Class 12 Physics Quantization of energy Notes

Chapter 23 – Quantization of Energy | Nepal eNotes
PHYSICS • CHAPTER 23

Quantization of Energy

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Quantization of Energy

Bohr’s Postulates

  • An electron revolving round the nucleus has angular momentum equal to integral multiple of h/2π, i.e.
L = n h / 2π
  • An electron revolving round the nucleus doesn’t emit the energy continuously. It emit energy when it jumps from higher orbit to lower orbit and absorb energy when it jumps from higher to lower orbit.
The second postulate above is preserved as written in the source, including the repeated “higher to lower” wording.

Bohr’s Theory of Hydrogen Atom

rₙ vₙ e⁻

Suppose an electron is revolving round the nucleus in an orbit of radius rn with velocity vn.

The electrostatic force between electron and nucleus provides centripetal force.

Electrostatic force = centripetal force

1/(4πε0) · e²/rn² = mvn²/rn
rn = e²/(4πε0mvn²)    …(1)

From Bohr’s postulate:

mvnrn = nh/2π
vn = nh/(2πmrn)    …(2)

Putting value of vn in equation (1):

rn = e² / [4πε0m(nh/2πmrn)²]
rn = e² × 4π²m²rn² / (4πε0mn²h²)
e²πmrn = ε0n²h²
rn = ε0n²h² / (πme²)

Velocity of Electron in nth Orbit

Putting rn value in equation (2):

vn = nh / [2πm(ε0n²h²/πme²)]
vn = e²/(2ε0nh)

Expression for Energy of Electron in nth Orbit

An electron revolving around the nucleus possesses K.E. due to its motion and P.E. due to force of attraction between electron and nucleus. So, total energy of electron is the sum of its K.E. and P.E.

T.E. = K.E. + P.E.
K.E. = 1/2 mvn²
= 1/2 m [e²/(2ε0nh)]²
K.E. = me⁴/(8ε0²n²h²)

And,

P.E. = −e²/(4πε0rn)
= −e² / [4πε0(ε0n²h²/πme²)]
P.E. = −me⁴/(4ε0²n²h²)

Therefore,

T.E. = K.E. + P.E.
= me⁴/(8ε0²n²h²) − me⁴/(4ε0²n²h²)
En = −me⁴/(8ε0²n²h²)
NOTE: When electron jumps from higher energy state E2 to lower energy state E1, the energy released is given by:
hf = E2 − E1
hf = −me⁴/(8ε0²h²n2²) + me⁴/(8ε0²h²n1²)
hf = me⁴/(8ε0²h²) [1/n1² − 1/n2²]
f = me⁴/(8ε0²h³) [1/n1² − 1/n2²]
c/λ = me⁴/(8ε0²h³) [1/n1² − 1/n2²]
1/λ = R [1/n1² − 1/n2²]

where R is Rydberg’s constant.

Spectral Series

n=1 n=2 n=3 n=4 n=5 n=6 n=∞ Lyman series Balmer series Paschen series Brackett series Pfund series

1) Lyman Series

This series is obtained when an electron jumps from higher energy state to ground state (n = 1).

For this series n1 = 1, n2 = 2, 3, 4, … ∞

1/λ = R [1/1² − 1/n2²]

2) Balmer Series

This series is obtained when an electron jumps from higher energy state to first excited state (n = 2).

For this series n1 = 2, n2 = 3, 4, 5, …

1/λ = R [1/4 − 1/n2²]

3) Paschen Series

This series is obtained when an electron jumps from higher energy state to second excited state (n = 3).

For this series n1 = 3, n2 = 4, 5, 6, …

1/λ = R [1/9 − 1/n2²]

4) Brackett Series

This series is obtained when an electron jumps from higher energy state to third excited state (n = 4).

For this series n1 = 4, n2 = 5, 6, 7, …

1/λ = R [1/16 − 1/n2²]

5) Pfund Series

This series is obtained when an electron jumps from higher energy state to fourth excited state.

For this series n1 = 5, n2 = 6, 7, 8, …

1/λ = R [1/25 − 1/n2²]

de-Broglie Theory

It states that a particle moving in an orbit has both wave and particle nature. The wave associated with matter is called matter wave or de-Broglie wave.

The wavelength of matter wave is given by:

λ = h/mv
Since, K.E. = 1/2 mv²
v = √(2E/m)
λ = h/√(2mE)

Heisenberg’s Uncertainty Principle

It states that it is impossible to measure position of the particle and momentum, energy and time accurately and simultaneously.

If Δx and Δp are uncertainty in position and momentum, then:

Δx · Δp ≥ h/2π

And if ΔE and Δt are uncertainty in energy and time, then:

ΔE · Δt ≥ h/2π

X-Rays

X-rays are electromagnetic radiation having very short wavelength in the order of 1 Å to 100 Å. X-rays are produced when fast moving electrons strike a metal having high atomic no and high melting point.

Properties

  • They are electromagnetic radiation having very short wavelength.
  • They are chargeless.
  • They are not deflected in electric and magnetic field.
  • They have high penetrating power.
  • They can ionize the gas through which they pass.
  • They can cast shadow of an object on which they fall.

Production of X-Rays (Coolidge Tube)

F C T cold water copper tube X-rays electrons

The experimental arrangement is as shown in fig. It consists of an evacuated glass tube having a filament (F) connected to low tension battery and a target metal (T) connected to a copper tube in which water is circulating continuously. A high tension battery is connected between filament and target metal.

The filament gets heated due to low tension battery and electrons are emitted. The emitted electrons are accelerated by high tension battery…

…and strike the target metal. On striking almost 98% of energy of electrons is converted into heat and remaining energy is converted into X-rays.

NOTE
  1. Control of intensity of X-rays: Intensity of X-rays depends on no. of electrons striking the target metal, so by adjusting filament current intensity of X-rays can be controlled.
  2. Control of penetrating power of X-rays (quality): The quality of X-rays depend on energy of electron striking the target, so by adjusting voltage between filament and target quality of X-rays can be improved.

Diffraction of X-Ray (Laue’s Experiment)

X-ray source slits ZnS photographic plate

The experimental arrangement is as shown in figure. The X-rays produced from X-ray source are collimated by using slits and are allowed to strike ZnS crystal. When photographic plate is sufficiently exposed to X-rays, a pattern is observed with central bright spot and many faint spots. This concluded that X-rays show diffraction and prove wave nature of X-rays.

Bragg’s Law

It states that the intensity of reflected X-rays from different layers will be maximum if path difference between the waves is integral multiple of wavelength of X-ray.

d θ θ

Consider two beams of X-rays are incident on two layers of crystal as shown in figure. The path difference between waves is MB′ + B′N.

MB′ = d sinθ
B′N = d sinθ
So, path difference = d sinθ + d sinθ
= 2d sinθ

For maximum intensity:

Path difference = nλ
2d sinθ = nλ

Diffraction

It is the phenomenon of the bending of light around the corners of an obstacle if the size is comparable to the wavelength of light.

Types of Diffraction

Fresnel Diffraction: The source and screen are at finite distance from the obstacle. In general one or both of the incident wavefront is spherical or cylindrical.

Fraunhofer Diffraction: The source and screen are at infinite distance. Lenses are used to make the incident light parallel and the wavefront is plane.

Fraunhofer Diffraction at Single Slit

A B O d screen P θ y

Consider a plane wavefront AB is incident on a slit of width d. According to Huygens’s theory each point on AB act as source of disturbance.

The point O is middle of wavefront whose perpendicular centre of screen is zero, so central maxima is obtained.

Let P is any point at angle θ at which position of maxima or minima is to be determined.

The path difference between waves from A and B on reaching P is BM.

In ΔABM:

sinθ = BM/AB
BM = AB sinθ
BM = d sinθ

Position of Minima

Path difference = nλ
d sinθ = nλ
For very small θ, sinθ ≈ θ
dθ = nλ
θn = nλ/d

Position of Maxima

Path difference = (2n + 1)λ/2
d sinθ = (2n + 1)λ/2
dθ = (2n + 1)λ/2
θn = (2n + 1)λ/(2d)

Width of Central Maxima

d 2y D θ

The separation between two 1st order minima gives width of central maxima.

Since, for 1st minima:

d sinθ = λ
For small θ, θ = λ/d    …(1)

And from figure:

tanθ = y/D
For very small θ, tanθ ≈ θ
θ = y/D    …(2)

From (1) and (2):

y/D = λ/d
y = λD/d
Width of central maxima (2y) = 2λD/d

Diffraction Grating

It is an arrangement having large number of equal sized slits separated by equal sized opaque space. If a represent size of slit and b represent size of opaque space and d is called grating element.

d = a + b

No. of lines per unit length of grating is given by N:

a + b = 1/N

In diffraction grating maxima is obtained if:

d sinθ = nλ

n = order of diffraction.

Numerical Problems

1)

N = 500 lines/mm
n = 2
θ = 30°
Here, d sinθ = nλ
(1/500) × (1/2) = 2λ
1/1000 = 2λ
1 × 10−3 = 2λ
λ = 5 × 10−4 mm

2)

N = 600 lines/mm = 600/10−3 = 600000 lines/m
λ = 700 nm = 700 × 10−9 m
n = 1

Using d sinθ = nλ:

(1/N) sinθ = 1 × 700 × 10−9
sinθ = 700 × 10−9 × 600000
sinθ = 0.42
θ = sin−1(0.42) = 24.83°

For maximum order, θ = 90°:

(1/N) sin90° = n × 700 × 10−9
1/600000 = n × 700 × 10−9
1.66 × 10−6 = n × 700 × 10−9
n = 2.37

3)

Width of central maxima: 2y = 2λD/d
= [2 × 5000 × 10−10 × 3.5] / [0.01 × 10−3]
= 0.35

4)

λ = 5880 Å = 5880 × 10−10 = 5.88 × 10−7 m
N = 6000/cm = 600000/m
n = 2
d sinθ = nλ
(1/N) sinθ = nλ
(1/600000) sinθ = 2 × 5.89 × 10−7
sinθ = 1.178 × 10−6 × 600000
sinθ ≈ 0.7068
θ ≈ 45°

Additional Numerical

N = ?
λ = 5.893 × 10−7 m
θ = 27° 42′ = 27 + 42/60 = 27.7°
d sinθ = nλ
(1/N) sin27.7° = 1 × 5.893 × 10−7
N = sin27.7° / (5.893 × 10−7)
N ≈ 780587.13
The numerical working above is transcribed from the final page. The handwritten intermediate sine value is faint, so the clearly written final value has been retained.

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