quantum physics · 2018-08-14 · quantum physics black body: a perfect black body is one which...
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QUANTUM PHYSICS
Black body: A perfect black body is one which absorbs all the radiation of heat falling on it
and emits all the radiation when heated in an isothermal enclosure. The heat radiation emitted by the black body is called black body radiation. Laws of Black body radiation: Wien’s displacement law:
This law states that the product of the maximum wavelength (Om) corresponding
to the maximum energy and the absolute temperature is a constant.
λmT=constant λm = constant
T λm ∝
1 T
i.e. when the temperature increases the wavelength corresponding to
the maximum energy decreases. Wien’s radiation law: The maximum energy radiated at peak wavelength is directly proportional to the
fifth power of the absolute temperature (T5).
i.e. ∝ 5 = Constant 5
The energy density is derived as: = − − . − 2 = 1 −5
, Where C1 and C2 are constants. 1 = ℎ , and 2 = ℎ .
Limitation: This law holds well only for the short wavelength and not for the longer wavelength. Raleigh – Jean’s Law: S.KALPANA, DEPT OF APPLIED PHYSICS SVCE
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This law states that the energy distribution is directly proportional to absolute
temperature (T) and inversely proportional to the fourth power of wavelength
(O4).
i.e. ∝ 4 =
Limitation: This law holds well only for the longer wavelength and not for
the shorter wavelength.
Planck’s quantum theory of black body radiation:
Planck’s theory:
1. A black body contains a large number of oscillating particles: 2. Each particle is vibrating with a characteristic frequency. 3. The frequency of radiation emitted by the oscillator is the same as the
oscillator frequency.
4. The oscillator can absorb energy in multiples of small unit called quantum. 5. This quantum of radiation is called photon. 6. The energy of a photon is directly proportional to the frequency of
radiation emitted. E∝ν o E=hν
7. An oscillator vibrating with frequency can only emit energy in integral multiples of hQ. n= hν , whe e = , , , ……. . is alled ua tu u e .
Planck’ law of radiation:
The energy density of radiations emitted by a black body at a temperature T
in the wavelength range O to O+dO is given by
O O =
(
− )
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h =6.625x10-34
Js-1
- Planck’s constant.
C= 3x108 ms-1 -velocity of light
K = 1.38x 10-23 J/K -Blotzmann constant
T is the absolute temperature in kelvin.
Derivation of Planck’s law:
x Consider a black body with a large number of atomic oscillators.
x Average energy per oscillator is E = NE − − − − −
x E is the total energy of all the oscillators and N is the number of oscillators.
x Let the number of oscillators in ground state is be N0.
x According to Maxwell’s law of distribution, the number of oscillators having
an energy value EN is givenby Nn = N0e−EkTn − − − − −
x T is the absolute temperature. K is the Boltzmann constant.
x Let N0 be the number of oscillators having energy E0,
o N1 be the number of oscillators having energy E1,
o N2 be the number of oscillators having energy E2 and so on.
x Then N = N0 + N1 + N1 + ⋯ . . − − − − − − −E
0 −E1 −E
2 N = N0e kT + N0e kT + N0e kT + ⋯ . . − − − − − −
x From Planck’s theory, E can take only integral values of hQ.
x Hence the possible energy are 0, hQ, 2hQ,3 hQ……. a d so o . x . . = ℎ , ℎ = , , ,
0 = 0, 1 = ℎQ , 2 = ℎQ, 3 = ℎQ, and so on.
ℎ ℎ ℎ
= 0 + −
+ −
+ −
… ..
0 0 0 0
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x = 0 + 0 −ℎ + 0 − ℎ + 0 − ℎ … . .
Putting ℎ = − (5)
= 0 + 0 + 0 2 + 0 3 … . . = 0[1 + + 2 + 3 … . ].
= −0 ) … … … … ( using Binomial expansion)
x The total energy E = E0 N0 + E1 N1 + E2 N2 + E3 N3 …
x Substituting the value for 0 1 2 3and + + … … ( )
− −
−
=
+ Q
+ Q
+ Q … . +
= Q −
+ Q −
+ ⋯ . ± − − − − −
x Putting = −( )
= Q + Q + ⋯ . ± − − − = Q [ + + … +]
= Q [ + + … +] =
Q − − − − − −
{ − = − − = + + ⋯ }
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x Substituting (11) and (7) in (1)
Q
(
)
−
=
−
Q
= −
Q
=
(
− )
Q
=
(
− )
x Substituting the value for x
Q
=
( − )
−
Q = − − − −
( − )
x The number of oscillators per unit volume in the wavelength range O and
O+dO is given by
O
− − − − − − −
O
x Hence the energy density of radiation between the wavelength range O
and O+dO is EOdO= No. of oscillator per unit volume in the range O and
O+dO X Average energy.
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O = O Q − − − −
O O
( − )
O ⁄
O O =
O
O (
− )
O O = O
O
( − )
O =O
O
O
( − )
O = O − − − − −
O O
( − )
O = − − − − − − −
O
( − )
x The equation (16) represents Planck’s law of radiation.
x Planck’s law can also be represented in terms of frequencies.
Q = Q Q − − − − − − −
Q
( − )
Deductions of Radiation laws from Planck’s Law:
I. Wien’s Law:
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1) Wien’s law holds good when wavelength O is small.( Q is large)
2) Therefore > 1 and
is very large compared to 1.
3) Neglecting 1 in equation (16)
O = − − − −
O
4) Equation (18) represents Wien’s law. 5) Thus Planck’s law reduces to Wien’s law at shorter wavelength.
II. Raleigh- Jean’s law:
1) Raleigh- jean’s law holds good when wavelength O is large.( Q is small). ℎ
2) Therefore< 1 and expanding we get (1+ )
3) Substituting in (16)
O =
O
( +
− )
O =
O
= = S
OQ
=
= S O /O
=
S
O = O
4) Thus Planck’s law reduces to Raleigh- jean’s law at longer wavelength.
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Scattering of X-Rays:
Two Kinds:
1. Coherent scattering or classical scattering or
Thomson scattering
2. Incoherent scattering or Compton scattering
Coherent scattering:
1. X rays are scattered without any change in wavelength. 2. Obeys classical electromagnetic theory
Compton scattering:
1. Scattered beam consists of two wavelengths. 2. One is having same wavelength as the incident beam 3. The other is having a slightly longer wavelength called modified beam.
Compton scattering:
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1. When a beam of high frequency radiation is scattered by a substance of low
atomic number the scattered radiations consists of two lines.
2. One is having the same wavelength O as the incident beam 3. The other is having slightly longer wavelength. 4. This change in wavelength of the scattered X rays is known as the Compton
shift.
5. This effect is called Compton Effect.
Theory of Compton Effect:
1. Compton treated this scattering as the interaction between X ray and
the matter as a particle collision between X ray photon and loosely
bound electron in the matter.
2. Consider an X ray photon of frequency Q striking an electron at rest.
3. This Photon is scattered through an angle T to x-axis.
4. Let the frequency of the scattered photon be Q’.
5. During collision the photon gives energy to the electron. 6. This electron moves with a velocity V at an angle I to x axis. 7. Total energy before collision: Energy of the incident photon = Q
Energy of the electron at rest = = where m0 is the rest mass of electron and C the velocity of light.
Therefore total energy before collision = = Q + Total energy after collision: Final energy of the photon= = Q′ Final energy of the scattered photon = 2, where m is the mass of electron and C the velocity of light. Therefore total energy after collision = ℎQ′ + 2
By the law of conservation of energy, Total energy before collision = Total energy after collision i.e. Q + = Q′ +
2 = ℎQ − ℎQ′ + 02
= (Q − Q′) + − − − − S.KALPANA, DEPT OF APPLIED PHYSICS SVCE
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8. Total momentum along X axis before collision: Q
Initial momentum of photon along x axis = =
Initial momentum of electron along x axis = 0
Total momentum before collision along x axis= = Q
9. Total momentum along x axis after collision: The momentum is resolved along x axis and y axis.
Final momentum of momentum along x axis =
Q′ T Final momentum of electron along x axis = = I
Total final momentum along s axis = = Q′ T + I
Applying the law of conservation of momentum,
Momentum before collision = momentum after collision
Q = Q′ T +I … … .
Q − Q′
T =I
(Q − Q′ T) =I
(Q − Q′ T) = I I = (Q − Q′ T … … … .
10. Total momentum along y axis before collision: Initial momentum of photon along y axis = 0 Initial momentum of electron along y axis = 0 Total momentum before collision along y axis= 0
11. Total momentum along y axis after collision:
Final momentum of photon along y axis = = Q′ T
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Final momentum of electron along y axis == − I ( along the negative Y direction)
Total momentum after collision along y axis= = Q′
T − I
Applying the law of conservation of momentum,
Momentum before collision = momentum after collision Q′
= T − I
Squaring (3) and (4) and adding,
I + I = (Q − Q′ T) + Q′T … . .
LHS of the equation is: =I +I
= ( I + I)
= 2 2 2 since ( 2 I + 2 I ) = 1 ℎ2(Q − Q′ T)2 + ℎ2Q′ 2T
RHS of the equation: = ℎ2(Q − Q′ T)2 + ℎ2Q′ 2T = ℎ2(Q2 − QQ′ T + Q′ T + ℎ2Q′ T ℎ2(Q2 − QQ′ T + Q′ T + Q′ T) ℎ2(Q2 − QQ′ T + Q′ )
Equating LHS and RHS, 2 2 2 = ℎ2(Q2 − QQ′ T + Q′ − − − −
Squaring (1) , 2 4 = ℎ Q − Q′) + 0
2)2 − − − − 2 4 = ℎ2(Q − Q′)2 + ℎ Q − Q′) 0 2 + 0
2 4)
2 4 = ℎ2(Q2 − QQ′ + Q′ + ℎ Q − Q′) 2 + 2 4)— (8)
0 0
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(8)-(6) 2 4 − 2 2 2
= ℎ2(Q2 − QQ′ T + Q′ + ℎ Q − Q′) 0 2 + 02 4 − ℎ2(Q2 − QQ′ T + Q′ ) 2 4 − 2 2 2 = − ℎ2QQ′ + ℎ2QQ′ T + ℎ Q − Q′) 0 2 + 2 4 0
From the theory the variation of mass with velocity is given by 0
= − − √ 2 − 2
Squaring (10)
2 2 2 2
2 = 0
=
0
= 0
−
2 2 − 2
2 − 2
2 2
. . 2( 2 − 2) = 02 2 Multiplying on both sides by C2
2 2( 2 − 2) = 2 4 − − −
0
Substituting in (10)
2 4 = − ℎ2QQ′ − T + ℎ Q − Q′) 2 + 2 4
0 ℎ2QQ′ − T = ℎ Q − Q′) 0 0
2
(Q − Q′) 0
= ℎ − T )
(QQ′) 2
Q
Q
0
′ ℎ
− T )
−
=
QQ′ QQ′ 0 2
1 1 ℎ − T )
−
=
Q′ Q 2
0
Multiplying by C on both the sides,
ℎ
−
=
− T )
Q′ Q
0
O′ − O−=
ℎ − T )
0
Therefore the change in wavelength is given by
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QUANTUM PHYSICS ℎ O−= 0 − T − − − −
x The change in wavelength dO does not depend on the
i. wavelength of the incident photon ii. Nature of the scattering material.
x It depends only on the scattering angle.
Case (1) When T=0 then,
O−=
ℎ − = 0
0
Case (2) When T=0 then,
O−=
ℎ − = ℎ
0 0
Substituting the values for h,m0 and C
O−= ℎ
= 6.63 10−34
= 0.0243 0
0 9.11 10−31 3 108
ℎ )
Case(3) When T=180 then, O−= 0 − −
O−=
ℎ 2 = 0.485 0
0
The change in wavelength is maximum at 1800
Experimental verification of Compton Effect.
1. The experimental set up is as shown in the fig. 2. A beam of mono chromatic X ray beam is allowed to fall on the
scattering material.
3. The scattered beam is received by a Bragg spectrometer. 4. The intensity of the scattered beam is measured for various angles
of scattering. 5. A graph is plotted between the intensity and the wavelength. 6. Two peaks were found. 7. One belongs to unmodified and the other belongs to the modified beam. 8. The difference between the two peaks gives the shift in wavelength.
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9. When the scattering angle is increased the shift also gets increased in
accordance with ℎ
O−= 0 − T − − − −
10. The experimental values were found to be in god agreement with
that found by the formula.
Matter WAVES:
De Broglie’s Hypothesis:
1. Waves and particles are the modes of energy propagation. 2. Universe of composed of matter and radiations. 3. Since matter loves symmetry matter and waves must be symmetric. 4. If radiation like light which is a wave can act like particle some time, then
materials like particles can also act like wave some time.
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De Broglie waves and wavelength: x From Planck’s theory = ℎQ − − − − x According to Einstein’s theory, = 2 − − −
Equation (1) and (2) ℎQ = 2 − − −
x ℎ O = 2
x Therefore O = ℎ
2 O = ℎ
O = ℎ
= ℎ − − − − −
Where p is the momentum of the particle.
De Broglie wavelength in terms of energy
We know that Kinetic energy 1
= 2 2 − − −
Multiplying by m on both sides 1
= 2 2 2 2 2 = 2 √ 2 2 = √ = √ − − −
Substituting in (4) ℎ O == − − − − − √
Schrodinger wave equation
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Schrodinger equation is basic equation of matter waves.
The two forms of the wave equation are:
1. Time independent wave equation 2. Time dependent wave equation
I. Schrodinger time independent equation:
x Consider a wave associated with a particle.
x Let x,y,z be the coordinates of the particle.
x Let \be the displacement for the de Broglie wave at any time,
x The 3D wave equation for wave motion is given by:
2
+
2
+
2
=
1 2 − − −
2 2 2 2
x is the velocity of the wave.
x Equation is rewritten as 1 2
2 = − − −
x Where 2 = 2 + 2 + 2
2 2 2
x is a Laplacian’s operator.
x The solution equation of this equation is of the form ( , , , ) = 0( , , , ) − Z = 0 − Z − − − −
x Ψ0(x, y, z, t) is a function of x,y,z,t and gives the amplitude with respect to time t.
x Differentiating twice with respect to t
Ψ = − ZΨ e−iZt
0
2 = − Z − Z) 0 − Z
2 = ( 2Z2) 0 − Z 2
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QUANTUM PHYSICS 2 = −Z2 0 − Z
2 2 = −Z2 − − −
Since 0 −
Z = 0
x Substituting (4) in (2), Z2
2 = − 2 Z2
2 + 2 = − − −
x Angular frequency is Z = 2 = 2 O
x Therefore Z 2 = O − − −
x Squaring (6)
Z2 4 2
=
− − −
2 O2
2
x Substituting (7) in (5), ∇2Ψ + π Ψ = − − − −
2
O
x Putting O = ℎ
, in equation (8)
2
4 2
+
= − − −
ℎ2
2 2
2 +
4 2 2 2 = − − −
ℎ2
x If E is the energy of the particle and V is the potential energy and ½ mv2 is
the kinetic energy, then Total energy (E)= potential energy (V) + kinetic
energy (½ mv2).
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QUANTUM PHYSICS 1
= + 2 2 1 − = 2 2 − = 2
x Multiplying by on both sides, 2 2 = − − − −
x Substituting (10) in (9), ∇2Ψ + π2 − Ψ = ℎ2 π2 − ∇2Ψ + Ψ = − − − −
x This equation is called Schrodinger Time independent equation. x Substituting ℏ = 2ℎ
S , ℎ2 ℏ2 = 4S2 − − − , where ℏ is called reduced Planck’s constant. x Substituting (12) in (11), − ∇2Ψ + Ψ = − − − −
x Equation (13) has no term representing time and hence it is called Time
independent Schrodinger equation.
Special case:
The one-dimensional equation is represented as
2Ψ
+
− Ψ = − − −
2 ℏ2
II. Schrodinger time independent equation:
x Consider a wave associated with a particle.
x Let x,y,z be the coordinates of the particle.
x Let \be the displacement for the de Broglie wave at any time,
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x The 3D wave equation for wave motion is given by: 2
+ 2
+ 2
=
2 2 2
1 2
− − −
2
is the velocity of the wave. The solution equation of this equation is of the
form ( , , , ) = 0( , , , ) − Z
= 0 − Z − − − −
x Differentiating twice with respect to t
Ψ = − ZΨ e−iZt
0
= − SQ) − Z
0
= − SQ − − −
= − S
{ = ℎQ , ℎ =
}
ℎ Q
= − S
ℎ
= −
ℎ
2S
= −
− − −
ℏ
Multiplying by i on both sides
= − = − 2
ℏ ℏ
=
ℏ
= ℏ
− − −
Schrodinger time independent equation(6)
∇2Ψ + − Ψ = − − −
ℏ2
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QUANTUM PHYSICS 2 ∇2Ψ + ℏ2 Ψ − Ψ =
Substituting ( 5) in the above equaton
∇2Ψ + 2 { ℏ − Ψ} =
ℏ2
∇2Ψ = −2 { ℏ − Ψ}
ℏ2
−ℏ2 ∇2Ψ = { ℏ − Ψ}
2
−ℏ2 ∇2Ψ + Ψ = ℏ
2
[−ℏ2 ∇2 + ] Ψ = ℏ − − − −
2 HΨ = E − −
→ [ −ℏ2 ∇2 + ] is Hamiltonian operator and → ℏ is energy
2
operator.
Equation (8) is called time dependent Schrodinger equation.
Physical significance of the wave function \
x The variable \ characterizes the de Broglie waves is called wave function.
x The wave function connects the particle nature and the associated wave
nature statistically.
x It gives the probability of finding the particle at any instant.
x The probability 1 corresponds to the certainty of finding the particle at
a point at any instant. ∭ ∗ = 1 , means the particle is present
x The probability 0 corresponds to the certainty of not finding the particle at
a point at any instant. ∭ ∗ = 0 , means the particle is not present.
x The wave function is a complex quantity that cannot be measured. x The probability density is given by , = | , |2 = ∗
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Particle in a box
1. Consider a particle of mass m moving inside a one dimensional box. 2. The walls of the box are between x=0 and x=a. 3. The potential energy (V) is assumed to be 0 inside the box. 4. The potential function is : ( ) = 00 < < and ( ) = ∞ ≥ ≥ 5. This function is known as the square well potential. 6. The value of \ the wave function of the particle is found by applying the
boundary conditions to the one dimensional Schrodinger equation.
7. The Schrodinger equation is
2
+
− = − − −
2 ℏ2
8. Since V is 0 between the walls ,
2 2
+
= − − −
2 ℏ2
9. Substituting
2
= 2 (2)
ℏ2
2 + 2 = − − −
10. The general solution is of the form ( ) = + − − −
11. A and B are unknown constants.
12. Applying the boundary condition (i) \=0 at x=0, to (4) 0 = + − − −
0 = 0 + 1 = 0
ℎ 0 =
That is either A=0 or sin ka=0.
14. Since B is already zero, A cannot be zero. Therefore sin ka = 0. S.KALPANA, DEPT OF APPLIED PHYSICS SVCE
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15. Hence ka = nS, whe e = , , ,…….. o ,
=
− − −
16. Substituting (5) in (4),
( ) =
− − − −
17. Squaring (5)
2 =
2 2 − − −
18. From (2)
2
2 2 8 2 ℎ2
2 =
=
= − − ℏ2 =
ℏ2 ℎ2 ℎ2 4
2
4 2
19. Equating (7) & (8) 8 2 2 2
=
= 2 ℎ2 − − −
8 2
20. For each value of n,( n=1,2,3..) there is an energy level.
21. Thus the particle in a box can have only a discrete energy level given by (9). 22. Each energy value is called Eigen value and the corresponding
wave function is called Eigen function.
"Eigenvalue" and "eigenvector" come from the meaning "inherent, characteristic"
Normalization of Wave function:
Probability density is = ∗
The eigen function is
( ) = (6)
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Therefore
∗ =
∗ = 2
2
[
] − − −
The probability of finding the particle anywhere inside the box is given by: ∫ ∗ = 1 0
Substituting the value from (1) in (2)
2
2 [
∫
] = 1
0
−
2
∫ 2 = 1
0
1 1
2 [∫0
−
∫0 2
] = 1
2 2
2
2
[
−
] = 1
2
2
0
2
[
]
= 1
2 0
2
= 1
2
Therefore
2 = 2 = √ 2 − − − −
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Substituting in (6) , the eigen function \n belonging to is eigen values En is
expressed as
2 = √ − − −
Equation (12) is called Normalized wave function.
Special Cases:
Case1: For n=1, ℎ2 1 = 8 2
2 1 = √
Case2: For n=2, ℎ2 2 = 8 2 = 4 1
2 2 2 = √
Case3: For n=3, ℎ2 2 = 8 2 = 9 1
2 3 2 = √
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Electron microscope
Types:
1. Transmission electron microscope (TEM) 2. Scanning electron microscope (SEM) 3. Scanning transmission electron microscope (STEM)
1. Transmission electron microscope (TEM)
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Principle:
1. Accelerated primary electrons are made to pass through the specimen. 2. The image formed by using the transmitted beam or diffracted beam. 3. The transmitted beam produces bright field image 4. The diffracted image is used to produce dark field image.
Construction:
1. Electron gun: i. Electron gun produces a high energy electron beam by
thermionic emission. ii. These electrons are accelerated by the anode towards the specimen
2. Magnetic condensing lens: 1. These are coils carrying current. 2. The beam of electrons passing between two can be made to
converge or diverge.
3. The focal length can be adjusted by varying the current in the coils 4. The electron beam can be focused to a fine point on the specimen.
3. Magnetic projector lens: i. The magnetic projector lens is a diverging lens.
ii. It is placed in before the fluorescent screen 4. Fluorescent ( Phosphor) screen or Charge coupled device (CCD)
1. The image can be recorded by using florescent or Phosphor or charge
coupled device.
Working:
1. The electron beam produced by the electron gun is focused on the
specimen. 2. Based on the angle of incident the beam is partially transmitted
and partially diffracted.
3. Both the beams are combined to form phase contrast image.
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4. The diffracted beam is eliminated to increase the contrast
and brightness.
5. The magnified image is recorded in the florescent screen or CCD.
Advantages:
1. High resolution image which cannot be obtained in optical
microscope.
2. It is easy to change the focal length of the lenses by adjusting the
current.
3. Different types of image processing are possible.
Limitations:
1. TEM requires an extensive sample preparation.
2. The penetration may change the structure of the sample.
3. Time consuming process.
4. The region of analysis is too small and may not give the characteristic
of the entire sample.
5. The sample may get damaged by the electrons.
Applications:
1. It is used in the investigation of atomic structure and in various field
of science. 2. In biological applications, TEM is used to create tomographic
reconstruction of small cells or a section of a large cell.
3. In material science it is used to find the dimensions of nanotubes. 4. Used to study the defects in crystals and metals. 5. High resolution TEM (HRTEM) is used to study the crystal structure
directly.
2. Scanning electron microscope (SEM)
Principle:
1. Accelerated primary electrons are made to strike the object. 2. The secondary electrons emitted from the objects are collected by the
detector to give the three dimensional image of the object. S.KALPANA, DEPT OF APPLIED PHYSICS SVCE
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QUANTUM PHYSICS
Construction:
1. Electron gun: i. This produces a high energy electron beam by thermionic emission
ii. These electrons are accelerated by the anode towards the specimen. 2. Magnetic condensing lens:
i. These are coils carrying current. ii. The beam of electrons can be made to converge or diverge.
iii. The focal length can be adjusted by varying the current in the coils iv. The electron beam can be focused to a fine point on the specimen.
3. Scanning coil: i. This coil is placed between the condensing lens and the specimen.
ii. This is energized by a varying voltage. iii. This produces a time varying magnetic field. iv. This field deflects the beam and the specimen can be scanned
point by point.
4. Scintillator: i. This collects secondary electrons and converts into light signal.
5. Photomultiplier: i. The light signal is further amplified by photomultiplier.
6. CRO. i. Cathode ray oscilloscope produces the final image.
Working:
1. The primary electrons from the electron gun are incident on the
sample after passing through the condensing lenses and scanning coil. 2. These high speed primary electrons falling on the sample
produce secondary electrons.
3. The secondary electrons are collected by the Scintillator produces photons. 4. Photomultiplier converts these photons into electrical signal. 5. This signal after amplification is fed to the CRO, which in turn produces
the amplified image of the specimen.
S.KALPANA, DEPT OF APPLIED PHYSICS SVCE Page 29
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QUANTUM PHYSICS
Advantages:
1. SEM can be used to examine large specimen. 2. It has a large depth of focus. 3. Three dimensional image can be obtained using SEM. 4. Image can be viewed directly to study the structural details.
Limitations:
1. The resolution of the image is limited to 10-20 nm and hence it is
poor Applications:
1. SEM is used to study the structure virus, and find the method to destroy it. 2. It is used in the study of bacteria. 3. Chemical structure and composition of alloys, metals and semiconductors
can be studied.
4. SEM is used in the study of the surface structure of the material.
S.KALPANA, DEPT OF APPLIED PHYSICS SVCE Page 30
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DEPARTMENT OF APPLIED PHYSICS
IMPORTANT PROBLEMS QUESTIONS
SUBJECT CODE/NAME : PH2111/ENGINEERING PHYSICS-I
UNIT III -QUANTUM PHYSICS 1. An electron is accelerated by a potential difference of 150V. What is the wavelength of that electron wave?
(Anna University, April, 2004)
2. Calculate the De-Broglie wavelength of an electron of energy 100 eV.
(Anna University, November, 2000)
3. Find the energy of an electron moving in one dimension in an infinitely high potential box of width 0.1 nm.
(Anna University, November, 2003)
4. A neutron of mass 1.675×10−27 Kg is moving with a kinetic energy 10 KeV.
Calculate the De-Broglie wavelength associated with it. (Anna University, April, 2003)
5. An electron at rest in accelerated through a potential of 5000 V. Calculate De-Broglie wavelength of matter
wave associated with it.
6. Find the change in wavelength of an X -ray photon when it is scattered through an angle of E0 by an free
electron? (Anna University, January, 2004)
7. In a Compton scattering experiment, the incident photons have a wavelength of 3 × 10−10 m. Calculate the
wavelength of scattering photons if they are viewed at an angle of 60◦ to the direction of incidence.
(Anna University, April, 2003 and January & June 2010)
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8. Calculate the minimum energy of an electron can posses in an infinitely deep potential well of width 4 nm.
(Anna University, April, 2003)
9. An electron is trapped in a one-dimensional box of length 0.1 nm. Calculate the energy required to excite
the electron from its ground state to the sixth excited state. (Anna University, April, 2003)
10. An electron in a cubical box of metal is subjected to the influence of a magnetic Field such that its length l
increases by dl, while its breadth ‘b’ & thickness ‘d’ remains constant. Find the energy difference.
11. Calculate the energy of the electron in the energy level immediately after the lowest energy level, confined in a
cubical box of side 0.1 nm. Also find the temperature at which the average energy of the molecules of a perfect gas
would be equal to the average energy of the electron in the above said level.
(Anna University, April, 2002)
12. In Compton scattering the incident photon have wavelength 0.5 nm. Calculate the wavelength of scattered
radiation if they are viewed at an angle of 45◦ to the direction of incidence. (Anna University, June, 2009)
13. X -rays of wavelength 0.1 nm are scattered from a carbon block. Find the wavelength of the scattered beam in a
direction making an angle of 90owith the incident beam.
14. An electron is trapped in a cubical box of side 1Å. Find energy for the ground state and the first excited state
same solution for ground state.
15. Calculate the energy of an electron moving in one-dimensional is an infinitely high potential box of width 0.3 nm. If
the mass of the electron is 9.11×10−34 J.(Anna University, June, 2011)
16. Find the lowest energy of an electron confined in a box of length 0.2 nm.(Anna University, January, 2011)
17. An electron is confined to one dimensional box of side 10−10 m, obtain the first three values of the electron.
(Anna University, January, 2010)
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18. Calculate the energy in eV of a photon of wavelength 1.2 Å (Planck’s constant = 6.626 × 10−34Js and speed of
light = 3 × 108 m/s) (Anna University, January, 2011)
19. Calculate the De-Broglie wavelength associated with a photon moving with a velocity equal to 1/20th of the
velocity of the light. (Anna University, January, 1998)
20. Calculate De-Broglie wavelength of an electron accelerated to a potential of 100 volts. 21. X -ray of wavelength 0.1 nm are scattered from a carbon block. Find the wavelength of the scattered beam in the
direction making an angle of 60o with the incident beam. (Anna University, January, 2011)
22. Calculate the least energy that an electron can posses in a one directional potential box of width 0.5 nm and
infinite height. (Anna University, January, 2006)
23. Calculate Compton wavelength for an electron.
Given : Planck’s constant h = 6.626 × 10−34 J
Mass of electron = 9.11 × 10−31 Kg
Velocity of light = 3 × 108 m/s (Anna University, December, 2002
24. Find the change in wavelength of an X -ray photon when it is scattered through an angle of 135o by a free electron.
(Anna University, May, 2007)
25. The wavelength of X -ray photon is doubled when it is scattered through an angle of 90o by a target material.
Find the incident wavelength. (Anna University, January, 2009)
26. Calculate the equivalent wavelength of electron moving with a velocity of 3 × 107 m/s
(Anna University, January, 2010)
27. The De-Broglie wavelength of an electron is 1.226Å. What is the accelerating potential?
(Anna University, January, 2009) 28. What is De-Broglie wavelength of aN electron which has accelerated from rest through a potential difference
of 10 v? 29. Find energy of electron hits in units of Newton whose De-Brogle wavelength is10Å. Given mass
of neutron =1.674 × 10−27 kg. Plank’s constant h = 6.60 ×10−34 J-S.
30. Calculate De-Broglie wavelength of 100 KeV Newton. Mass of Newton may be taken as 1.675 × 10−27 kg.
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31. Compute the De-Broglie wavelength of proton whose kinetic energy is equal to half of the rest energy of electron.
Mass of proton is 1836 times that of electron. 32. A beam of 1 keV electron is passed through a thin metallic sheet whose interplanar spacing is 0.5Å. Calculate angle
of deviation of first order diffraction maximum. 33. A photon recoils back after striking an electron at rest. What is the change in the wavelength of the photon.
(0.048Å)
34. X -rays of wavelength 10−11 m are scattered from a target, find (i) wavelength of X -rays scattered through 45O. (ii) maximum wavelength present in scattered X -rays. (λ = 0.1070 & λ_ = 0.148)
35. Find the energy of an electron in one dimensional potential box of width 1000×10−13 m. (38 n2eV, E1 = 38 eV, E2 = 152 eV)
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