Light and Quantum Physics: Black Body Radiation, Photoelectric Effect & Compton Scattering

House of Physics October 02, 2025
Light and Quantum Physics: Black Body Radiation, Photoelectric Effect & Compton Scattering | HRK Chapter 49

Light and Quantum Physics: Complete Guide

Black Body Radiation, Planck's Quantum Theory, Photoelectric Effect, Einstein's Photon Theory, and Compton Effect
Quantum Physics Black Body Radiation Photoelectric Effect Compton Effect Planck's Quantum Theory Reading Time: 30 min

📋 Table of Contents

📜 Historical Background: The Birth of Quantum Physics

The development of quantum physics in the early 20th century revolutionized our understanding of light and matter:

  • Max Planck (1900): Proposed quantum hypothesis to explain black body radiation
  • Albert Einstein (1905): Explained photoelectric effect using photon concept
  • Arthur Compton (1923): Demonstrated particle nature of light through X-ray scattering
  • Niels Bohr (1913): Developed quantum model of the atom

These discoveries marked a fundamental shift from classical physics and established the foundation of quantum mechanics.

Introduction to Light and Quantum Physics

🔬 The Dual Nature of Light

Light exhibits both wave-like and particle-like properties, a concept known as wave-particle duality:

  • Wave Behavior: Reflection, refraction, interference, diffraction, and polarization
  • Particle Behavior: Black body radiation, photoelectric effect, Compton scattering

This chapter explores experiments that can only be understood by considering electromagnetic radiation as a stream of particles called photons.

📝 The Quantum Revolution

The early 20th century witnessed a revolution in physics with the development of quantum theory, which:

  • Challenged classical physics' continuous energy concept
  • Introduced discrete energy quanta
  • Explained phenomena that classical physics failed to address
  • Laid the foundation for modern physics and technology

Thermal Radiations and Black Body

🔥 What are Thermal Radiations?

Thermal radiations are electromagnetic waves emitted by a body due to its temperature. All bodies:

  • Emit thermal radiation due to their temperature
  • Absorb thermal radiation from their surroundings
  • Reach thermal equilibrium when emission rate equals absorption rate

The characteristics of thermal radiation depend on the body's temperature, material, shape, and surface nature.

Cavity Radiator or Black Body

⚫ Black Body Definition

A black body is an ideal object that:

  • Absorbs all radiation incident upon it
  • Has maximum emission and absorption rates
  • Has emission spectrum dependent only on temperature

While no perfect black body exists, a small hole in a cavity with blackened inner walls closely approximates one.

⚙️ Black Body Approximation

BLACKENED CAVITY
↓ Small Hole
APPROXIMATE BLACK BODY

Working Principle: Radiation entering the small hole undergoes multiple reflections inside the cavity, with minimal chance of escaping. This makes the hole behave like a nearly perfect black body.

Key Property: For an ideal black body, the emission spectrum depends only on temperature, not on the material, surface nature, size, or shape of the body.

Radiant Intensity and Stefan-Boltzmann Law

📊 Radiant Intensity

Radiant intensity \( I(t) \) is defined as the energy emitted per unit area per unit time over all wavelengths. It represents the power radiated per unit area.

🧮 Stefan-Boltzmann Law

Law Statement

The radiant intensity of a black body is directly proportional to the fourth power of its absolute temperature:

\[ I(t) \propto T^4 \]
\[ I(t) = \sigma T^4 \]

Stefan-Boltzmann Constant

Where \( \sigma \) is the Stefan-Boltzmann constant:

\[ \sigma = 5.67 \times 10^{-8} \, \text{W/m}^2\text{K}^4 \]

Spectral Radiancy and Wien's Displacement Law

🌈 Spectral Radiancy

Spectral radiancy \( R(\lambda) \) is defined as the radiant intensity per unit wavelength at a given temperature:

\[ R(\lambda) = \frac{dI(t)}{d\lambda} \]

The product \( R(\lambda)d\lambda \) gives the energy emitted per unit area per unit time in the wavelength range from \( \lambda \) to \( \lambda + d\lambda \).

🧮 Total Radiant Intensity

Integration Over Wavelengths

The total radiant intensity at a particular temperature is obtained by integrating spectral radiancy over all wavelengths:

\[ I(t) = \int_{0}^{\infty} R(\lambda)d\lambda \]

🧮 Wien's Displacement Law

Law Statement

The wavelength corresponding to maximum spectral radiancy is inversely proportional to the absolute temperature:

\[ \lambda_{max} \propto \frac{1}{T} \]
\[ \lambda_{max} T = \text{Constant} \]

Constant Value

The constant in Wien's displacement law has the value:

\[ \lambda_{max} T = 2.89 \times 10^{-3} \, \text{m·K} \]
Sample Problem 1: Star Temperatures and Radiant Intensity

Calculate the surface temperatures, radiant intensity, and total radiated power for the following stars:

Star \(\lambda_{max}\) Radius
Sirius 240 nm Not given
Sun 500 nm \(7 \times 10^8\) m
Betelgeuse 850 nm \(4 \times 10^{11}\) m
Using Wien's displacement law: \( \lambda_{max} T = 2.89 \times 10^{-3} \)
For Sirius:
\[ T = \frac{2.89 \times 10^{-3}}{240 \times 10^{-9}} \]
\[ = \frac{2.89 \times 10^{-3}}{2.4 \times 10^{-7}} \]
\[ = 1.204 \times 10^4 \, \text{K} \]
For Sun:
\[ T = \frac{2.89 \times 10^{-3}}{500 \times 10^{-9}} \]
\[ = \frac{2.89 \times 10^{-3}}{5 \times 10^{-7}} \]
\[ = 5780 \, \text{K} \]
For Betelgeuse:
\[ T = \frac{2.89 \times 10^{-3}}{850 \times 10^{-9}} \]
\[ = \frac{2.89 \times 10^{-3}}{8.5 \times 10^{-7}} \]
\[ = 3400 \, \text{K} \]
Using Stefan-Boltzmann law: \( I(t) = \sigma T^4 \), \( \sigma = 5.67 \times 10^{-8} \, \text{W/m}^2\text{K}^4 \)
For Sirius:
\[ I(t) = 5.67 \times 10^{-8} \times (1.204 \times 10^4)^4 \]
\[ = 5.67 \times 10^{-8} \times 2.10 \times 10^{16} \]
\[ = 1.19 \times 10^9 \, \text{W/m}^2 \]
For Sun:
\[ I(t) = 5.67 \times 10^{-8} \times (5780)^4 \]
\[ = 5.67 \times 10^{-8} \times 1.12 \times 10^{15} \]
\[ = 6.35 \times 10^7 \, \text{W/m}^2 \]
For Betelgeuse:
\[ I(t) = 5.67 \times 10^{-8} \times (3400)^4 \]
\[ = 5.67 \times 10^{-8} \times 1.34 \times 10^{14} \]
\[ = 7.60 \times 10^6 \, \text{W/m}^2 \]
Total radiated power: \( P = I(t) \times 4\pi R^2 \)
For Sun:
\[ P = 6.35 \times 10^7 \times 4\pi (7 \times 10^8)^2 \]
\[ = 6.35 \times 10^7 \times 4\pi \times 4.9 \times 10^{17} \]
\[ = 3.91 \times 10^{26} \, \text{W} \]
For Betelgeuse:
\[ P = 7.60 \times 10^6 \times 4\pi (4 \times 10^{11})^2 \]
\[ = 7.60 \times 10^6 \times 4\pi \times 1.6 \times 10^{23} \]
\[ = 1.53 \times 10^{31} \, \text{W} \]

Failure of Classical Physics

❌ The Ultraviolet Catastrophe

Classical physics predicted that black body radiation intensity should increase without bound at shorter wavelengths, a phenomenon called the "ultraviolet catastrophe." This contradicted experimental observations that showed intensity reaching a maximum and then decreasing.

Rayleigh-Jeans Formula

🧮 Rayleigh-Jeans Derivation

Classical Approach

Based on classical physics and the equipartition theorem, Rayleigh and Jeans derived:

\[ R(\lambda) = \frac{2\pi c k T}{\lambda^4} \]

where \( k \) is Boltzmann's constant and \( c \) is the speed of light.

Problem with the Formula

As \( \lambda \to 0 \), \( R(\lambda) \to \infty \), which contradicts experimental results. This divergence at short wavelengths is the ultraviolet catastrophe.

Wien's Formula

🧮 Wien's Radiation Formula

Empirical Formula

Wien proposed an empirical formula based on thermodynamic arguments:

\[ R(\lambda) = \frac{A}{\lambda^5} e^{-B/\lambda T} \]

where A and B are constants.

Limitations

Wien's formula worked well at short wavelengths but failed at longer wavelengths, where it didn't match experimental data.

📈 Comparison of Radiation Formulas

[Graph: Black body radiation spectrum showing experimental data, Rayleigh-Jeans formula, Wien's formula, and Planck's formula]

The graph shows how classical formulas failed to match experimental data across all wavelengths, while Planck's quantum theory provided a perfect fit.

Planck's Quantum Theory

💡 Planck's Revolutionary Hypothesis

In 1900, Max Planck proposed a radical idea to resolve the black body radiation problem:

  • Energy is not continuous but quantized
  • Oscillators in the cavity walls can only have discrete energy values
  • The energy of an oscillator is an integer multiple of a fundamental quantum

Planck's Radiation Law

🧮 Planck's Radiation Formula

Quantum Hypothesis

Planck assumed that the energy of oscillators is quantized:

\[ E = nh\nu \]

where \( n = 0, 1, 2, \ldots \), \( h \) is Planck's constant, and \( \nu \) is frequency.

Planck's Radiation Law

Using statistical mechanics with this quantum hypothesis, Planck derived:

\[ R(\lambda) = \frac{2\pi h c^2}{\lambda^5} \cdot \frac{1}{e^{hc/\lambda kT} - 1} \]

This formula perfectly matched experimental data across all wavelengths.

\( h = 6.626 \times 10^{-34} \, \text{J·s} \)
Planck's Constant - The Fundamental Quantum of Action

Derivation of Classical Formulas from Planck's Law

🧮 Classical Limits of Planck's Law

Rayleigh-Jeans Limit

For long wavelengths (\( \lambda \to \infty \)) or high temperatures:

\[ e^{hc/\lambda kT} \approx 1 + \frac{hc}{\lambda kT} \]
\[ R(\lambda) \approx \frac{2\pi h c^2}{\lambda^5} \cdot \frac{1}{\frac{hc}{\lambda kT}} \]
\[ = \frac{2\pi c k T}{\lambda^4} \]

This is exactly the Rayleigh-Jeans formula.

Wien's Limit

For short wavelengths (\( \lambda \to 0 \)) or low temperatures:

\[ e^{hc/\lambda kT} \gg 1 \]
\[ R(\lambda) \approx \frac{2\pi h c^2}{\lambda^5} e^{-hc/\lambda kT} \]

This matches Wien's radiation formula with \( A = 2\pi h c^2 \) and \( B = hc/k \).

💡 Planck's Legacy

Planck's quantum hypothesis marked the beginning of quantum physics. Though initially proposed as a mathematical trick, it soon became clear that energy quantization was a fundamental property of nature, revolutionizing our understanding of the microscopic world.

Photoelectric Effect

⚡ What is the Photoelectric Effect?

The photoelectric effect is the emission of electrons from a metal surface when light of sufficiently high frequency falls on it. Key observations:

  • Electrons are emitted only when light frequency exceeds a threshold
  • Emission is instantaneous, regardless of light intensity
  • Kinetic energy of emitted electrons depends on frequency, not intensity
  • Number of emitted electrons depends on light intensity

Experimental Setup

⚙️ Photoelectric Effect Apparatus

LIGHT SOURCE
↓ Monochromatic Light
PHOTOCATHODE
↓ Emitted Electrons
ANODE
To Ammeter

Working Principle: When light of sufficient frequency strikes the photocathode, electrons are emitted and travel to the anode, creating a measurable current.

Key Components:

  • Monochromatic light source with variable frequency and intensity
  • Photocathode (metal surface)
  • Anode to collect emitted electrons
  • Voltage source to create stopping potential
  • Ammeter to measure photocurrent

Experimental Results

📊 Key Observations

1. Threshold Frequency

For each metal, there exists a minimum frequency \( \nu_0 \) below which no electrons are emitted, regardless of light intensity.

2. Instantaneous Emission

Electrons are emitted as soon as light strikes the surface, with no measurable time delay.

3. Kinetic Energy vs. Frequency

The maximum kinetic energy of emitted electrons increases linearly with frequency:

\[ K_{max} \propto (\nu - \nu_0) \]

4. Photocurrent vs. Intensity

The number of emitted electrons (photocurrent) is directly proportional to light intensity.

Failure of Classical Wave Theory

⚠️ Classical Physics Predictions vs. Experimental Results

Classical Prediction Experimental Result
Emission should occur at all frequencies given sufficient intensity Emission occurs only above threshold frequency
Kinetic energy should increase with intensity Kinetic energy depends on frequency, not intensity
There should be a time delay for electron emission Emission is instantaneous

These contradictions between classical predictions and experimental results could not be resolved within the framework of classical physics.

Einstein's Photon Theory

🌟 Einstein's Photon Hypothesis

In 1905, Albert Einstein extended Planck's quantum idea to explain the photoelectric effect:

  • Light consists of discrete packets of energy called photons
  • Each photon has energy \( E = h\nu \)
  • Energy is transferred in discrete quanta during light-matter interactions

This revolutionary concept earned Einstein the Nobel Prize in 1921.

Work Function and Threshold Frequency

⚡ Work Function

The work function \( \phi \) is the minimum energy required to remove an electron from a metal surface. It varies for different metals.

The threshold frequency \( \nu_0 \) is related to the work function by:

\[ \phi = h\nu_0 \]

Einstein's Photoelectric Equation

🧮 Einstein's Explanation

Energy Conservation

When a photon strikes a metal surface:

\[ \text{Photon Energy} = \text{Work Function} + \text{Maximum Kinetic Energy} \]
\[ h\nu = \phi + K_{max} \]

Einstein's Photoelectric Equation

\[ K_{max} = h\nu - \phi \]
\[ = h\nu - h\nu_0 \]
\[ = h(\nu - \nu_0) \]

Stopping Potential

The stopping potential \( V_0 \) is related to maximum kinetic energy:

\[ K_{max} = eV_0 \]
\[ eV_0 = h\nu - \phi \]
\[ V_0 = \frac{h}{e}\nu - \frac{\phi}{e} \]

📈 Stopping Potential vs. Frequency

[Graph: Linear relationship between stopping potential and frequency with slope h/e and intercept -φ/e]

The graph shows a straight line with slope \( h/e \), providing a method to determine Planck's constant experimentally.

Sample Problem 2: Photoelectric Effect Calculations

Light of wavelength 2000 Å falls on an aluminum surface (work function = 4.2 eV). Calculate:

(a) The kinetic energy of the fastest emitted photoelectrons

(b) The stopping potential

(c) The threshold wavelength for aluminum

Given:
\[ \lambda = 2000 \, \text{Å} = 2000 \times 10^{-10} \, \text{m} \]
\[ \phi = 4.2 \, \text{eV} \]
\[ h = 6.626 \times 10^{-34} \, \text{J·s} \]
\[ c = 3 \times 10^8 \, \text{m/s} \]
\[ 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \]
(a) Energy of incident photon:
\[ E = \frac{hc}{\lambda} \]
\[ = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{2000 \times 10^{-10}} \]
\[ = \frac{1.9878 \times 10^{-25}}{2 \times 10^{-7}} \]
\[ = 9.939 \times 10^{-19} \, \text{J} \]
\[ = \frac{9.939 \times 10^{-19}}{1.6 \times 10^{-19}} = 6.212 \, \text{eV} \]
Maximum kinetic energy:
\[ K_{max} = E - \phi \]
\[ = 6.212 - 4.2 \]
\[ = 2.012 \, \text{eV} \]
(b) Stopping potential:
\[ V_0 = \frac{K_{max}}{e} \]
\[ = 2.012 \, \text{V} \]
(c) Threshold wavelength:
\[ \lambda_0 = \frac{hc}{\phi} \]
\[ = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{4.2 \times 1.6 \times 10^{-19}} \]
\[ = \frac{1.9878 \times 10^{-25}}{6.72 \times 10^{-19}} \]
\[ = 2.957 \times 10^{-7} \, \text{m} \]
\[ = 2957 \, \text{Å} \]

Compton Effect

🔍 What is the Compton Effect?

The Compton effect is the increase in wavelength of X-rays when scattered by electrons in matter. Discovered by Arthur Compton in 1923, this phenomenon provided direct evidence for the particle nature of light.

Key observations:

  • Scattered X-rays have longer wavelength than incident X-rays
  • The wavelength shift depends on scattering angle
  • The effect cannot be explained by classical wave theory

Experimental Setup

⚙️ Compton Scattering Apparatus

X-RAY SOURCE
↓ Monochromatic X-rays
SCATTERING TARGET
↓ Scattered X-rays at angle θ
X-RAY DETECTOR

Working Principle: Monochromatic X-rays strike a target containing loosely bound electrons. The scattered X-rays are analyzed at various angles to measure their wavelength.

Key Findings:

  • The scattered X-rays consist of two components: one with the original wavelength and one with increased wavelength
  • The wavelength shift \( \Delta\lambda \) increases with scattering angle θ
  • The effect is explained by treating X-rays as particles (photons) colliding with electrons

Compton Shift Derivation

🧮 Compton Wavelength Shift

Step 1: Energy Conservation

Consider a photon colliding with an electron at rest:

\[ h\nu + m_0c^2 = h\nu' + mc^2 \]

where \( \nu \) and \( \nu' \) are frequencies of incident and scattered photons, \( m_0 \) is electron rest mass, and \( m \) is relativistic mass.

Step 2: Momentum Conservation

In x-direction:

\[ \frac{h\nu}{c} = \frac{h\nu'}{c} \cos\theta + p\cos\phi \]

In y-direction:

\[ 0 = \frac{h\nu'}{c} \sin\theta - p\sin\phi \]

Step 3: Relativistic Energy-Momentum Relation

\[ (mc^2)^2 = (pc)^2 + (m_0c^2)^2 \]

Step 4: Compton Shift Formula

Solving these equations gives:

\[ \lambda' - \lambda = \frac{h}{m_0c}(1 - \cos\theta) \]
\[ \Delta\lambda = \lambda_c(1 - \cos\theta) \]

Step 5: Compton Wavelength

Where \( \lambda_c \) is the Compton wavelength of the electron:

\[ \lambda_c = \frac{h}{m_0c} \]
\[ = \frac{6.626 \times 10^{-34}}{9.11 \times 10^{-31} \times 3 \times 10^8} \]
\[ = 2.426 \times 10^{-12} \, \text{m} \]
\[ = 0.02426 \, \text{Å} \]
\( \lambda_c = 0.02426 \, \text{Å} \)
Compton Wavelength of Electron - Fundamental Length Scale in Quantum Physics
Sample Problem 3: Compton Scattering

X-rays of wavelength 0.240 nm are scattered from a carbon target. The scattered radiation is viewed at 60° to the incident beam. Calculate:

(a) The Compton shift

(b) The energy of the scattered X-ray

(c) The kinetic energy of the recoiling electron

Given:
\[ \lambda = 0.240 \, \text{nm} = 2.40 \times 10^{-10} \, \text{m} \]
\[ \theta = 60^\circ \]
\[ \lambda_c = 2.426 \times 10^{-12} \, \text{m} \]
(a) Compton shift:
\[ \Delta\lambda = \lambda_c(1 - \cos\theta) \]
\[ = 2.426 \times 10^{-12}(1 - \cos 60^\circ) \]
\[ = 2.426 \times 10^{-12}(1 - 0.5) \]
\[ = 2.426 \times 10^{-12} \times 0.5 \]
\[ = 1.213 \times 10^{-12} \, \text{m} \]
\[ = 0.001213 \, \text{nm} \]
(b) Wavelength of scattered X-ray:
\[ \lambda' = \lambda + \Delta\lambda \]
\[ = 0.240 + 0.001213 \]
\[ = 0.241213 \, \text{nm} \]
Energy of scattered X-ray:
\[ E' = \frac{hc}{\lambda'} \]
\[ = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{0.241213 \times 10^{-9}} \]
\[ = \frac{1.9878 \times 10^{-25}}{2.41213 \times 10^{-10}} \]
\[ = 8.242 \times 10^{-16} \, \text{J} \]
\[ = \frac{8.242 \times 10^{-16}}{1.6 \times 10^{-19}} = 5151 \, \text{eV} \]
(c) Energy of incident X-ray:
\[ E = \frac{hc}{\lambda} \]
\[ = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{0.240 \times 10^{-9}} \]
\[ = \frac{1.9878 \times 10^{-25}}{2.40 \times 10^{-10}} \]
\[ = 8.283 \times 10^{-16} \, \text{J} \]
\[ = \frac{8.283 \times 10^{-16}}{1.6 \times 10^{-19}} = 5177 \, \text{eV} \]
Kinetic energy of recoiling electron:
\[ K = E - E' \]
\[ = 5177 - 5151 \]
\[ = 26 \, \text{eV} \]

Solved Problems

Problem 1: Black Body Radiation

A black body is at a temperature of 527°C. To radiate twice as much energy per second, what should be its temperature?

Using Stefan-Boltzmann law: \( I(t) = \sigma T^4 \)
Initial temperature: \( T_1 = 527^\circ C = 527 + 273 = 800 \, K \)
We want \( I_2 = 2I_1 \)
\[ \sigma T_2^4 = 2\sigma T_1^4 \]
\[ T_2^4 = 2T_1^4 \]
\[ T_2 = T_1 \times 2^{1/4} \]
\[ = 800 \times 2^{0.25} \]
\[ = 800 \times 1.189 \]
\[ = 951.2 \, K \]
\[ = 678.2^\circ C \]
Problem 2: Photoelectric Effect

The threshold wavelength for photoelectric emission from a metal surface is 3800 Å. Calculate the maximum kinetic energy of photoelectrons emitted when radiation of wavelength 2600 Å falls on it.

Given:
\[ \lambda_0 = 3800 \, \text{Å} \]
\[ \lambda = 2600 \, \text{Å} \]
Work function:
\[ \phi = \frac{hc}{\lambda_0} \]
Photon energy:
\[ E = \frac{hc}{\lambda} \]
Maximum kinetic energy:
\[ K_{max} = E - \phi \]
\[ = hc\left(\frac{1}{\lambda} - \frac{1}{\lambda_0}\right) \]
\[ = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{10^{-10}} \left(\frac{1}{2600} - \frac{1}{3800}\right) \]
\[ = 1.9878 \times 10^{-15} \left(\frac{3800 - 2600}{2600 \times 3800}\right) \]
\[ = 1.9878 \times 10^{-15} \times \frac{1200}{9.88 \times 10^6} \]
\[ = 1.9878 \times 10^{-15} \times 1.215 \times 10^{-4} \]
\[ = 2.415 \times 10^{-19} \, \text{J} \]
\[ = \frac{2.415 \times 10^{-19}}{1.6 \times 10^{-19}} = 1.509 \, \text{eV} \]
Problem 3: Compton Effect

In a Compton scattering experiment, the incident radiation has wavelength 0.0709 nm. Calculate the wavelength of radiation scattered through 180°.

Given:
\[ \lambda = 0.0709 \, \text{nm} \]
\[ \theta = 180^\circ \]
\[ \lambda_c = 0.002426 \, \text{nm} \]
Compton shift:
\[ \Delta\lambda = \lambda_c(1 - \cos\theta) \]
\[ = 0.002426(1 - \cos 180^\circ) \]
\[ = 0.002426(1 - (-1)) \]
\[ = 0.002426 \times 2 \]
\[ = 0.004852 \, \text{nm} \]
Wavelength of scattered radiation:
\[ \lambda' = \lambda + \Delta\lambda \]
\[ = 0.0709 + 0.004852 \]
\[ = 0.075752 \, \text{nm} \]

Frequently Asked Questions

❓ Why is the photoelectric effect important in quantum physics?

The photoelectric effect is crucial in quantum physics because:

  • It provided the first direct evidence for the particle nature of light
  • It demonstrated that energy is quantized in discrete packets (photons)
  • It showed that light intensity affects the number of photons, not their individual energy
  • It contradicted classical wave theory predictions, forcing a paradigm shift in physics
  • It led to Einstein's Nobel Prize and established the foundation of quantum mechanics

The photoelectric effect continues to have practical applications in photodetectors, solar cells, and image sensors.

❓ What is the difference between the photoelectric effect and Compton scattering?

While both demonstrate the particle nature of light, there are key differences:

Photoelectric Effect Compton Scattering
Occurs with visible/UV light Occurs with X-rays and gamma rays
Photon is completely absorbed Photon is scattered with reduced energy
Involves bound electrons Involves free or loosely bound electrons
Demonstrates energy quantization Demonstrates both energy and momentum quantization
Electron receives all photon energy Energy and momentum are shared between photon and electron

Both effects are complementary evidence for the quantum nature of electromagnetic radiation.

❓ How did Planck's quantum hypothesis resolve the ultraviolet catastrophe?

Planck's quantum hypothesis resolved the ultraviolet catastrophe by introducing energy quantization:

  • Classical Physics: Assumed oscillators could have any energy value (continuous)
  • Planck's Hypothesis: Oscillators can only have discrete energy values \( E = nh\nu \)

This quantization has two crucial consequences:

  1. At high frequencies (short wavelengths), the energy quantum \( h\nu \) is large, making it difficult to excite oscillators, thus reducing radiation at these wavelengths
  2. At low frequencies (long wavelengths), the energy quantum is small, and classical behavior is recovered

The mathematical result is that Planck's radiation formula:

\[ R(\lambda) = \frac{2\pi h c^2}{\lambda^5} \cdot \frac{1}{e^{hc/\lambda kT} - 1} \]

correctly predicts that spectral radiancy approaches zero as \( \lambda \to 0 \), eliminating the ultraviolet catastrophe while matching experimental data across all wavelengths.

📚 Master Quantum Physics

Understanding light and quantum physics is fundamental to modern physics, chemistry, and engineering. The concepts of black body radiation, photoelectric effect, and Compton scattering laid the foundation for quantum mechanics and revolutionized our understanding of the microscopic world.

Read More: Physics HRK Notes of Modern Physics

© House of Physics | HRK Physics Chapter 49: Light and Quantum Physics

Based on Halliday, Resnick, and Krane's "Physics" with additional insights from university physics curriculum

House of Physics | Contact: aliphy2008@gmail.com

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