𝑬
𝒉
𝒅𝝍
𝒅𝒕
=
− i 2π E 𝝍
𝒉
▽ 2
ψ +
𝟐 𝒎𝑬 𝝍
ħ 2 −
𝟐 𝒎𝑽 𝝍
ħ 2 = 𝟎 (1)
[= - i ω 𝝍0 𝒆−𝒊𝝎𝒕
]

19. ħ =
𝒉
𝟐𝝅
𝐝𝛙
𝐝𝐭
=
− i E 𝝍
ħ
Multiply and divide by i
𝐝𝛙
𝐝𝐭
=
− i2 E 𝝍
i ħ
𝐝𝛙
𝐝𝐭
=
E 𝝍
i ħ
i2 = - 1
E 𝝍 = i ħ
𝐝𝛙
𝐝𝐭
Substituting for E 𝝍 in equation (1)
▽ 2
ψ +
𝟐 𝒎
ħ 2 i ħ
𝐝𝛙
𝐝𝐭
−
𝟐 𝒎𝑽 𝝍
ħ 2 =
𝟎
As,
Multiplying by
ħ 2
𝟐𝒎
▽ 2
ψ +
𝟐 𝒎
ħ 2 i ħ
𝐝𝛙
𝐝𝐭
−
𝟐 𝒎𝑽 𝝍
ħ 2 = 𝟎
ħ 2
𝟐𝒎
▽ 2
ψ + i ħ
𝐝𝛙
𝐝 𝐭
− 𝑽 ψ = 𝟎
ħ 2
𝟐𝒎
▽ 2
ψ − 𝑽 ψ = − i ħ
𝐝𝛙
𝐝 𝐭
−
ħ 2
𝟐𝒎
▽ 2
ψ + 𝑽 ψ = i ħ
𝐝𝛙
𝐝 𝐭
(V −
ħ 2
𝟐𝒎
▽ 2
)ψ = i ħ
𝐝𝛙
𝐝 𝐭
This is Schrodinger's time dependent wave equation.

20. (V −
ħ 2
𝟐𝒎
▽
2
) = H H is called as Hamiltonian Operator
i ħ
𝐝𝛙
𝐝 𝐭
= 𝑬𝛙
E is the Energy Operator
𝑯 𝛙 = 𝑬 𝛙
This is another form of Schrodinger's time dependent wave equation.
Here, potential energy depends on time.
(V −
ħ 2
𝟐𝒎
▽ 2
)ψ = i ħ
𝐝𝛙
𝐝 𝐭
So the above equation can be written as

21. Application of Schrodinger's equation
Particle in a box or in a one dimensional potential well
Consider a particle of mass m confined to a box or potential well and the particle motion is restricted to
one direction motion only (say x - direction).
The width of box is L and particle can move within the box back and forth only along x – direction.
Inside the box, the potential energy (V) of a particle is zero and at the walls and outside the box, the
potential energy is infinite.
V = 0, 0 < x < L
V = Infinity, 0 ≥ x ≥ L
The wave function ψ
IψI 2 ≠ 0 0 < x < L
IψI 2 = 0 0 ≥ x ≥ L
IψI 2 is valid only if the particle is definitely present and as the particle is there within the box it has a
finite value within the box (0 < x < L ) and it is zero outside the box and at the walls of the box as the
particle is not present outside the box (0 ≥ x ≥ L)

22. The Schrodinger's equation for the particle will be of the form
▽ 2
ψ +
𝟖𝝅𝟐
𝒎
𝒉𝟐 𝑬 − 𝑽 𝝍 = 𝟎
But as the particle is a free particle its potential energy(V) is equal to zero and as the movement is restricted
to one direction only (x direction), only
𝒅𝟐 𝝍
𝒅𝒙𝟐 term will be there. So the equation becomes
𝒅𝟐 𝝍
𝒅𝒙𝟐 +
𝟖𝝅𝟐
𝒎
𝒉𝟐 𝑬 𝝍 = 𝟎
Let,
𝟖𝝅𝟐
𝒎
𝒉𝟐 𝑬 = 𝒌𝟐
𝒅𝟐 𝝍
𝒅𝒙𝟐 + 𝐤𝟐𝝍 = 𝟎
Solution of the above equation will be of the form
ψ = A sin k x + B cos k x
Where A and B can be evaluated by applying the boundary conditions x = 0 and L and A is the wave amplitude.

23. 1. When x = 0, IψI 2 = 0 and hence ψ = 0
ψ = A sin k x + B cos k x
0 = A sin k 0 + B cos k 0
B = 0
2. When x = L, ψ = 0
ψ = A sin k x + B cos k x
0 = A sin k L + 0 cos k L
0 = A sin k L
As the particle is present inside the box, A is not equal to zero, so sin k L = 0
sin k L = 0
k L= nπ (n = 1,2,3,4 …….)

24. k =
𝒏𝝅
𝑳
𝟖𝝅𝟐𝒎
𝒉𝟐
𝑬 = 𝒌𝟐
𝟖𝝅𝟐𝒎
𝒉𝟐 𝑬 =
𝒏𝟐𝝅𝟐
𝑳𝟐
𝟖𝒎
𝒉𝟐 𝑬 =
𝒏𝟐
𝑳𝟐
𝐄 =
𝐧𝟐𝐡𝟐
𝟖𝐦𝐋𝟐
E is the energy of the particle moving in a one dimensional potential well or box
When n = 1, E = E1
𝐄𝟏 =
𝐡𝟐
𝟖𝐦𝐋𝟐
When n = 2, E = E2
𝐄𝟏 =
𝟒𝐡𝟐
𝟖𝐦𝐋𝟐
When n = 3, E = E3
𝐄𝟏 =
𝟗𝐡𝟐
𝟖𝐦𝐋𝟐
As,

25. So the energy of the particle is not continuous, it cannot have all possible energy values. It can have
only discrete energy values. Hence, the energy of the particle moving in an one dimensional
potential well or box is quantized.
The electron or particle moving in a box or one dimensional potential well can have energy according
to the equation
𝐄𝐧 =
𝐧𝟐𝐡𝟐
𝟖𝐦𝐋𝟐
The energy values are discrete and quantized and these allowed or permissible energy values which
the particle can have are called as eigen values.
ψ = A sin
𝒏𝝅𝒙
𝑳
The wave function is given by
Here, B = 0 and k =
𝒏𝝅
𝑳
ψ = A sin k x + B cos k x
The value of A can be evaluated by using the normalisation condition
− ∞
∞
𝐈𝟁𝐥𝟐𝒅𝒙 = 𝟏

26. 𝟎
𝑳
A2 sin2 (
𝒏𝝅𝒙
𝑳
) 𝒅𝒙 = 𝟏
𝐀𝐬, 𝐬𝐢𝐧𝟐ɵ =
𝟏 −𝒄𝒐𝒔 𝟐ɵ
𝟐
L
0
∫A2
𝟏 − 𝐜𝐨𝐬(
𝟐𝐧𝛑𝐱
𝐋
)
𝟐
𝒅𝒙 = 𝟏
L
0
∫ 𝒅𝒙 = 𝟏
𝐀𝟐
𝟐
𝟏 − 𝐜𝐨𝐬(
𝟐𝐧𝛑𝐱
𝐋
)
L
0
= 𝟏
𝐀𝟐
𝟐
x −
𝒔𝒊𝒏(
𝟐𝐧𝛑𝐱
𝑳
)
(
𝟐𝐧𝛑
𝐋
)
= 𝟏
𝐀𝟐
𝟐
L −
𝒔𝒊𝒏(𝟐𝒏π)
(
𝟐𝐧𝛑
𝐋
)

27. 𝑨𝟐
𝟐
L = 1
𝑨𝟐
=
𝟐
𝑳
𝐀 =
𝟐
𝐋
So the wave function can be written as (after substituting for A)
ψ =
𝟐
𝑳
sin
𝒏𝝅𝒙
𝑳
This is the wave function corresponding to the particle moving in one dimensional potential well or box.
When n = 1, 𝟁 = 𝟁1
ψ1 =
𝟐
𝑳
sin
𝝅𝒙
𝑳
When n = 2, 𝟁 = 𝟁2
ψ3 =
𝟐
𝑳
sin
𝟑𝝅𝒙
𝑳
When n = 3, 𝟁 = 𝟁3
ψ2 =
𝟐
𝑳
sin
𝟐𝝅𝒙
𝑳

28. ψn =
𝟐
𝑳
sin
𝒏 𝝅 𝒙
𝑳
In general for a particle or electron moving in a one
dimensional potential well or box the wave function is
These wave functions are discrete in nature and these
allowed or permitted wave functions of the particle or the
electron are known as eigen functions.
ψ5
ψ4
ψ3
ψ2
ψ1
x

29. Calculate the first three allowed energy levels of an electron confined to an one dimensional box
of width 1 Å
𝐄𝐧 =
𝐧𝟐𝐡𝟐
𝟖𝐦𝐋𝟐
1. First energy level, n =1
𝐄𝟏 =
𝟏x (𝟔. 𝟔𝟑 x 𝟏𝟎
− 𝟑𝟒
)
𝟐
𝟖 x 𝟗 x 𝟏𝟎 − 𝟑𝟏 (
𝟏x𝟏𝟎 − 𝟏𝟎)𝟐
𝐄𝟏 =
𝟒𝟑. 𝟗𝟓 x 𝟏𝟎
− 𝟔𝟖
𝟕𝟐 x 𝟏𝟎 − 𝟓𝟏
E1= 0.610 x 10-17 J
𝐄𝟏 =
𝟎. 𝟔𝟏𝟎x 𝟏𝟎
− 𝟏𝟕
𝟏. 𝟔 x 𝟏𝟎 − 𝟏𝟗
E1= 38.12 eV

30. 2. Second energy level, n = 2
En = n2 E1
E2 = 22 E1
E2 = 4 x E1
E2 = 4 x 38.12
E2 = 152.48 eV
3. Third energy level, n = 3
En = n2 E1
E2 = 32 E1
E2 = 9 x E1
E2 = 9 x 38.12
E2 = 343.08 eV

31. Electron
Detector
Photo-
multiplier
Video
amplifier
Scan
generator
Secondary
electrons
Scanning coil
Electron beam
Sample CRO
(magnetic)
It is used to study the surface morphology
of materials. In SEM accelerated electrons
are made to scan the material to be studied.
Secondary electrons are emitted from the
material and these electrons are used to
produce the magnified image of the object.
Magnification is up to the order of more
than 105 times.
Scanning Electron Microscope (SEM)
Electrons emitted from the electron
gun are accelerated and then
focused and are made to fall on
sample by using magnetic lens
assembly. A scanning coil is present
in the path of electrons and this
scanning coil makes the electrons to
scan each and every point of sample.
In response to incident electrons the sample emits secondary electrons.

32. Applications
• To study surface morphology of materials.
• To study about atomic or molecular
arrangement in solids (crystal structures).
• To study the texture of textile fibres.
• Also, to study micro-organism like virus
bacteria etc.
The secondary electrons have with them the information about the point from which they are emitted.
These secondary electrons are collected by an electron detector and fed to a scintillator device which
produces large number of photons as output in proportional to received electrons. The scintillator output
is given to a photomultiplier which gives an amplified electron signal which carries with it information
about sample. The electron signal is fed to the CRT and 3D image of the object is reproduced on the
fluorescent screen of CRT. The image can also be recorded and stored in computer.

33. It is used to study the crystal structure of materials
and also about micro-organisms. TEM can produce
a magnified image of the object which has been
magnified by the order of 2 x 105 times.
Transmission Electron Microscope (TEM)
Electrons emitted from the electron gun are
suitably accelerated and then made to fall on the
object by using appropriate magnetic lens
assembly.
Magnetic
condensing
lens
Projection
lens
Objective
lens
Final magnified
image
Electron gun
Intermediate image
The electron passing through the object depends on
the nature of the medium of travel. More electrons
traverse through a region of object, if that region is
of less density and less number of electrons pass
through the region having higher density. The
electrons coming out of the sample carry with them
the information about the region through which
they have traversed. This image can be recorded in
computer.

34. The electrons passing through the object are used to produce a magnified image (intermediate
image) by using objective lens. Then with the help of projection lens the intermediate image
obtained is further magnified to a greater extent and hence a highly magnified image is
obtained on screen. This image can also be recorded in a computer.

35. THANK YOU