15.6 Compton Scattering AP Physics 2 MCQ Style Questions - New Syllabus

15.6 Compton Scattering AP Physics 2 MCQ

Unit 15: Modern Physics

Weightage : 15–18%

AP Physics 2 Exam Style Questions – All Topics

Question

In the Compton effect, an X-ray photon collides with a stationary free electron and scatters. Which of the following best describes what happens to the photon’s wavelength after the collision?

A) The wavelength decreases because the photon gains energy from the electron.
B) The wavelength remains unchanged because energy is conserved in all collisions.
C) The wavelength increases because the photon transfers some energy to the electron.
D) The wavelength change depends on the mass of the target atom, not the scattering angle.

▶️ Answer/Explanation

Ans: C

In Compton scattering, the incident photon collides with a free electron (at rest) and transfers some of its energy and momentum to the electron. Since photon energy \(E = hc/\lambda\), a loss of energy means a longer wavelength. The Compton shift formula confirms this:

\[\Delta\lambda = \frac{h}{m_e c}(1 – \cos\theta) \geq 0\]

The right-hand side is always ≥ 0 for any angle θ, so \(\lambda_f \geq \lambda_i\) always. The scattered photon always has a longer wavelength (lower energy) than the incident photon.

Option A is incorrect — the photon cannot gain energy from a stationary electron in this type of collision. Option D is incorrect — the Compton shift depends on \(m_e\) (electron mass), not the nuclear mass.

Question

What is the maximum possible wavelength shift \(\Delta\lambda_{max}\) in a Compton scattering event, and at what scattering angle does it occur?

A) \(2.43 \times 10^{-12}\) m at \(\theta = 90°\)
B) \(4.85 \times 10^{-12}\) m at \(\theta = 180°\)
C) \(2.43 \times 10^{-12}\) m at \(\theta = 180°\)
D) \(4.85 \times 10^{-12}\) m at \(\theta = 90°\)

▶️ Answer/Explanation

Ans: B

The Compton shift: \(\Delta\lambda = \dfrac{h}{m_e c}(1 – \cos\theta)\)

The factor \((1 – \cos\theta)\) is maximized when \(\cos\theta\) is minimized. \(\cos 180° = -1\), so:

\[\Delta\lambda_{max} = \frac{h}{m_e c}(1-(-1)) = \frac{2h}{m_e c} = 2 \times 2.43 \times 10^{-12} = 4.85 \times 10^{-12}\,\text{m}\]

This occurs at \(\theta = 180°\) — backscattering, where the photon bounces directly back toward its source. At \(\theta = 90°\), \(\Delta\lambda = h/m_e c = 2.43 \times 10^{-12}\) m (the Compton wavelength).

Question

Why did Compton use X-rays rather than visible light in his scattering experiments?

A) X-rays travel faster than visible light, making the experiment easier to observe.
B) The wavelength shift \(\Delta\lambda \approx 2.43 \times 10^{-12}\) m is a measurable fraction of X-ray wavelengths (~10⁻¹⁰ m), but negligibly small compared to visible light wavelengths (~10⁻⁷ m).
C) X-ray photons have less momentum than visible light photons, so the electron recoil is larger.
D) Visible light photons cannot scatter off electrons; only X-rays can.

▶️ Answer/Explanation

Ans: B

The Compton wavelength shift \(\Delta\lambda_{max} = 4.85 \times 10^{-12}\) m is fixed regardless of the incident wavelength. For X-rays with \(\lambda \sim 10^{-10}\) m, this represents about a 5% shift — detectable with the spectrometers of Compton’s era.

For visible light with \(\lambda \sim 5 \times 10^{-7}\) m, the same shift is only about 0.001% — far too small to measure and would appear as no shift at all.

All photons, regardless of wavelength, can scatter off electrons. X-rays don’t travel faster (all EM radiation travels at c). X-ray photons actually have more momentum than visible light photons (shorter wavelength).

Question

The magnitude of the momentum of a photon with wavelength \(\lambda\) is given by \(p = h/\lambda\). An X-ray photon has a wavelength of \(1.45 \times 10^{-10}\) m. What is the magnitude of its momentum?

\((h = 6.63 \times 10^{-34}\,\text{J·s})\)

A) \(9.61 \times 10^{-44}\) kg·m/s
B) \(4.57 \times 10^{-24}\) kg·m/s
C) \(4.57 \times 10^{-24}\) N·s
D) Both B and C — they are the same quantity expressed the same way.

▶️ Answer/Explanation

Ans: D

\[ p = \frac{h}{\lambda} = \frac{6.63 \times 10^{-34}}{1.45 \times 10^{-10}} = 4.57 \times 10^{-24}\,\text{kg·m/s} \]

Note: kg·m/s and N·s are identical units for momentum, since 1 N = 1 kg·m/s². Both B and C give the correct numerical value with equivalent units, making D the best answer. This tests unit awareness, a key AP skill.

15.6 Compton Scattering AP Physics 2 MCQ Style Questions - New Syllabus