Physics 2 Quiz: Photon Momentum
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Photon MomentumQuestion 1 of 8

A beam of monochromatic light with wavelength λ\lambda passes through a narrow slit of width aa, producing a single-slit diffraction pattern on a distant screen. A student claims: 'The uncertainty in the photon's transverse momentum after passing through the slit is on the order of h/ah/a, consistent with the Heisenberg uncertainty principle.'

Which of the following best evaluates the student's claim in terms of photon momentum concepts?

The claim is incorrect. The correct transverse momentum uncertainty is Δpx≈hλ/a2\Delta p_x \approx h\lambda/a^2, because the diffraction angle θ≈λ/a\theta \approx \lambda/a must be multiplied by the photon's wavelength λ\lambda (rather than its momentum h/λh/\lambda) to properly convert the angular spread into a transverse momentum spread.
The claim is incorrect. The photon's transverse momentum uncertainty is Δpx≈h/λ\Delta p_x \approx h/\lambda, not h/ah/a, because the photon's total momentum magnitude is fixed at h/λh/\lambda by its wavelength, and diffraction only redirects this momentum without introducing any new uncertainty beyond what the wavelength itself sets.
The claim is correct. Confining the photon to width aa gives Δx≈a\Delta x \approx a, so by the uncertainty principle Δpx∼h/a\Delta p_x \sim h/a. This is confirmed by diffraction: the central maximum spans angles up to sin⁡θ≈λ/a\sin\theta \approx \lambda/a, giving transverse momentum spread Δpx≈(h/λ)(λ/a)=h/a\Delta p_x \approx (h/\lambda)(\lambda/a) = h/a.
The claim is valid only when a≫λa \gg \lambda. For a≲λa \lesssim \lambda, diffraction is so strong that the photon spreads into all angles, its transverse momentum uncertainty saturates at h/λh/\lambda, and the uncertainty principle in the form Δpx∼h/a\Delta p_x \sim h/a breaks down and can no longer be applied to predict the diffraction spread.
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Physics 2 Quiz

Physics 2 Quiz: Photon Momentum

Practice Photon Momentum in Physics 2 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 Photon Momentum, giving you a quick way to practice the rules, question types, and explanations that matter most for Physics 2.

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

A beam of monochromatic light with wavelength λ\lambda passes through a narrow slit of width aa, producing a single-slit diffraction pattern on a distant screen. A student claims: 'The uncertainty in the photon's transverse momentum after passing through the slit is on the order of h/ah/a, consistent with the Heisenberg uncertainty principle.'

Which of the following best evaluates the student's claim in terms of photon momentum concepts?

  1. The claim is incorrect. The correct transverse momentum uncertainty is Δpx≈hλ/a2\Delta p_x \approx h\lambda/a^2, because the diffraction angle θ≈λ/a\theta \approx \lambda/a must be multiplied by the photon's wavelength λ\lambda (rather than its momentum h/λh/\lambda) to properly convert the angular spread into a transverse momentum spread.
  2. The claim is incorrect. The photon's transverse momentum uncertainty is Δpx≈h/λ\Delta p_x \approx h/\lambda, not h/ah/a, because the photon's total momentum magnitude is fixed at h/λh/\lambda by its wavelength, and diffraction only redirects this momentum without introducing any new uncertainty beyond what the wavelength itself sets.
  3. The claim is correct. Confining the photon to width aa gives Δx≈a\Delta x \approx a, so by the uncertainty principle Δpx∼h/a\Delta p_x \sim h/a. This is confirmed by diffraction: the central maximum spans angles up to sin⁡θ≈λ/a\sin\theta \approx \lambda/a, giving transverse momentum spread Δpx≈(h/λ)(λ/a)=h/a\Delta p_x \approx (h/\lambda)(\lambda/a) = h/a. (correct answer)
  4. The claim is valid only when a≫λa \gg \lambda. For a≲λa \lesssim \lambda, diffraction is so strong that the photon spreads into all angles, its transverse momentum uncertainty saturates at h/λh/\lambda, and the uncertainty principle in the form Δpx∼h/a\Delta p_x \sim h/a breaks down and can no longer be applied to predict the diffraction spread.
Explanation: When a question connects diffraction patterns to the Heisenberg uncertainty principle, your job is to link the spatial confinement of the photon to its momentum spread using two parallel lines of reasoning that should agree. Here's the core logic that makes C correct. When a photon passes through a slit of width aa, its transverse position is confined to Δx≈a\Delta x \approx a. The uncertainty principle then demands Δpx∼h/a\Delta p_x \sim h/a. You can verify this independently through diffraction: the first minimum occurs at sin⁡θ≈λ/a\sin\theta \approx \lambda/a, so the angular spread of the central maximum is θ≈λ/a\theta \approx \lambda/a. The photon's total momentum is p=h/λp = h/\lambda, so the transverse component spread is Δpx≈psin⁡θ=(h/λ)(λ/a)=h/a\Delta p_x \approx p\sin\theta = (h/\lambda)(\lambda/a) = h/a. Both approaches give the same result — a satisfying consistency check. A is wrong because it multiplies the diffraction angle by λ\lambda instead of the photon's momentum h/λh/\lambda. To convert an angular spread into a momentum spread, you multiply by the total momentum magnitude, not by wavelength. This produces a dimensionally incorrect result. B is wrong because it confuses the total momentum h/λh/\lambda with the transverse momentum uncertainty. Diffraction absolutely introduces a spread in transverse momentum — that spread depends on slit width aa, not just on wavelength. D is wrong because the uncertainty principle doesn't "break down" for small slits. When a≲λa \lesssim \lambda, the formula Δpx∼h/a\Delta p_x \sim h/a still holds — it simply predicts very large angular spread, which is exactly what's observed. Study tip: Always verify uncertainty principle results using two methods — direct application of Δx Δp∼h\Delta x \, \Delta p \sim h and physical geometry. If they agree, you're on solid ground.

Question 2

A particle physicist considers a single photon with momentum p=h/λp = h/\lambda. She wants to double the photon's de Broglie momentum. She has two options: (I) halve the wavelength, or (II) double the frequency. Which of the following correctly evaluates these options?

  1. Both options double the photon momentum and are physically equivalent, because halving the wavelength is identical to doubling the frequency for electromagnetic radiation, and either change alone is sufficient to double pp. (correct answer)
  2. Option I doubles the momentum but Option II does not, because momentum depends on wavelength via p=h/λp = h/\lambda directly, whereas frequency enters only through energy E=hfE = hf and affects momentum only after an additional relativistic correction is applied.
  3. Option II doubles the momentum but Option I does not, because p=hf/cp = hf/c shows momentum is proportional to frequency, while halving the wavelength changes the photon's spatial extent without necessarily increasing its momentum unless the medium's index of refraction is also considered.
  4. Neither option alone guarantees doubled momentum without specifying the medium, because both λ\lambda and ff depend on the index of refraction nn, and only the vacuum wavelength determines photon momentum via p=h/λ0p = h/\lambda_0.
Explanation: When dealing with photon momentum, the key is recognizing that wavelength and frequency are not independent properties — they're locked together by the wave relationship c=fλc = f\lambda. This means any change to one automatically changes the other, and you cannot treat them as separate handles on momentum. The momentum formula p=h/λp = h/\lambda tells you directly that halving λ\lambda doubles pp. That's Option I. But now consider Option II: if you double the frequency, then since c=fλc = f\lambda must hold for a photon in vacuum (cc is fixed), doubling ff forces λ\lambda to halve. Substituting into p=h/λp = h/\lambda, a halved wavelength doubles the momentum. You can also see this through the equivalent form p=hf/cp = hf/c — doubling ff directly doubles pp. Both options lead to exactly the same physical result: a photon with half the original wavelength and double the original frequency. Answer A is correct. Answer B is wrong because it treats frequency as somehow secondary to momentum, requiring an "extra relativistic step." For photons, E=hfE = hf and p=E/c=hf/cp = E/c = hf/c are directly connected — no additional correction is needed. Answer C is wrong because it claims halving the wavelength doesn't increase momentum unless the medium's index of refraction is considered. In vacuum (the standard assumption here), p=h/λp = h/\lambda holds cleanly, and spatial extent is irrelevant. Answer D is wrong because in vacuum there is no index of refraction ambiguity — λ\lambda and ff are uniquely related through cc, so either one fully determines momentum. Study tip: Whenever a photon question separates λ\lambda and ff as if they're independent, immediately invoke c=fλc = f\lambda — they always move together in vacuum.

Question 3

A photon of wavelength λ0\lambda_0 is absorbed by a hydrogen atom in its ground state, promoting the electron to the n=2n = 2 level (E2−E1=10.2 eVE_2 - E_1 = 10.2 \text{ eV}). Immediately after absorption, which statement about the atom's center-of-mass momentum is most accurate?

  1. The atom's center-of-mass momentum increases by h/λ0h/\lambda_0 minus the electron's orbital momentum in the n=2n=2 state, because part of the photon's momentum goes into the electron's increased angular momentum and only the remainder contributes to center-of-mass translation.
  2. The atom's center-of-mass momentum does not change, because the photon's momentum is transferred to the electron specifically, not to the atom as a whole, and the electron is bound inside the atom with its momentum quantized by the orbital quantum numbers.
  3. The atom's center-of-mass momentum increases by h/λ0h/\lambda_0 in the direction of the photon's propagation, because by conservation of momentum the absorbed photon's momentum is transferred entirely to the atom as a whole, regardless of the internal electronic transition. (correct answer)
  4. The atom's center-of-mass momentum increases by h/λ0h/\lambda_0 only if the photon's energy exactly matches the transition energy; if there is any detuning, the momentum transfer is reduced by the ratio of the transition energy to the photon energy.
Explanation: When a photon is absorbed by any system, the fundamental law governing the outcome is conservation of total momentum — a law that applies to the atom as a whole, not just to individual subparticles inside it. This is the key lens for this question. Before absorption, the photon carries momentum p=h/λ0p = h/\lambda_0 in its direction of travel. After absorption, that photon no longer exists — its momentum must go somewhere. Since the atom (electron + nucleus together) is the absorbing system, the entire momentum h/λ0h/\lambda_0 is transferred to the atom's center of mass. The internal rearrangement of the electron into the n=2n = 2 orbital is exactly that — internal. Internal changes do not affect the total momentum of the system. This confirms C as correct. A is wrong because angular momentum and linear (center-of-mass) momentum are completely different physical quantities. The electron's orbital angular momentum in the n=2n = 2 state cannot "consume" a portion of the photon's linear momentum — they don't subtract from each other in any physical sense. B is wrong because it treats the electron as somehow separate from the atom. The electron is bound to the nucleus; together they form the atom. Momentum is conserved for the entire atom, not distributed selectively to one constituent. D is wrong because momentum conservation holds regardless of energy matching conditions. If the photon is absorbed at all, its full momentum h/λ0h/\lambda_0 is transferred — the transition energy is irrelevant to the momentum bookkeeping. Study tip: Whenever you see absorption or emission problems, keep two conservation laws separate in your mind: energy determines whether a transition occurs (resonance condition), while momentum conservation determines recoil — and recoil always belongs to the whole atom.

Question 4

A laser emits light of wavelength λ=500 nm\lambda = 500 \text{ nm} at a power of P=2.0 mWP = 2.0 \text{ mW}. The beam is directed perpendicularly onto a perfectly absorbing surface for a time interval Δt=1.0 s\Delta t = 1.0 \text{ s}.

What is the total momentum delivered to the absorbing surface during this interval? (Use h=6.626×10−34 J⋅sh = 6.626 \times 10^{-34} \text{ J}\cdot\text{s} and c=3.00×108 m/sc = 3.00 \times 10^8 \text{ m/s})

  1. 6.7×10−12 kg⋅m/s6.7 \times 10^{-12} \text{ kg}\cdot\text{m/s}, obtained by dividing the total energy deposited by the speed of light, consistent with the relation ptotal=Etotal/cp_{\text{total}} = E_{\text{total}}/c (correct answer)
  2. 1.3×10−11 kg⋅m/s1.3 \times 10^{-11} \text{ kg}\cdot\text{m/s}, obtained by treating the surface as a perfect reflector and doubling the single-photon momentum, then multiplying by the photon flux
  3. 2.65×10−27 kg⋅m/s2.65 \times 10^{-27} \text{ kg}\cdot\text{m/s}, the momentum of a single 500 nm photon, without accounting for the number of photons arriving during the full time interval
  4. 3.3×10−12 kg⋅m/s3.3 \times 10^{-12} \text{ kg}\cdot\text{m/s}, obtained by computing the single-photon momentum and multiplying by the number of photons per second but forgetting to account for the full power conversion to energy
Explanation: When a laser beam strikes a perfectly absorbing surface, you're dealing with radiation pressure and photon momentum. The key insight is that electromagnetic radiation carries momentum, and for an absorbing surface, the total momentum delivered equals the total energy deposited divided by the speed of light: ptotal=Etotal/cp_{\text{total}} = E_{\text{total}}/c. Here's the calculation for choice A. The total energy deposited is simply power times time: Etotal=P⋅Δt=(2.0×10−3 W)(1.0 s)=2.0×10−3 JE_{\text{total}} = P \cdot \Delta t = (2.0 \times 10^{-3}\ \text{W})(1.0\ \text{s}) = 2.0 \times 10^{-3}\ \text{J}. Dividing by cc gives ptotal=2.0×10−33.00×108≈6.7×10−12 kg⋅m/sp_{\text{total}} = \frac{2.0 \times 10^{-3}}{3.00 \times 10^8} \approx 6.7 \times 10^{-12}\ \text{kg}\cdot\text{m/s}, confirming A is correct. Choice B is the trap for perfect reflectors, not absorbers. When light reflects, the momentum change doubles (the photon reverses direction), giving p=2E/cp = 2E/c. The problem explicitly states the surface is perfectly absorbing, so this doubling doesn't apply. Choice C gives only the momentum of a single photon (p=h/λ≈1.33×10−27 kg⋅m/sp = h/\lambda \approx 1.33 \times 10^{-27}\ \text{kg}\cdot\text{m/s}) without multiplying by the enormous number of photons arriving over the full interval. This ignores the cumulative effect of the beam entirely. Choice D suggests a unit/conversion error — computing single-photon momentum times photons-per-second but failing to correctly incorporate the full power, essentially dropping a factor that accounts for the complete energy delivered. Study tip: Always distinguish absorbing from reflecting surfaces on radiation pressure problems — the factor of 2 separating those cases is one of the most common traps on Physics 2 exams.

Question 5

A sodium atom at rest emits a photon of wavelength λ=589 nm\lambda = 589 \text{ nm} during a spontaneous emission event. The mass of a sodium atom is mNa=3.82×10−26 kgm_{\text{Na}} = 3.82 \times 10^{-26} \text{ kg}.

Immediately after emission, what is the recoil speed of the sodium atom, and in which direction does it move relative to the emitted photon?

  1. v≈2.95×10−2 m/sv \approx 2.95 \times 10^{-2} \text{ m/s}, and the atom moves in the same direction as the emitted photon because the photon carries energy away, which by the work-energy theorem pushes the atom forward in the direction of energy flow.
  2. v≈2.95×10−2 m/sv \approx 2.95 \times 10^{-2} \text{ m/s}, and the atom moves in the opposite direction to the emitted photon, consistent with conservation of momentum for an initially stationary system whose total momentum must remain zero. (correct answer)
  3. v≈5.90×10−2 m/sv \approx 5.90 \times 10^{-2} \text{ m/s}, and the atom moves in the opposite direction to the emitted photon, obtained by applying a factor of 2 from the photon's relativistic momentum doubling upon emission into vacuum.
  4. v≈2.95×10−2 m/sv \approx 2.95 \times 10^{-2} \text{ m/s}, and the direction cannot be determined without knowing the polarization of the emitted photon, since the momentum direction of the photon depends on its polarization state at the moment of emission.
Explanation: When a system starts at rest, its total momentum must remain zero at all times — this is conservation of momentum, and it's the core concept being tested here. When the sodium atom emits a photon, the photon carries momentum pphoton=h/λp_{\text{photon}} = h/\lambda, so the atom must recoil with equal and opposite momentum to keep the total at zero. The calculation is straightforward. The photon's momentum is: pphoton=hλ=6.626×10−34589×10−9≈1.125×10−27 kg⋅m/sp_{\text{photon}} = \frac{h}{\lambda} = \frac{6.626 \times 10^{-34}}{589 \times 10^{-9}} \approx 1.125 \times 10^{-27} \text{ kg}\cdot\text{m/s} Setting patom=pphotonp_{\text{atom}} = p_{\text{photon}} and solving for the recoil speed: v=pm=1.125×10−273.82×10−26≈2.95×10−2 m/sv = \frac{p}{m} = \frac{1.125 \times 10^{-27}}{3.82 \times 10^{-26}} \approx 2.95 \times 10^{-2} \text{ m/s} The atom moves opposite to the photon — confirming answer B. Answer A gets the speed right but flips the direction, confusing energy flow with momentum transfer. The work-energy theorem doesn't determine the direction of recoil here; momentum conservation does, and the atom must go the opposite way to cancel the photon's momentum. Answer C doubles the speed without justification. There is no "relativistic momentum doubling" upon emission — photon momentum is simply h/λh/\lambda, period. Answer D invents a false dependency on polarization. A photon's polarization describes the oscillation of its electric field, not the direction of its propagation or linear momentum, which is always along the direction of travel. Study tip: Any time a particle at rest emits or ejects something, immediately invoke conservation of momentum — the two pieces always move in opposite directions, and the magnitudes of their momenta are equal.

Question 6

Two photons travel in opposite directions along the x-axis. Photon 1 has wavelength λ1=200 nm\lambda_1 = 200 \text{ nm} and travels in the +x+x direction. Photon 2 has wavelength λ2=600 nm\lambda_2 = 600 \text{ nm} and travels in the −x-x direction.

What is the magnitude of the total momentum of this two-photon system?

  1. h200 nm+h600 nm≈4.42×10−27 kg⋅m/s\dfrac{h}{200 \text{ nm}} + \dfrac{h}{600 \text{ nm}} \approx 4.42 \times 10^{-27} \text{ kg}\cdot\text{m/s}, obtained by adding the magnitudes of both photon momenta without regard to direction, treating momentum as a positive scalar quantity for each photon
  2. h200 nm−h600 nm≈2.21×10−27 kg⋅m/s\dfrac{h}{200 \text{ nm}} - \dfrac{h}{600 \text{ nm}} \approx 2.21 \times 10^{-27} \text{ kg}\cdot\text{m/s}, obtained by assigning opposite signs to the momenta due to opposite travel directions and taking the magnitude of the net vector sum (correct answer)
  3. h400 nm≈1.66×10−27 kg⋅m/s\dfrac{h}{400 \text{ nm}} \approx 1.66 \times 10^{-27} \text{ kg}\cdot\text{m/s}, obtained by averaging the two wavelengths and computing momentum from the mean wavelength, since the system's effective wavelength is the arithmetic mean
  4. (h200 nm)2+(h600 nm)2≈3.49×10−27 kg⋅m/s\sqrt{\left(\dfrac{h}{200 \text{ nm}}\right)^2 + \left(\dfrac{h}{600 \text{ nm}}\right)^2} \approx 3.49 \times 10^{-27} \text{ kg}\cdot\text{m/s}, obtained by adding the momentum magnitudes in quadrature as one would for perpendicular vectors, treating the antiparallel directions as orthogonal components
Explanation: Whenever you see a question about the total momentum of a multi-particle system, remember that momentum is a vector — direction always matters. For photons, the momentum magnitude is p=h/λp = h/\lambda, but the sign depends on the direction of travel. Here, Photon 1 travels in the +x+x direction, so p1=+h/λ1=+h/200 nmp_1 = +h/\lambda_1 = +h/200\text{ nm}. Photon 2 travels in the −x-x direction, so p2=−h/λ2=−h/600 nmp_2 = -h/\lambda_2 = -h/600\text{ nm}. The total momentum is the vector sum: ptotal=h200 nm−h600 nm=h(3−1600 nm)=2h600 nm=h300 nm≈2.21×10−27 kg⋅m/sp_\text{total} = \frac{h}{200\text{ nm}} - \frac{h}{600\text{ nm}} = h\left(\frac{3-1}{600\text{ nm}}\right) = \frac{2h}{600\text{ nm}} = \frac{h}{300\text{ nm}} \approx 2.21 \times 10^{-27}\text{ kg}\cdot\text{m/s} Taking the magnitude confirms answer B is correct. Answer A is the classic trap: adding both momenta as positive scalars ignores direction entirely. This would only be valid if both photons traveled the same direction. Discarding sign information violates the rules of vector addition. Answer C uses an "average wavelength" — there is no physical principle that allows you to combine two particles' momenta by averaging their wavelengths. This is a fabricated shortcut with no basis in physics. Answer D applies Pythagorean (quadrature) addition, which is correct for perpendicular vectors. These photons travel along the same axis in opposite directions, not at 90° to each other, so quadrature addition is completely inappropriate here. Your takeaway: always assign signs to momenta based on direction before summing, then take the magnitude at the end — never add magnitudes first.

Question 7

A spaceship moves away from Earth at speed v=0.5cv = 0.5c. It emits a laser pulse of wavelength λ0=400 nm\lambda_0 = 400 \text{ nm} (as measured in the ship's rest frame) directed toward Earth.

An observer on Earth measures the momentum of each photon in the laser pulse. Compared to the momentum p0=h/λ0p_0 = h/\lambda_0 that would be measured if the ship were stationary, the Earth observer measures a photon momentum that is:

  1. greater than p0p_0, because the ship is moving away from Earth and the relativistic Doppler effect blue-shifts the photons, shortening their wavelength and increasing their momentum as measured by the Earth observer.
  2. less than p0p_0 by exactly a factor of 1/21/2, because at v=0.5cv = 0.5c the relativistic Doppler formula gives λobs=2λ0\lambda_{\text{obs}} = 2\lambda_0, doubling the wavelength and halving the photon momentum. This follows directly from substituting β=0.5\beta = 0.5 into the approximation λobs≈λ0(1+β)\lambda_{\text{obs}} \approx \lambda_0(1+\beta).
  3. equal to p0p_0, because photon momentum is a Lorentz invariant — since the speed of light is the same in all inertial frames, every observer must measure the same photon wavelength and therefore the same photon momentum, regardless of the relative motion between source and observer.
  4. less than p0p_0, because the ship moves away from Earth, so the photons are red-shifted to a longer wavelength by the relativistic Doppler effect. With β=0.5\beta = 0.5, the observed wavelength is λobs=λ0(1+β)/(1−β)=4003≈693 nm\lambda_{\text{obs}} = \lambda_0\sqrt{(1+\beta)/(1-\beta)} = 400\sqrt{3} \approx 693 \text{ nm}, giving p=h/λobs<p0p = h/\lambda_{\text{obs}} < p_0. (correct answer)
Explanation: When a source moves away from an observer, the relativistic Doppler effect red-shifts the emitted light — stretching the wavelength and reducing photon momentum. Since p=h/λp = h/\lambda, a longer wavelength means lower momentum. This question tests whether you can correctly apply the relativistic (not classical) Doppler formula and connect it to photon momentum. For a source receding at β=v/c=0.5\beta = v/c = 0.5, the observed wavelength is: λobs=λ01+β1−β=4001.50.5=4003≈693 nm\lambda_{\text{obs}} = \lambda_0\sqrt{\frac{1+\beta}{1-\beta}} = 400\sqrt{\frac{1.5}{0.5}} = 400\sqrt{3} \approx 693 \text{ nm} Since λobs>λ0\lambda_{\text{obs}} > \lambda_0, the photon momentum p=h/λobsp = h/\lambda_{\text{obs}} is less than p0p_0, confirming D is correct. A has the physics exactly backwards — a receding source red-shifts light, not blue-shifts it. Blue-shifting occurs when the source approaches the observer. B uses the classical Doppler approximation λobs≈λ0(1+β)\lambda_{\text{obs}} \approx \lambda_0(1+\beta), which gives λobs=600 nm\lambda_{\text{obs}} = 600\text{ nm}, not 2λ0=800 nm2\lambda_0 = 800\text{ nm}. Even if the formula were applied correctly, using the non-relativistic approximation at β=0.5\beta = 0.5 introduces significant error — you must use the full relativistic formula at such speeds. C reflects a common misconception: while the speed of light is invariant across frames, wavelength and frequency — and therefore photon momentum and energy — are not invariant. They transform between frames via the Doppler effect. Study tip: Always identify whether the source is approaching or receding before applying the Doppler formula, and default to the relativistic version (1±β)/(1∓β)\sqrt{(1\pm\beta)/(1\mp\beta)} whenever β\beta is not negligibly small.

Question 8

In a proposed solar sail design, a perfectly reflective sail of area A=1000 m2A = 1000 \text{ m}^2 is oriented perpendicular to sunlight at a distance of r=1.0 AUr = 1.0 \text{ AU} from the Sun. The solar intensity at 1 AU is I=1361 W/m2I = 1361 \text{ W/m}^2. The sail has mass m=10 kgm = 10 \text{ kg}.

Which of the following correctly computes the acceleration of the solar sail due to radiation pressure, and which key factor distinguishes this from the case of a perfectly absorbing sail of identical area and mass?

  1. The acceleration is a=IA/(mc)≈4.54×10−4 m/s2a = IA/(mc) \approx 4.54 \times 10^{-4} \text{ m/s}^2 for the reflective sail, the same as for an absorbing sail, because in both cases each photon transfers its momentum h/λh/\lambda to the sail upon impact, and reflection merely changes the photon's direction without altering the magnitude of momentum exchanged with the surface.
  2. The acceleration is a=IA/(2mc)≈2.27×10−4 m/s2a = IA/(2mc) \approx 2.27 \times 10^{-4} \text{ m/s}^2 for the reflective sail, which is half the acceleration of an absorbing sail because the reflection process splits the photon's momentum symmetrically between the incoming and outgoing beams, with each beam contributing only h/(2λ)h/(2\lambda) of impulse to the sail.
  3. The acceleration is a=2IA/(mc)≈9.07×10−4 m/s2a = 2IA/(mc) \approx 9.07 \times 10^{-4} \text{ m/s}^2 for the reflective sail, which equals the acceleration of the absorbing sail because the factor of 2 from reflection is exactly cancelled by the fact that reflected photons carry energy away from the sail, reducing the net momentum deposited to the same value as for pure absorption.
  4. The acceleration is a=2IA/(mc)≈9.07×10−4 m/s2a = 2IA/(mc) \approx 9.07 \times 10^{-4} \text{ m/s}^2 for the reflective sail. This is twice the acceleration of a perfectly absorbing sail because reflection reverses the photon's momentum, so the momentum transferred per photon is 2h/λ2h/\lambda rather than h/λh/\lambda, doubling the force compared to absorption. (correct answer)
Explanation: When light hits a surface, it exerts a force because photons carry momentum. The key is using impulse-momentum reasoning: how much momentum does the sail gain per photon? For an absorbing sail, a photon arrives with momentum p=h/λp = h/\lambda and is absorbed, so the sail gains Δp=h/λ\Delta p = h/\lambda. For a perfectly reflective sail, that same photon bounces back with momentum −h/λ-h/\lambda, so the change in the photon's momentum is −2h/λ-2h/\lambda, meaning the sail gains Δp=2h/λ\Delta p = 2h/\lambda — exactly double. This is the same principle as a ball bouncing elastically off a wall versus sticking to it. Translating this into macroscopic force: the radiation pressure on an absorbing sail is Pabs=I/cP_{abs} = I/c, giving force F=IA/cF = IA/c and acceleration a=IA/(mc)a = IA/(mc). For a reflective sail, the pressure doubles to Pref=2I/cP_{ref} = 2I/c, giving a=2IA/(mc)=2(1361)(1000)/[(10)(3×108)]≈9.07×10−4 m/s2a = 2IA/(mc) = 2(1361)(1000)/[(10)(3\times10^8)] \approx 9.07\times10^{-4} \text{ m/s}^2. Answer D captures this correctly. Answer A is wrong because it claims both sails produce the same acceleration — it incorrectly ignores the direction reversal of the reflected photon. Answer B gets the reflective formula backwards, claiming reflection halves the force; it invents a fictitious "momentum splitting" that has no physical basis. Answer C arrives at the right number but for a completely wrong reason — reflected photons do not cancel out their own momentum contribution; carrying energy away and transferring momentum are independent effects. Remember: reflection doubles the momentum transfer compared to absorption. Whenever a question involves radiation pressure, always ask yourself whether the surface absorbs or reflects — it's the single most important factor.