Photon Momentum - Physics 2

Which of the following best evaluates the student's claim in terms of photon momentum concepts?
Answer: The claim is correct. Confining the photon to width $a$ gives $\Delta x \approx a$ , so by the uncertainty principle $\Delta p_x \sim h/a$ . This is confirmed by diffraction: the central maximum spans angles up to $\sin\theta \approx \lambda/a$ , giving transverse momentum spread $\Delta p_x \approx (h/\lambda)(\lambda/a) = h/a$ . 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 $a$ , its transverse position is confined to $\Delta x \approx a$ . The uncertainty principle then demands $\Delta p_x \sim h/a$ . You can verify this independently through diffraction: the first minimum occurs at $\sin\theta \approx \lambda/a$ , so the angular spread of the central maximum is $\theta \approx \lambda/a$ . The photon's total momentum is $p = h/\lambda$ , so the transverse component spread is $\Delta p_x \approx p\sin\theta = (h/\lambda)(\lambda/a) = h/a$ . Both approaches give the same result — a satisfying consistency check. A is wrong because it multiplies the diffraction angle by $\lambda$ instead of the photon's momentum $h/\lambda$ . 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 $h/\lambda$ with the transverse momentum uncertainty. Diffraction absolutely introduces a spread in transverse momentum — that spread depends on slit width $a$ , not just on wavelength. D is wrong because the uncertainty principle doesn't "break down" for small slits. When $a \lesssim \lambda$ , the formula $\Delta p_x \sim h/a$ 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 $\Delta x \, \Delta p \sim h$ and physical geometry. If they agree, you're on solid ground.
What is the magnitude of the total momentum of this two-photon system?
Answer: $\dfrac{h}{200 \text{ nm}} - \dfrac{h}{600 \text{ nm}} \approx 2.21 \times 10^{-27} \text{ kg}\cdot\text{m/s}$ , obtained by assigning opposite signs to the momenta due to opposite travel directions and taking the magnitude of the net vector sum 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 $p = h/\lambda$ , but the sign depends on the direction of travel. Here, Photon 1 travels in the $+x$ direction, so $p_1 = +h/\lambda_1 = +h/200\text{ nm}$ . Photon 2 travels in the $-x$ direction, so $p_2 = -h/\lambda_2 = -h/600\text{ nm}$ . The total momentum is the vector sum: $p_\text{total} = \frac{h}{200\text{ nm}} - \frac{h}{600\text{ nm}} = h\left(\frac{3-1}{600\text{ nm}}\right) = \frac{2h}{600\text{ nm}} = \frac{h}{300\text{ nm}} \approx 2.21 \times 10^{-27}\text{ kg}\cdot\text{m/s}$ 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.
A photon of wavelength $\lambda_0$ is absorbed by a hydrogen atom in its ground state, promoting the electron to the $n = 2$ level ( $E_2 - E_1 = 10.2 \text{ eV}$ ). Immediately after absorption, which statement about the atom's center-of-mass momentum is most accurate?
Answer: The atom's center-of-mass momentum increases by $h/\lambda_0$ 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. 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 $p = h/\lambda_0$ 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 $h/\lambda_0$ is transferred to the atom's center of mass. The internal rearrangement of the electron into the $n = 2$ 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 $n = 2$ 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 $h/\lambda_0$ 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.
A particle physicist considers a single photon with momentum $p = h/\lambda$ . 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?
Answer: 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 $p$ . 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 $c = f\lambda$ . This means any change to one automatically changes the other, and you cannot treat them as separate handles on momentum. The momentum formula $p = h/\lambda$ tells you directly that halving $\lambda$ doubles $p$ . That's Option I. But now consider Option II: if you double the frequency, then since $c = f\lambda$ must hold for a photon in vacuum ( $c$ is fixed), doubling $f$ forces $\lambda$ to halve. Substituting into $p = h/\lambda$ , a halved wavelength doubles the momentum. You can also see this through the equivalent form $p = hf/c$ — doubling $f$ directly doubles $p$ . 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, $E = hf$ and $p = E/c = hf/c$ 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), $p = h/\lambda$ holds cleanly, and spatial extent is irrelevant. Answer D is wrong because in vacuum there is no index of refraction ambiguity — $\lambda$ and $f$ are uniquely related through $c$ , so either one fully determines momentum. Study tip: Whenever a photon question separates $\lambda$ and $f$ as if they're independent, immediately invoke $c = f\lambda$ — they always move together in vacuum.
What is the total momentum delivered to the absorbing surface during this interval? (Use $h = 6.626 \times 10^{-34} \text{ J}\cdot\text{s}$ and $c = 3.00 \times 10^8 \text{ m/s}$ )
Answer: $6.7 \times 10^{-12} \text{ kg}\cdot\text{m/s}$ , obtained by dividing the total energy deposited by the speed of light, consistent with the relation $p_{\text{total}} = E_{\text{total}}/c$ 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: $p_{\text{total}} = E_{\text{total}}/c$ . Here's the calculation for choice A. The total energy deposited is simply power times time: $E_{\text{total}} = P \cdot \Delta t = (2.0 \times 10^{-3}\ \text{W})(1.0\ \text{s}) = 2.0 \times 10^{-3}\ \text{J}$ . Dividing by $c$ gives $p_{\text{total}} = \frac{2.0 \times 10^{-3}}{3.00 \times 10^8} \approx 6.7 \times 10^{-12}\ \text{kg}\cdot\text{m/s}$ , 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 $p = 2E/c$ . 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 ( $p = h/\lambda \approx 1.33 \times 10^{-27}\ \text{kg}\cdot\text{m/s}$ ) 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.
Immediately after emission, what is the recoil speed of the sodium atom, and in which direction does it move relative to the emitted photon?
Answer: $v \approx 2.95 \times 10^{-2} \text{ m/s}$ , 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. 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 $p_{\text{photon}} = h/\lambda$ , 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: $p_{\text{photon}} = \frac{h}{\lambda} = \frac{6.626 \times 10^{-34}}{589 \times 10^{-9}} \approx 1.125 \times 10^{-27} \text{ kg}\cdot\text{m/s}$ Setting $p_{\text{atom}} = p_{\text{photon}}$ and solving for the recoil speed: $v = \frac{p}{m} = \frac{1.125 \times 10^{-27}}{3.82 \times 10^{-26}} \approx 2.95 \times 10^{-2} \text{ m/s}$ 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 $h/\lambda$ , 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.
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?
Answer: The acceleration is $a = 2IA/(mc) \approx 9.07 \times 10^{-4} \text{ m/s}^2$ 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 $2h/\lambda$ rather than $h/\lambda$ , doubling the force compared to absorption. 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 $p = h/\lambda$ and is absorbed, so the sail gains $\Delta p = h/\lambda$ . For a perfectly reflective sail, that same photon bounces back with momentum $-h/\lambda$ , so the change in the photon's momentum is $-2h/\lambda$ , meaning the sail gains $\Delta p = 2h/\lambda$ — 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 $P_{abs} = I/c$ , giving force $F = IA/c$ and acceleration $a = IA/(mc)$ . For a reflective sail, the pressure doubles to $P_{ref} = 2I/c$ , giving $a = 2IA/(mc) = 2(1361)(1000)/[(10)(3\times10^8)] \approx 9.07\times10^{-4} \text{ m/s}^2$ . 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.
An observer on Earth measures the momentum of each photon in the laser pulse. Compared to the momentum $p_0 = h/\lambda_0$ that would be measured if the ship were stationary, the Earth observer measures a photon momentum that is:
Answer: less than $p_0$ , because the ship moves away from Earth, so the photons are red-shifted to a longer wavelength by the relativistic Doppler effect. With $\beta = 0.5$ , the observed wavelength is $\lambda_{\text{obs}} = \lambda_0\sqrt{(1+\beta)/(1-\beta)} = 400\sqrt{3} \approx 693 \text{ nm}$ , giving $p = h/\lambda_{\text{obs}} < p_0$ . 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 $p = h/\lambda$ , 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 $\beta = v/c = 0.5$ , the observed wavelength is: $\lambda_{\text{obs}} = \lambda_0\sqrt{\frac{1+\beta}{1-\beta}} = 400\sqrt{\frac{1.5}{0.5}} = 400\sqrt{3} \approx 693 \text{ nm}$ Since $\lambda_{\text{obs}} > \lambda_0$ , the photon momentum $p = h/\lambda_{\text{obs}}$ is less than $p_0$ , 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 $\lambda_{\text{obs}} \approx \lambda_0(1+\beta)$ , which gives $\lambda_{\text{obs}} = 600\text{ nm}$ , not $2\lambda_0 = 800\text{ nm}$ . Even if the formula were applied correctly, using the non-relativistic approximation at $\beta = 0.5$ 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 $\sqrt{(1\pm\beta)/(1\mp\beta)}$ whenever $\beta$ is not negligibly small.

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