Photoelectric Effect vs Compton Effect — Comparison of Particle Nature Evidence

Question

How do the photoelectric effect and Compton effect both demonstrate the particle nature of light, and what are the key differences between them?

Solution — Step by Step

When light of frequency $\nu \geq \nu_0$ (threshold frequency) hits a metal surface, electrons are ejected.

Einstein’s equation:

KE_{\max} = h\nu - \phi

where $\phi = h\nu_0$ is the work function.

Key observations that wave theory could NOT explain:

No emission below threshold frequency, regardless of intensity
KE depends on frequency, not intensity
Emission is instantaneous (no time lag)

The photon is completely absorbed — it transfers all its energy to one electron.

When X-rays (high-energy photons) scatter off loosely bound electrons, the scattered photon has a longer wavelength (lower energy) than the incident one.

\Delta\lambda = \lambda' - \lambda = \frac{h}{m_e c}(1 - \cos\theta)

where $\theta$ is the scattering angle and $\frac{h}{m_e c} = 0.0243$ angstrom is the Compton wavelength of the electron.

The photon is not fully absorbed — it bounces off like a billiard ball, losing some energy to the electron.

Feature	Photoelectric Effect	Compton Effect
Photon fate	Completely absorbed	Partially scattered
Electron source	Bound electrons in metal	Loosely bound/free electrons
Photon energy	UV/visible range	X-ray range
Energy transfer	Total photon energy	Partial (depends on angle)
Key equation	$KE = h\nu - \phi$	$\Delta\lambda = \frac{h}{m_e c}(1-\cos\theta)$
What it proves	Photon has energy $E = h\nu$	Photon has momentum $p = h/\lambda$

graph TD
    A[Particle Nature of Light] --> B[Photoelectric Effect]
    A --> C[Compton Effect]
    B --> B1[Proves: E = h nu - photon energy is quantised]
    B --> B2[Photon absorbed completely by electron]
    B --> B3[Low energy photons - UV/visible]
    C --> C1[Proves: p = h/lambda - photon has momentum]
    C --> C2[Photon scattered like a billiard ball]
    C --> C3[High energy photons - X-rays]
    B1 --> D[Together: photons behave as particles with E and p]
    C1 --> D

Why This Works

The photoelectric effect proved that light energy comes in discrete packets ( $E = h\nu$ ). But it did not directly show that photons carry momentum. Compton’s experiment filled this gap — the wavelength shift of scattered X-rays can only be explained if the photon carries momentum $p = h/\lambda$ and the collision follows conservation of energy AND momentum, just like two billiard balls.

Together, these two experiments established that light has both energy and momentum — it behaves as a particle (photon) in certain interactions.

JEE Main 2023 asked: “Which effect demonstrates that photons carry momentum?” Answer: Compton effect. The photoelectric effect shows energy quantisation; the Compton effect shows momentum quantisation. Know this distinction clearly.

Alternative Method

We can also understand Compton scattering through relativistic energy-momentum conservation. Treating the photon as a massless particle with $E = pc$ and the electron relativistically, the algebra yields the Compton formula directly. This derivation appears in JEE Advanced level problems.

Common Mistake

Students think the Compton effect happens with visible light. It does not produce any measurable effect with visible light because $\Delta\lambda$ is at most 0.0486 angstrom (at 180 degree scattering). For visible light (wavelength ~5000 angstrom), this shift is negligible. Only X-rays (wavelength ~1 angstrom) show a measurable fractional change. This is why Compton used X-rays, not visible light.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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