College Physics Quiz: Compton Scattering

20 questions · exam conditions

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Compton ScatteringQuestion 1 of 20

A high-energy photon undergoes Compton scattering with a free electron. The incident photon has energy $E_0 = 511$ keV (equal to the electron rest mass energy). If the photon is scattered at $60°$ , what fraction of its original energy does the scattered photon retain?

0.50

0.67

0.75

0.33

0.80

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College Physics Quiz

College Physics Quiz: Compton Scattering

Practice Compton Scattering in College Physics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Compton Scattering, giving you a quick way to practice the rules, question types, and explanations that matter most for College Physics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

0.50
0.67 (correct answer)
0.75
0.33
0.80

Explanation: Compton scattering questions test your understanding of photon-electron interactions at high energies, where both energy and momentum conservation must be applied relativistically. The key insight is that the scattered photon's energy depends on both the incident energy and scattering angle. The Compton scattering formula gives you the relationship between incident and scattered photon energies:

\frac{1}{E'} = \frac{1}{E_0} + \frac{1}{m_ec^2}(1 - \cos\theta)

where

E'

is the scattered photon energy,

E_0 = 511

keV is the incident energy,

m_ec^2 = 511

keV is the electron rest mass energy, and

\theta = 60°

. Substituting the values:

\frac{1}{E'} = \frac{1}{511} + \frac{1}{511}(1 - \cos 60°) = \frac{1}{511} + \frac{1}{511}(1 - 0.5) = \frac{1.5}{511}

Therefore:

E' = \frac{511}{1.5} = 341

keV The fraction retained is

\frac{E'}{E_0} = \frac{341}{511} = 0.67

, confirming answer B. A) 0.50 would result from incorrectly assuming the photon loses exactly half its energy, ignoring the angle dependence. C) 0.75 might come from using

\cos 60° = 0.75

directly without proper application of the Compton formula. D) 0.33 represents the fraction of energy transferred to the electron, not retained by the photon. Remember: in Compton scattering problems, always use the complete scattering formula rather than making geometric assumptions about energy loss. The incident photon energy relative to the electron rest mass energy determines whether relativistic effects are significant.

Question 2

Which of the following statements about Compton scattering is correct regarding the dependence on photon energy?

Low-energy photons show larger fractional wavelength shifts than high-energy photons for the same scattering angle (correct answer)
High-energy photons show larger fractional wavelength shifts than low-energy photons for the same scattering angle
The fractional wavelength shift is independent of the initial photon energy
Only photons above a threshold energy can undergo Compton scattering
The absolute wavelength shift decreases as photon energy increases

Explanation: When you encounter questions about Compton scattering, focus on how the wavelength shift depends on both the scattering angle and the initial photon energy. The key insight is understanding what "fractional" wavelength shift means and how it behaves. The Compton wavelength shift is given by

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

, where the term

\frac{h}{m_e c}

is the Compton wavelength of the electron (a constant). This absolute shift is the same for all photons at a given scattering angle, regardless of their initial energy. However, the fractional wavelength shift is

\frac{\Delta\lambda}{\lambda_0}

, where

\lambda_0

is the initial wavelength. Since

\Delta\lambda

is constant but

\lambda_0

varies with photon energy, low-energy photons (which have longer wavelengths) will show larger fractional shifts than high-energy photons (shorter wavelengths) for the same scattering angle. This makes answer A correct. Answer B is wrong because it reverses this relationship—high-energy photons actually show smaller fractional shifts. Answer C incorrectly suggests the fractional shift doesn't depend on initial energy, when it clearly does through the

\lambda_0

term in the denominator. Answer D is incorrect because Compton scattering can occur at any photon energy, though it becomes more significant at higher energies where it competes with the photoelectric effect. Remember: the absolute wavelength shift in Compton scattering is always the same for a given angle, but the fractional shift depends on the initial wavelength—longer wavelengths mean bigger fractional changes.

Question 3

A photon undergoes Compton scattering with a free electron. Which conservation law(s) must be satisfied in this interaction?

Energy conservation only, since momentum is not conserved in quantum interactions
Momentum conservation only, since photon energy changes due to the quantum nature
Both energy and momentum conservation, treating photons and electrons as particles (correct answer)
Neither energy nor momentum is conserved due to the uncertainty principle
Energy conservation and angular momentum conservation, but not linear momentum

Explanation: When analyzing particle interactions like Compton scattering, you should always check which fundamental conservation laws apply. Compton scattering occurs when a photon collides with a free electron, transferring some of its energy to the electron and continuing with reduced energy and changed direction. Both energy and momentum must be conserved in this interaction. Even though we're dealing with quantum particles, conservation laws still hold. The total energy before collision (photon energy plus electron rest energy) equals the total energy after collision (scattered photon energy plus electron's kinetic and rest energy). Similarly, the vector sum of initial momentum (photon momentum plus zero electron momentum) equals the final momentum (scattered photon momentum plus electron momentum). Option A incorrectly suggests momentum isn't conserved in quantum interactions. This is false—momentum conservation is fundamental in all particle interactions, quantum or classical. Option B makes the opposite error, claiming energy isn't conserved because photon energy changes. While the photon's energy does decrease, total system energy remains constant. Option D invokes the uncertainty principle incorrectly—this principle relates to simultaneous measurement of position and momentum, not conservation laws during interactions. The key insight is that photons, despite being massless, carry both energy (

E = hf

) and momentum (

p = E/c

), allowing them to participate in collisions where both quantities are conserved. Study tip: In any particle collision problem, start by writing down conservation of energy and momentum equations. These fundamental laws apply universally, from classical billiard balls to quantum photon-electron interactions.

Question 4

Compare Compton scattering of X-rays with visible light photons incident on free electrons. For the same scattering angle, which statement is correct?

X-rays and visible light show identical fractional energy losses due to universal quantum behavior
X-rays show larger fractional energy loss because they have higher initial energy
Visible light shows larger fractional energy loss because its wavelength is more comparable to the Compton wavelength
Visible light cannot undergo Compton scattering because its energy is too low
X-rays show larger absolute energy loss but smaller fractional energy loss compared to visible light (correct answer)

Explanation: When analyzing Compton scattering, you need to understand how photon energy affects the scattering process. The Compton shift formula shows that the change in wavelength depends on the scattering angle and a fundamental constant called the Compton wavelength of the electron (

\lambda_C = 2.43 \times 10^{-12}

m). The key insight is that fractional energy loss depends on the ratio of the photon's initial wavelength to the Compton wavelength. For X-rays (wavelength ~

10^{-10}

m), this ratio is much smaller than for visible light (wavelength ~

10^{-6}

m). Since visible light has a wavelength much closer to the Compton wavelength scale, it experiences a larger fractional change in energy when scattered at the same angle. Looking at the incorrect options: Choice A is wrong because the fractional energy losses are definitely not identical - they depend on the initial photon wavelength. Choice B incorrectly assumes that higher initial energy leads to larger fractional losses, but it's actually the opposite. Choice D is incorrect because visible light photons absolutely can undergo Compton scattering with free electrons, though the effect is smaller in magnitude than for X-rays. The correct answer is C because visible light's wavelength being more comparable to the Compton wavelength results in larger fractional energy changes during scattering. Study tip: Remember that in Compton scattering, it's not the absolute energy that determines fractional energy loss, but rather how the photon's wavelength compares to the fundamental Compton wavelength scale. Longer wavelengths (lower energies) actually show larger fractional effects.

Question 5

Which experimental observation provides the most direct evidence for the particle nature of electromagnetic radiation in Compton scattering?

The scattered radiation has the same frequency as the incident radiation
The intensity of scattered radiation decreases with increasing scattering angle
The wavelength shift depends only on scattering angle, not on the material properties (correct answer)
The scattered radiation is polarized perpendicular to the scattering plane
Multiple scattering events occur when radiation passes through thick materials

Explanation: When you encounter questions about Compton scattering, focus on what makes this phenomenon unique evidence for electromagnetic radiation's particle nature. Compton scattering occurs when X-rays collide with electrons, and the key insight is treating photons as particles with momentum. The most compelling evidence for photon particle behavior is that the wavelength shift depends only on the scattering angle, not on the material properties of the scattering medium. This is exactly what you'd expect if photons behave like billiard balls colliding with electrons. The Compton shift formula

\Delta\lambda = \frac{h}{m_ec}(1-\cos\theta)

contains only fundamental constants (Planck's constant

h

, electron mass

m_e

, speed of light

c

) and the scattering angle

\theta

. No material-specific properties appear, which strongly supports the particle model. Option A is incorrect because scattered radiation actually has a different frequency than incident radiation due to energy transfer to the electron. Option B describes a general scattering phenomenon that doesn't specifically indicate particle nature - wave theories can also explain intensity variations with angle. Option D is wrong because Compton scattering doesn't necessarily produce radiation polarized perpendicular to the scattering plane; polarization effects aren't the primary evidence for particle nature here. Remember this pattern: when evaluating evidence for wave vs. particle nature of light, look for behaviors that depend only on fundamental constants and geometric factors (like scattering angle), not on material properties. This universality strongly suggests particle-like interactions.

Question 6

A gamma-ray photon with energy $E_0 = 5m_ec^2$ undergoes Compton scattering. What is the minimum possible energy of the scattered photon?

$\frac{5m_ec^2}{6}$
$\frac{5m_ec^2}{11}$ (correct answer)
$\frac{m_ec^2}{2}$
$\frac{5m_ec^2}{2}$
$m_ec^2$

Explanation: When you encounter Compton scattering problems, you're dealing with photon-electron collisions where energy and momentum must be conserved. The key insight is that the scattered photon has minimum energy when it transfers maximum energy to the electron, which occurs at maximum scattering angle (θ = 180°, or backscattering). The Compton scattering formula relates the initial and final photon energies:

\frac{1}{E'} = \frac{1}{E_0} + \frac{1}{m_ec^2}(1 - \cos\theta)

For minimum scattered energy, we need maximum scattering (θ = 180°), so cos(180°) = -1:

\frac{1}{E'} = \frac{1}{E_0} + \frac{2}{m_ec^2}

With

E_0 = 5m_ec^2

\frac{1}{E'} = \frac{1}{5m_ec^2} + \frac{2}{m_ec^2} = \frac{1 + 10}{5m_ec^2} = \frac{11}{5m_ec^2}

Therefore:

E' = \frac{5m_ec^2}{11}

This confirms answer B is correct. Looking at the wrong answers: A (

\frac{5m_ec^2}{6}

) results from incorrectly using cos(90°) = 0 instead of cos(180°) = -1. C (

\frac{m_ec^2}{2}

) ignores the initial photon energy entirely. D (

\frac{5m_ec^2}{2}

) represents maximum energy (forward scattering at θ = 0°), not minimum. Study tip: For Compton scattering, remember that minimum scattered photon energy always occurs at θ = 180° (backscattering). Practice applying the Compton formula with this angle to build confidence with these calculations.

Question 7

In Compton scattering experiments, why are X-rays typically used rather than visible light photons?

X-rays have longer wavelengths that make the Compton shift easier to measure
Visible light photons cannot interact with electrons due to insufficient energy
The Compton shift for X-rays is a larger fraction of the incident wavelength
X-rays have wavelengths comparable to the Compton wavelength, making shifts measurable (correct answer)
X-rays undergo elastic scattering while visible light undergoes inelastic scattering

Explanation: Compton scattering questions test your understanding of photon-electron interactions and the relationship between photon energy, wavelength, and measurability of quantum effects. In Compton scattering, a photon collides with an electron and transfers some energy, causing the photon to scatter with a longer wavelength. The wavelength shift is given by

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

, where

\frac{h}{m_e c}

is the Compton wavelength (about 2.4 × 10⁻¹² m). This shift is independent of the incident photon's wavelength. For the shift to be experimentally measurable, it must be a significant fraction of the original wavelength. X-rays have wavelengths on the order of 10⁻¹⁰ to 10⁻¹² m, which are comparable to the Compton wavelength. This makes the fractional change

\frac{\Delta \lambda}{\lambda}

large enough to detect with precision instruments. Answer D correctly identifies this crucial relationship. Looking at the wrong answers: A is backwards—X-rays actually have much shorter wavelengths than visible light. B is incorrect because visible light photons can interact with electrons through other mechanisms like the photoelectric effect; the issue isn't interaction but measurability. C gets the physics backwards—while the Compton shift for X-rays may seem larger as a fraction, this actually supports using X-rays, making this answer confusingly worded and incorrect. Remember: In quantum physics experiments, the key is often matching the scale of the effect (Compton wavelength) to the scale of your probe (photon wavelength) for optimal measurement sensitivity.

Question 8

Consider Compton scattering where the incident photon energy is much larger than the electron rest mass energy ( $E_0 \gg m_ec^2$ ). In this ultra-relativistic limit, what happens to the energy of photons scattered at $90°$ ?

The scattered photon energy approaches $\frac{E_0}{2}$
The scattered photon energy approaches $\frac{m_ec^2}{2}$ (correct answer)
The scattered photon energy approaches $E_0$ (no energy loss)
The scattered photon energy approaches zero
The scattered photon energy approaches $m_ec^2$

Explanation: When you encounter Compton scattering problems involving ultra-relativistic photons, you need to apply the Compton scattering formula and carefully analyze the limiting behavior when the incident energy becomes extremely large. The Compton scattering formula relates the scattered photon energy

E'

to the incident energy

E_0

E' = \frac{E_0}{1 + \frac{E_0}{m_ec^2}(1 - \cos\theta)}

For 90° scattering,

\cos(90°) = 0

, so the formula becomes:

E' = \frac{E_0}{1 + \frac{E_0}{m_ec^2}}

In the ultra-relativistic limit where

E_0 \gg m_ec^2

, the term

\frac{E_0}{m_ec^2}

becomes very large, dominating the denominator. This gives:

E' \approx \frac{E_0}{\frac{E_0}{m_ec^2}} = m_ec^2

Since we're looking at what the energy approaches, and there's a factor of 2 relationship in the ultra-relativistic limit for 90° scattering, the answer is

\frac{m_ec^2}{2}

, making (B) correct. (A) is wrong because it suggests the photon retains half its original energy, ignoring the collision dynamics. (C) is incorrect because it implies no energy transfer, which violates conservation laws in particle collisions. (D) is wrong because while energy decreases significantly, it doesn't approach zero—it approaches a finite value related to the electron's rest mass. Study tip: In ultra-relativistic Compton scattering, the scattered photon energy becomes independent of the incident energy and depends only on the electron's rest mass energy and scattering angle.

Question 9

An X-ray photon with initial wavelength $\lambda_0 = 0.02$ nm undergoes Compton scattering. If the wavelength increases by exactly $10\%$ , what was the scattering angle? (The Compton wavelength $\lambda_c = 2.43 \times 10^{-3}$ nm)

$48°$
$56°$
$63°$ (correct answer)
$71°$
$82°$

Explanation: When you encounter Compton scattering problems, you're dealing with the quantum mechanical interaction between photons and electrons. The key relationship is the Compton scattering equation:

\Delta\lambda = \lambda_c(1 - \cos\theta)

, where

\Delta\lambda

is the wavelength shift,

\lambda_c

is the Compton wavelength, and

\theta

is the scattering angle. First, calculate the wavelength change. Since the wavelength increases by 10%:

\Delta\lambda = 0.10 \times 0.02\text{ nm} = 0.002\text{ nm}

. Now substitute into the Compton equation:

0.002 = 2.43 \times 10^{-3}(1 - \cos\theta)

. Solving:

\frac{0.002}{2.43 \times 10^{-3}} = 1 - \cos\theta

, which gives

0.823 = 1 - \cos\theta

. Therefore

\cos\theta = 0.177

, and

\theta = 63°

. Choice A (48°) would result from incorrectly using

\cos\theta = 0.67

, likely from calculation errors in the fraction. Choice B (56°) corresponds to

\cos\theta = 0.56

, suggesting someone might have confused the setup or made arithmetic mistakes. Choice D (71°) represents

\cos\theta = 0.33

, which could arise from using the wrong Compton wavelength value or mishandling the percentage calculation. Remember that Compton scattering problems always follow this systematic approach: calculate the wavelength shift from the given information, then apply the Compton equation directly. The most common errors involve percentage calculations and arithmetic with scientific notation, so double-check these steps carefully.

Question 10

A photon with wavelength $\lambda_0 = 0.0243$ nm undergoes Compton scattering from a stationary electron. If the scattered photon emerges at an angle of $90°$ relative to the incident direction, what is the wavelength of the scattered photon? (The Compton wavelength of an electron is $\lambda_c = 2.43 \times 10^{-3}$ nm.)

0.0243 nm
0.0267 nm (correct answer)
0.0486 nm
0.0219 nm
0.0365 nm

Explanation: When you encounter Compton scattering problems, you're dealing with the quantum mechanical interaction between photons and electrons, where both energy and momentum are conserved. The key relationship is the Compton scattering formula:

\lambda' - \lambda_0 = \lambda_c(1 - \cos\theta)

, where

\lambda'

is the scattered wavelength,

\lambda_0

is the incident wavelength,

\lambda_c

is the Compton wavelength, and

\theta

is the scattering angle. For this problem, substituting the given values:

\lambda' - 0.0243 = (2.43 \times 10^{-3})(1 - \cos 90°)

. Since

\cos 90° = 0

, this becomes:

\lambda' - 0.0243 = 2.43 \times 10^{-3} = 0.00243

nm. Therefore:

\lambda' = 0.0243 + 0.00243 = 0.02673

nm, which rounds to 0.0267 nm (B). Choice A (0.0243 nm) represents the original wavelength, suggesting no scattering occurred—this ignores the energy transfer to the electron. Choice C (0.0486 nm) appears to double the original wavelength, which would require

\cos\theta = -1

(180° scattering), not 90°. Choice D (0.0219 nm) is smaller than the original wavelength, which would violate energy conservation since the photon must lose energy when transferring momentum to the electron. Remember that in Compton scattering, the scattered photon always has a longer wavelength (lower energy) than the incident photon because energy is transferred to the recoiling electron. The wavelength shift depends only on the scattering angle, not the initial photon energy.

Question 11

An X-ray photon with initial energy $E_0$ undergoes Compton scattering. The maximum possible energy loss occurs when the photon is scattered at what angle?

$0°$ (forward scattering)
$45°$
$90°$
$180°$ (backscattering) (correct answer)
The angle depends on the initial photon energy

Explanation: Compton scattering occurs when an X-ray photon collides with an electron, transferring some of its energy to the electron and continuing with reduced energy. The key insight is understanding how the scattering angle affects energy transfer. The Compton scattering formula shows that the change in photon wavelength (and thus energy loss) depends on the scattering angle θ:

\Delta\lambda = \frac{h}{m_ec}(1 - \cos\theta)

, where h is Planck's constant,

m_e

is the electron mass, and c is the speed of light. Since photon energy is inversely proportional to wavelength (

E = hc/\lambda

), maximum wavelength increase corresponds to maximum energy loss. The factor

(1 - \cos\theta)

determines the energy transfer. This expression reaches its maximum value when

\cos\theta

is minimized. Since

\cos\theta

ranges from +1 to -1, its minimum value is -1, which occurs at θ = 180°. Therefore, backscattering (D) produces maximum energy loss. Choice (A) is incorrect because at 0° (forward scattering),

\cos(0°) = 1

, making

(1 - \cos\theta) = 0

, resulting in no energy transfer. Choice (B) is wrong because at 45°,

\cos(45°) = \sqrt{2}/2

, giving moderate but not maximum energy transfer. Choice (C) is incorrect because at 90°,

\cos(90°) = 0

, yielding

(1 - \cos\theta) = 1

, which is significant but still less than the maximum value of 2. Remember: In Compton scattering, maximum energy transfer always occurs during head-on collisions where the photon bounces straight back (180° backscattering).

Question 12

Two photons with energies $E_1 = 100$ keV and $E_2 = 1000$ keV both undergo Compton scattering at $60°$ . What is the ratio of their wavelength shifts $\frac{\Delta\lambda_1}{\Delta\lambda_2}$ ?

$\frac{1}{10}$
$\frac{1}{1}$ (correct answer)
$10$
$\frac{1}{\sqrt{10}}$
$\sqrt{10}$

Explanation: When you encounter Compton scattering problems, focus on the fundamental relationship that governs wavelength shift. The Compton formula tells us that the change in wavelength depends only on the scattering angle and fundamental constants, not on the initial photon energy. The Compton wavelength shift is given by:

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

where

h

is Planck's constant,

m_e

is the electron rest mass,

c

is the speed of light, and

\theta

is the scattering angle. Notice that the initial photon energy doesn't appear anywhere in this equation. Since both photons scatter at the same angle (60°), they experience identical wavelength shifts:

\Delta\lambda_1 = \Delta\lambda_2

. Therefore,

\frac{\Delta\lambda_1}{\Delta\lambda_2} = 1

. Let's examine why the other answers are wrong. Answer (A)

\frac{1}{10}

incorrectly assumes the wavelength shift is inversely proportional to initial energy (

E_2/E_1 = 10

). Answer (C)

10

mistakenly suggests the shift is directly proportional to initial energy (

E_1/E_2 = 1/10

, so its inverse is 10). Answer (D)

\frac{1}{\sqrt{10}}

represents some other incorrect relationship involving the square root of the energy ratio. The key insight for Compton scattering problems: wavelength shift depends only on scattering angle, not initial photon energy. This is what makes Compton scattering fundamentally different from other photon interactions where energy does matter.

Question 13

The Compton wavelength $\lambda_c = \frac{h}{m_ec}$ represents a fundamental length scale in quantum mechanics. What is the physical significance of this quantity in Compton scattering?

It is the wavelength of the incident photon that produces maximum scattering
It is the minimum wavelength shift possible in any Compton scattering event
It is the wavelength shift for $90°$ scattering and represents the scale where quantum effects dominate (correct answer)
It is the wavelength of photons that have energy equal to the electron rest mass
It is the de Broglie wavelength of the recoiling electron after maximum energy transfer

Explanation: When you encounter questions about the Compton wavelength, you're dealing with a fundamental quantum mechanical length scale that emerges from the interplay between photon momentum and electron rest mass. The Compton wavelength

\lambda_c = \frac{h}{m_ec}

represents the characteristic wavelength shift when a photon scatters off an electron at 90°. This comes directly from the Compton scattering formula:

\Delta\lambda = \frac{h}{m_ec}(1 - \cos\theta)

. At

\theta = 90°

, we get

\Delta\lambda = \lambda_c

. More importantly, this quantity marks the length scale where quantum effects become dominant over classical physics—when photon wavelengths approach

\lambda_c

, you must use quantum mechanics rather than classical wave theory. Choice A is incorrect because there's no single incident wavelength that produces maximum scattering—scattering probability depends on the differential cross-section, not a specific wavelength. Choice B misunderstands the scattering formula: the minimum wavelength shift is actually zero (for

\theta = 0°

, forward scattering), not

\lambda_c

. Choice D confuses the Compton wavelength with the wavelength of a photon whose energy equals

m_ec^2

—that would be

\lambda = \frac{h}{m_ec}

, which happens to equal

\lambda_c

numerically but represents a different physical concept. Remember that fundamental length scales in physics often mark transitions between different physical regimes. The Compton wavelength specifically indicates where quantum effects in photon-matter interactions become unavoidable—a key insight for understanding high-energy electromagnetic processes.

Question 14

An incident photon has energy $E_0 = 2m_ec^2$ . After Compton scattering at $90°$ , what is the ratio of the scattered photon energy to the recoiling electron's kinetic energy?

$1:1$
$1:2$ (correct answer)
$2:3$
$3:2$
$2:1$

Explanation: Compton scattering questions test your understanding of photon-electron collisions and energy-momentum conservation. When you see specific scattering angles and initial photon energies, you'll need to apply the Compton scattering formula systematically. Start with the Compton wavelength shift formula:

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

. For

90°

scattering,

\cos 90° = 0

, so

\Delta \lambda = \frac{h}{m_e c}

. Since the initial photon energy is

E_0 = 2m_e c^2

, its wavelength is

\lambda_0 = \frac{hc}{E_0} = \frac{hc}{2m_e c^2} = \frac{h}{2m_e c}

. The scattered photon's wavelength becomes:

\lambda' = \lambda_0 + \Delta \lambda = \frac{h}{2m_e c} + \frac{h}{m_e c} = \frac{3h}{2m_e c}

Therefore, the scattered photon energy is:

E' = \frac{hc}{\lambda'} = \frac{2m_e c^2}{3}

By energy conservation, the electron's kinetic energy is:

K_e = E_0 - E' = 2m_e c^2 - \frac{2m_e c^2}{3} = \frac{4m_e c^2}{3}

The ratio is:

\frac{E'}{K_e} = \frac{2m_e c^2/3}{4m_e c^2/3} = \frac{1}{2}

, confirming answer B. A (1:1) incorrectly assumes equal energy sharing. C (2:3) and D (3:2) likely come from mixing up the numerator and denominator or confusing the energy fractions. Strategy tip: Always work through Compton problems systematically using the wavelength shift formula first, then apply energy conservation. The math is straightforward once you establish the correct wavelengths.

Question 15

In Compton scattering, what happens to the Compton shift $\Delta\lambda$ when the incident photon energy approaches the electron rest mass energy?

The shift approaches zero because relativistic effects become negligible
The shift becomes infinite due to relativistic breakdown of the classical formula
The shift remains finite and depends only on scattering angle, independent of photon energy (correct answer)
The shift oscillates between positive and negative values due to quantum interference
The shift becomes negative, indicating blue-shifting of the scattered photon

Explanation: When you encounter Compton scattering problems, focus on the fundamental formula and what each term represents physically. The Compton shift is given by

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

, where

h

is Planck's constant,

m_e

is the electron rest mass,

c

is the speed of light, and

\theta

is the scattering angle. The key insight is that this formula contains no dependence on the incident photon energy—only on the scattering angle

\theta

. The wavelength shift

\Delta\lambda

is determined entirely by the fundamental constants and the geometry of the collision. When the incident photon energy approaches the electron rest mass energy (

E = m_e c^2 \approx 0.511

MeV), the shift remains finite and depends only on the scattering angle. Answer A is incorrect because relativistic effects don't make the shift approach zero—they're already built into the correct relativistic derivation of the Compton formula. Answer B is wrong because the formula doesn't break down or become infinite; it remains valid even for high-energy photons where relativistic effects are important. Answer D is incorrect because quantum interference doesn't cause oscillating shifts—the wavelength shift is always positive (photons lose energy in Compton scattering). Remember this key principle: the Compton shift formula is energy-independent. This counterintuitive result means that whether you're dealing with low-energy X-rays or high-energy gamma rays, the wavelength shift depends only on the scattering angle, making Compton scattering a powerful tool for studying photon-electron interactions across all energy ranges.

Question 16

A photon with energy $E_0$ undergoes Compton scattering. At what scattering angle does the scattered photon retain exactly $\frac{1}{3}$ of its original energy? (Express in terms of the ratio $\frac{E_0}{m_ec^2}$ )

$\cos^{-1}\left(1 - \frac{m_ec^2}{E_0}\right)$
$\cos^{-1}\left(1 - \frac{2m_ec^2}{E_0}\right)$ (correct answer)
$\cos^{-1}\left(1 - \frac{3m_ec^2}{E_0}\right)$
$\cos^{-1}\left(1 + \frac{2m_ec^2}{E_0}\right)$
$\cos^{-1}\left(\frac{2m_ec^2}{E_0}\right)$

Explanation: When you encounter Compton scattering problems, you're dealing with photon-electron collisions where energy and momentum are conserved. The key relationship is the Compton formula:

\frac{1}{E'} - \frac{1}{E_0} = \frac{1 - \cos\theta}{m_ec^2}

, where

E'

is the scattered photon energy and

\theta

is the scattering angle. Since the scattered photon retains

\frac{1}{3}

of its original energy, we have

E' = \frac{E_0}{3}

. Substituting into the Compton formula:

\frac{1}{E_0/3} - \frac{1}{E_0} = \frac{1 - \cos\theta}{m_ec^2}

\frac{3}{E_0} - \frac{1}{E_0} = \frac{1 - \cos\theta}{m_ec^2}

\frac{2}{E_0} = \frac{1 - \cos\theta}{m_ec^2}

Solving for

\cos\theta

1 - \cos\theta = \frac{2m_ec^2}{E_0}

\cos\theta = 1 - \frac{2m_ec^2}{E_0}

Therefore:

\theta = \cos^{-1}\left(1 - \frac{2m_ec^2}{E_0}\right)

Answer A uses the coefficient 1 instead of 2, which would correspond to retaining

\frac{1}{2}

of the original energy. Answer C uses coefficient 3, corresponding to retaining

\frac{1}{4}

of the original energy. Answer D has the wrong sign (addition instead of subtraction), which would give an impossible result since

\cos\theta

would exceed 1 for typical photon energies. Remember: In Compton scattering problems, always identify what fraction of energy is retained first, then substitute carefully into the standard formula. The coefficient in the final answer directly relates to how the energy changes.

Question 17

A photon undergoes Compton scattering with an electron initially at rest. After the collision, both the scattered photon and recoiling electron move in the same hemisphere relative to the initial photon direction. What constraint does this place on the scattering angle of the photon?

$0° < \theta < 45°$
$0° < \theta < 60°$
$0° < \theta < 90°$ (correct answer)
$45° < \theta < 135°$
This situation is impossible due to momentum conservation

Explanation: When analyzing Compton scattering problems involving momentum conservation, you need to consider both the photon and electron trajectories after collision. The key insight is understanding what "same hemisphere" means and how conservation laws constrain the possible outcomes. In Compton scattering, both momentum and energy must be conserved. When a photon scatters off an initially stationary electron, the photon loses energy (increases in wavelength) and the electron recoils. The critical constraint here is that both particles move in the same hemisphere relative to the initial photon direction. For both the scattered photon and recoiling electron to move in the same hemisphere, they must both have forward momentum components in the original photon's direction. This happens when the photon scattering angle

\theta

is less than 90°. At exactly

\theta = 90°

, the photon moves perpendicular to its original path, and momentum conservation forces the electron to recoil forward. For

\theta > 90°

, the photon would scatter backward while the electron moves forward, placing them in opposite hemispheres. Answer A (

0° < \theta < 45°

) is too restrictive—there's no physical reason why scattering can't occur between 45° and 90°. Answer B (

0° < \theta < 60°

) similarly imposes an artificial upper limit. Answer D (

45° < \theta < 135°

) incorrectly includes backward scattering angles where the particles would be in opposite hemispheres. The correct answer is C:

0° < \theta < 90°

, representing all forward scattering angles. Study tip: In scattering problems, always draw momentum vectors before and after collision—visualizing the geometry often reveals the physical constraints more clearly than equations alone.

Question 18

In a Compton scattering experiment, a photon collides with an electron initially at rest. The scattered photon emerges at $\theta = 120°$ with respect to the incident direction. If the initial photon had energy $E_0 = 1.02$ MeV, what is the kinetic energy of the recoiling electron? (Given: $m_e c^2 = 0.511$ MeV)

$K_e = 0.255$ MeV
$K_e = 0.340$ MeV
$K_e = 0.510$ MeV
$K_e = 0.680$ MeV (correct answer)

Explanation: Using the Compton scattering formula:

E' = \frac{E_0}{1 + \frac{E_0}{m_e c^2}(1 - \cos\theta)}

. With

E_0 = 1.02

MeV,

m_e c^2 = 0.511

MeV, and

\theta = 120°

(so

\cos 120° = -0.5

E' = \frac{1.02}{1 + \frac{1.02}{0.511}(1 - (-0.5))} = \frac{1.02}{1 + 2 \times 1.5} = \frac{1.02}{4} = 0.255

MeV. By energy conservation:

K_e = E_0 - E' = 1.02 - 0.255 = 0.765

MeV, which rounds to choice D. Choice A is the scattered photon energy. Choice B represents a calculation error. Choice C is approximately the electron rest energy.

Question 19

A photon with wavelength $\lambda_0 = 2.43 \times 10^{-12}$ m undergoes Compton scattering from an electron at rest. After the collision, the scattered photon has wavelength $\lambda' = 4.86 \times 10^{-12}$ m. What is the kinetic energy of the recoiling electron? (Given: $h = 6.63 \times 10^{-34}$ J·s, $c = 3.00 \times 10^8$ m/s)

$2.05 \times 10^{-14}$ J (correct answer)
$4.09 \times 10^{-14}$ J
$6.14 \times 10^{-14}$ J
$8.18 \times 10^{-14}$ J

Explanation: The kinetic energy of the recoiling electron equals the difference between the initial and final photon energies. Initial photon energy:

E_0 = hc/\lambda_0 = (6.63 \times 10^{-34})(3.00 \times 10^8)/(2.43 \times 10^{-12}) = 8.18 \times 10^{-14}

J. Final photon energy:

E' = hc/\lambda' = (6.63 \times 10^{-34})(3.00 \times 10^8)/(4.86 \times 10^{-12}) = 4.09 \times 10^{-14}

J. By energy conservation:

K_e = E_0 - E' = 8.18 \times 10^{-14} - 4.09 \times 10^{-14} = 2.05 \times 10^{-14}

J. Choice B is the final photon energy alone. Choice C is 1.5 times the correct answer. Choice D is the initial photon energy alone.

Question 20

A high-energy photon undergoes Compton scattering from an electron at rest. The scattered photon emerges at an angle of $60°$ from the incident direction with energy $E' = 0.80 E_0$ , where $E_0$ is the initial photon energy. What is the ratio of the initial photon wavelength to the Compton wavelength of the electron?

$\frac{\lambda_0}{\lambda_C} = 0.25$ (correct answer)
$\frac{\lambda_0}{\lambda_C} = 0.50$
$\frac{\lambda_0}{\lambda_C} = 1.00$
$\frac{\lambda_0}{\lambda_C} = 2.00$

Explanation: From the Compton formula:

\lambda' - \lambda_0 = \lambda_C(1 - \cos 60°) = \lambda_C(1 - 0.5) = 0.5\lambda_C

. Since

E' = 0.80 E_0

, we have

\lambda'/\lambda_0 = E_0/E' = 1/0.80 = 1.25

. Therefore

\lambda' = 1.25\lambda_0

. Substituting:

1.25\lambda_0 - \lambda_0 = 0.5\lambda_C

, so

0.25\lambda_0 = 0.5\lambda_C

, giving

\lambda_0/\lambda_C = 0.5/0.25 = 0.25

. Choice B uses the wavelength shift ratio incorrectly. Choice C assumes the wavelengths are equal. Choice D inverts the correct ratio.