Physics 2 Quiz: Photon Energy

10 questions · exam conditions

0:00

Photon EnergyQuestion 1 of 10

A photon with energy $E_\gamma = 1.100 \text{ MeV}$ undergoes pair production near an atomic nucleus, creating an electron-positron pair. The rest mass energy of an electron (or positron) is $m_e c^2 = 0.511 \text{ MeV}$ . The electron and positron are created with equal kinetic energies. The positron subsequently slows down and annihilates with a separate stationary electron. What is the energy of each annihilation photon produced?

0.550 MeV, because the positron retains its kinetic energy from pair production, and when it annihilates with an electron at rest, each photon carries half the total energy of the positron (rest mass plus kinetic energy).

0.511 MeV, because the positron loses its kinetic energy to the surrounding medium before annihilating essentially at rest with a stationary electron, so each photon carries only the rest-mass energy of one electron.

0.511 MeV, because the original pair-production photon energy equals exactly twice the electron rest-mass energy, so no kinetic energy is available and both annihilation photons carry only rest-mass energy.

0.275 MeV, because the total energy of the original photon (1.100 MeV) is shared equally among all four photons — two from pair production and two from annihilation.

← Back to quizzes

Physics 2 Quiz

Physics 2 Quiz: Photon Energy

Practice Photon Energy 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 Energy, 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

0.550 MeV, because the positron retains its kinetic energy from pair production, and when it annihilates with an electron at rest, each photon carries half the total energy of the positron (rest mass plus kinetic energy).
0.511 MeV, because the positron loses its kinetic energy to the surrounding medium before annihilating essentially at rest with a stationary electron, so each photon carries only the rest-mass energy of one electron. (correct answer)
0.511 MeV, because the original pair-production photon energy equals exactly twice the electron rest-mass energy, so no kinetic energy is available and both annihilation photons carry only rest-mass energy.
0.275 MeV, because the total energy of the original photon (1.100 MeV) is shared equally among all four photons — two from pair production and two from annihilation.

Explanation: Pair production and annihilation questions require you to track two separate events carefully — they are sequential processes, not simultaneous ones. In pair production, the 1.100 MeV photon creates an electron-positron pair. Since each particle's rest mass energy is 0.511 MeV, the combined rest mass costs

2 \times 0.511 = 1.022 \text{ MeV}

, leaving

1.100 - 1.022 = 0.078 \text{ MeV}

of kinetic energy shared equally between the two particles. The positron then travels through matter and slows down, depositing that 0.078 MeV into the surrounding medium before coming essentially to rest. When it finally annihilates with a stationary electron, both particles are at rest, so the total energy available is just their combined rest mass energy:

2 \times 0.511 = 1.022 \text{ MeV}

. By conservation of momentum, two photons of equal energy are produced, each carrying

0.511 \text{ MeV}

. The correct answer is B. Choice A is wrong because it assumes the positron annihilates while still moving at its pair-production speed — the problem explicitly states it "slows down" first, meaning that kinetic energy is lost to the medium before annihilation occurs. Choice C reaches the right numerical answer (0.511 MeV per photon) but for a completely wrong reason — the original photon energy does not equal exactly twice the rest mass; there is leftover kinetic energy, it's just dissipated before annihilation. Choice D incorrectly treats all four photons as sharing the original energy equally, confusing the two independent events. Study tip: Always identify when and where energy is lost between sequential nuclear/atomic events — the phrase "slows down" is your signal that kinetic energy leaves the system before the next interaction.

Question 2

A radio station broadcasts at a frequency of $f = 100 \text{ MHz}$ with a radiated power of $P = 50 \text{ kW}$ . A quantum-mechanics student argues that because individual radio photons have very low energy, the station must emit an astronomically large number of photons per second, which is why classical electromagnetic theory — rather than quantum theory — adequately describes radio wave behavior.

Which of the following responses best evaluates the student's reasoning and correctly computes the relevant quantity?

The student's reasoning is correct. Each radio photon has energy $E = hf \approx 6.6 \times 10^{-26} \text{ J}$ , and the photon emission rate is $N = P/E \approx 7.6 \times 10^{29} \text{ photons/s}$ . The large photon number means quantum granularity is undetectable, and the field is well-described classically by the correspondence principle.
The student's reasoning is flawed. Each radio photon has energy $E = hf \approx 6.6 \times 10^{-26} \text{ J}$ , giving $N \approx 7.6 \times 10^{29} \text{ photons/s}$ , but large photon numbers do not justify classical treatment — quantum effects always dominate at the microscopic scale regardless of total photon count.
The student's conclusion is correct but the stated reason is incomplete. Each radio photon has energy $E = hf \approx 6.6 \times 10^{-26} \text{ J}$ and $N \approx 7.6 \times 10^{29} \text{ photons/s}$ . The deeper justification is that $hf \ll k_BT$ at room temperature ( $k_BT \approx 4 \times 10^{-21} \text{ J}$ ), placing the field in the classical Rayleigh-Jeans regime where quantum discreteness is negligible. (correct answer)
The student's reasoning is incorrect. Each radio photon has energy $E = hf \approx 6.6 \times 10^{-20} \text{ J}$ (incorrectly using $f = 10^{14} \text{ Hz}$ ), giving $N \approx 7.6 \times 10^{23} \text{ photons/s}$ , and classical theory applies because radio wavelengths are much longer than atomic dimensions.

Explanation: When a question asks you to evaluate physical reasoning — not just compute a number — you need to check both the math and the conceptual justification separately. Here, the calculation is the same across most choices; the real test is whether the reasoning is complete and correct. The photon energy is

E = hf = (6.626 \times 10^{-34})(10^8) \approx 6.6 \times 10^{-26} \text{ J}

, and the emission rate is

N = P/E = (5 \times 10^4)/(6.6 \times 10^{-26}) \approx 7.6 \times 10^{29} \text{ photons/s}

. That math is solid. But why does classical theory work? The strongest justification comes from the correspondence principle applied thermally: at room temperature,

k_BT \approx 4 \times 10^{-21} \text{ J} \gg hf \approx 6.6 \times 10^{-26} \text{ J}

. When

hf \ll k_BT

, the quantum Planck distribution reduces to the classical Rayleigh-Jeans law, meaning thermal fluctuations dominate quantum discreteness entirely. This is why C is correct — the student's conclusion stands, but the large-photon-number argument alone is incomplete. Choice A accepts the student's reasoning at face value without recognizing that "many photons ≈ classical" is an oversimplification that needs thermodynamic grounding. Choice B overcorrects by claiming quantum effects always dominate at microscopic scales — this is simply false; the correspondence principle exists precisely because quantum and classical descriptions must agree in appropriate limits. Choice D uses the wrong frequency (

10^{14}

Hz is optical, not radio), producing an incorrect energy and photon count, and then offers an irrelevant justification about wavelength versus atomic size. Study tip: On questions testing conceptual reasoning, always ask: is the conclusion right, and is the stated reason sufficient? A correct answer with incomplete justification is still a flawed argument — and the MCAT and physics exams love to test exactly that distinction.

Question 3

In a Compton scattering experiment, an X-ray photon with wavelength $\lambda_0$ scatters off a free electron at rest. The scattered photon is detected at an angle of $\theta = 90°$ relative to the incident beam. A student argues that since energy is conserved, the frequency of the scattered photon must satisfy $f' = f_0 - \Delta f$ , where $\Delta f$ is determined solely by the Compton wavelength shift formula $\Delta\lambda = \frac{h}{m_e c}(1-\cos\theta)$ . Which of the following correctly identifies the relationship between the energy lost by the photon and the kinetic energy gained by the electron?

The kinetic energy gained by the electron equals $hf_0 - hf'$ , where $f'$ is computed from $\lambda' = \lambda_0 + h/(m_e c)$ . This equals the recoil kinetic energy only when the electron's motion is treated non-relativistically, which is valid when $\lambda_0 \gg h/(m_e c)$ .
The kinetic energy gained by the electron equals $hf_0 - hf'$ , but the student's claim that $\Delta f$ is determined solely by $\Delta\lambda$ without reference to $\lambda_0$ is incomplete. Because $\Delta f = c/\lambda_0 - c/\lambda' = c\,\Delta\lambda/(\lambda_0\lambda')$ , the frequency shift depends on both $\Delta\lambda$ and $\lambda_0$ , so the energy transferred to the electron also depends on the incident wavelength. (correct answer)
The kinetic energy gained by the electron equals $hf_0 - hf'$ , and because $\Delta\lambda = h/(m_e c)$ at 90° is independent of the incident wavelength, the energy transferred is likewise independent of $\lambda_0$ and depends only on fundamental constants.
The kinetic energy gained by the electron equals $h(f_0 - f')/2$ , because half the photon's lost energy is radiated as bremsstrahlung during the collision, and the Compton formula accounts for only the total energy shift rather than the electron's net kinetic energy gain.

Explanation: Whenever you encounter Compton scattering problems, keep two separate ideas clearly distinguished: the wavelength shift formula and the energy transfer calculation. The Compton formula gives you a shift in wavelength, not directly a shift in frequency or energy — and that distinction is exactly what this question tests. The wavelength shift at

\theta = 90°

\Delta\lambda = \frac{h}{m_e c}(1 - \cos 90°) = \frac{h}{m_e c}

, which is indeed a fixed, universal constant independent of

\lambda_0

. However, the energy lost by the photon is

\Delta E = hf_0 - hf' = hc\left(\frac{1}{\lambda_0} - \frac{1}{\lambda'}\right) = hc\cdot\frac{\Delta\lambda}{\lambda_0 \lambda'}

. Because

\lambda' = \lambda_0 + \Delta\lambda

, this energy depends on

\lambda_0

. A large

\lambda_0

means a small fractional wavelength shift, so very little energy is transferred — even though

\Delta\lambda

is identical. By energy conservation, the electron's recoil kinetic energy equals exactly

hf_0 - hf'

, confirming B as correct. A is partially reasonable but misleading — it implies the energy equality holds only non-relativistically. In fact, the Compton derivation is fully relativistic, so no such restriction is needed for the energy-conservation statement itself. C commits the central error: confusing a constant wavelength shift with a constant energy shift. These are not the same, since energy goes as

1/\lambda

, not

\lambda

. D introduces bremsstrahlung radiation, which has no role in Compton scattering — that's a completely different physical process. Remember:

\Delta\lambda = \text{constant}

does not mean

\Delta E = \text{constant}

. Always convert to energy explicitly using

E = hc/\lambda

Question 4

A hydrogen atom undergoes a transition from the n=3 energy level to the n=1 energy level, emitting a photon. The energy levels of hydrogen are given by $E_n = -\frac{13.6 \text{ eV}}{n^2}$ . A student claims that a second photon with exactly half the frequency of the emitted photon could ionize a ground-state hydrogen atom if two such photons are absorbed simultaneously.

Which of the following best evaluates the student's claim?

The claim is correct because two photons each carrying half the required ionization energy can combine their energies to eject the electron, since the n=3 to n=1 transition emits a photon of 12.09 eV, making each half-frequency photon worth 6.045 eV, and 6.045 + 6.045 = 12.09 eV > 13.6 eV is false, but the combined absorption is still sufficient.
The claim is incorrect because each photon must independently exceed the 13.6 eV ionization threshold; a photon at half the frequency of the emitted photon carries only 6.045 eV, which is below threshold, so no ionization occurs regardless of how many such photons are present.
The claim is incorrect because the n=3 to n=1 transition emits a photon of 12.09 eV, so a photon at half that frequency carries only 6.045 eV. In standard single-photon absorption, each photon interacts independently with the electron; photon energies from separate absorption events do not add together, and 6.045 eV is insufficient to ionize hydrogen (which requires 13.6 eV). (correct answer)
The claim is correct because the n=3 to n=1 transition energy of 12.09 eV exceeds 6.8 eV, meaning a photon at half frequency still carries more than enough energy to reach an excited state from which thermal energy completes ionization under standard laboratory conditions.

Explanation: When evaluating claims about photon absorption and ionization, the key principle is how photons interact with electrons in quantum systems: in standard (single-photon) absorption, each photon is absorbed independently, and its energy either meets the threshold or it doesn't — partial energies from separate events don't accumulate. Start with the energy emitted in the n=3 → n=1 transition:

\Delta E = E_1 - E_3 = -13.6 + \frac{13.6}{9} = -12.09 \text{ eV}

So the emitted photon carries 12.09 eV. A photon at half that frequency carries half the energy:

E = 6.045 \text{ eV}

. Ionizing a ground-state hydrogen atom requires removing the electron from n=1, which demands at least 13.6 eV. Since 6.045 eV < 13.6 eV, a single such photon cannot ionize the atom — and crucially, two photons absorbed in separate events don't "pool" their energy for the electron. Answer C correctly identifies this: each absorption event is independent, 6.045 eV falls short of the 13.6 eV threshold, and the claim fails. Answer A is self-contradictory — it correctly notes that 12.09 eV < 13.6 eV, then reverses course without justification, embodying the exact misconception about energy pooling. Answer B reaches the right conclusion but overstates the rule: it isn't that a photon must exceed 13.6 eV in all contexts, but rather that in standard single-photon absorption, one photon must suffice. Answer D invents a "thermal completion" mechanism that has no basis in standard hydrogen ionization physics. On exam questions involving photon thresholds, always ask: does one photon carry enough energy? Quantum absorption doesn't work like a savings account — you can't deposit two small photons and withdraw one large ionization.

Question 5

The work function of cesium is $\phi = 2.0 \text{ eV}$ . A researcher illuminates a cesium surface simultaneously with two monochromatic light beams: Beam 1 has photon energy $E_1 = 1.5 \text{ eV}$ and Beam 2 has photon energy $E_2 = 3.0 \text{ eV}$ . Both beams have equal intensity $I$ .

Which of the following correctly describes the photoelectric emission from the cesium surface under simultaneous illumination by both beams?

Both beams contribute to photoelectric emission because the combined photon energy of $1.5 + 3.0 = 4.5 \text{ eV}$ exceeds the work function, and the maximum kinetic energy of emitted electrons is $2.5 \text{ eV}$ averaged over contributions from both beams.
Only Beam 2 causes photoelectric emission, producing electrons with maximum kinetic energy $KE_{max} = 1.0 \text{ eV}$ , while Beam 1 photons are individually too low in energy to eject electrons regardless of the presence of Beam 2 or the intensity of illumination. (correct answer)
Only Beam 2 causes photoelectric emission, but the presence of Beam 1 increases the total photoelectric current above what Beam 2 alone would produce, because Beam 1 photons elevate electrons to intermediate energy states from which Beam 2 photons can more efficiently complete the ionization.
Both beams independently cause photoelectric emission, with Beam 1 producing electrons of $KE_{max} = -0.5 \text{ eV}$ (no emission) and Beam 2 producing electrons of $KE_{max} = 1.0 \text{ eV}$ , so the net maximum kinetic energy observed is $1.0 \text{ eV}$ from Beam 2 alone, and Beam 1 slightly reduces the current by exciting surface phonons.

Explanation: Whenever you see a photoelectric effect question, anchor yourself to one fundamental rule: each photon acts alone. A single photon must carry enough energy by itself to overcome the work function — there is no "pooling" of photon energies. The photoelectric equation is

KE_{max} = E_{photon} - \phi

. For Beam 2:

KE_{max} = 3.0 - 2.0 = 1.0 \text{ eV}

. Since this is positive, Beam 2 ejects electrons. For Beam 1:

KE_{max} = 1.5 - 2.0 = -0.5 \text{ eV}

. A negative result simply means the photon lacks sufficient energy — no emission occurs, period. The presence of Beam 2 alongside Beam 1 changes nothing about Beam 1's inability to eject electrons. This confirms B is correct. A commits the classic "energy pooling" error — imagining that two photons can combine their energies to eject one electron. This does not happen in the standard photoelectric effect. Each absorption event involves exactly one photon and one electron. C introduces a fictional "intermediate energy state" mechanism, essentially describing a two-photon process that doesn't occur under normal photoelectric conditions (it would require extraordinarily high intensities, not relevant here). D gets the individual calculations right but then invents a "surface phonon excitation" effect from Beam 1 that reduces current — this has no basis in the photoelectric framework being tested and is a distractor designed to sound technical. Your go-to strategy: whenever a question mentions multiple light sources or high intensity, immediately ask yourself "does each individual photon exceed the work function?" Intensity affects the number of ejected electrons, not whether emission happens at all.

Question 6

Two light sources, X and Y, illuminate separate metal surfaces in a photoelectric experiment. Source X has intensity $I$ and frequency $f_X = 1.5 f_0$ , where $f_0$ is the threshold frequency of metal X. Source Y has intensity $2I$ and frequency $f_Y = 1.2 f_0$ , where $f_0$ is also the threshold frequency of metal Y (the same threshold frequency). Which statement correctly compares the maximum kinetic energy of photoelectrons and the photoelectric current from each metal?

Metal X produces photoelectrons with greater maximum kinetic energy and a smaller photoelectric current than metal Y, because the higher frequency of source X yields more energetic photons while the lower intensity means fewer photons per second strike the surface. (correct answer)
Metal Y produces photoelectrons with greater maximum kinetic energy and a larger photoelectric current than metal X, because intensity governs both the energy of individual photons and the total number of electrons ejected per unit time.
Metal X produces photoelectrons with greater maximum kinetic energy, and metal Y produces a larger photoelectric current, but the relative current comparison cannot be made without knowing the quantum efficiency of each metal surface.
Both metals produce photoelectrons with identical maximum kinetic energies because the threshold frequency is the same for both, and metal Y produces a larger photoelectric current because source Y has twice the intensity of source X.

Explanation: When tackling photoelectric effect questions, you need to track two completely separate quantities: the maximum kinetic energy of ejected electrons (controlled by photon frequency) and the photoelectric current (controlled by photon intensity, i.e., how many photons arrive per second). The maximum kinetic energy follows Einstein's photoelectric equation:

KE_{max} = hf - hf_0

. For metal X:

KE_X = h(1.5f_0) - hf_0 = 0.5hf_0

. For metal Y:

KE_Y = h(1.2f_0) - hf_0 = 0.2hf_0

. Since

0.5hf_0 > 0.2hf_0

, metal X produces more energetic photoelectrons — the higher frequency source wins, regardless of intensity. For current, intensity tells you the energy delivered per unit time. Since source Y delivers twice the intensity (

2I

I

) at similar frequencies, it delivers roughly twice as many photons per second, ejecting more electrons per second and producing a larger current. So answer A is correct: X wins on kinetic energy, Y wins on current. Answer B is wrong on two counts — intensity does not control photon energy (frequency does), and Y doesn't produce higher kinetic energy. Answer C is tempting because quantum efficiency is real, but the question gives you enough information to compare currents through intensity alone — don't overcomplicate it. Answer D is wrong because identical threshold frequencies do not mean identical kinetic energies; what matters is how far each source frequency exceeds

f_0

, and

1.5f_0

exceeds it more than

1.2f_0

. Study tip: Always mentally separate the photoelectric effect into two lanes — frequency lane (determines electron energy) and intensity lane (determines electron count/current). These two quantities are completely independent.

Question 7

A sodium vapor lamp emits photons primarily at two wavelengths: $\lambda_1 = 589.0 \text{ nm}$ and $\lambda_2 = 589.6 \text{ nm}$ (the sodium D-line doublet). A physicist uses this lamp to illuminate a photoelectric cell with a work function of $\phi = 1.82 \text{ eV}$ . The intensity of each line is equal.

Which of the following best describes the photoelectron spectrum — specifically the number of distinct maximum kinetic energy values — produced by this lamp?

Two distinct maximum kinetic energy values are produced, one for each wavelength, with the difference between them equal to $hc(1/\lambda_1 - 1/\lambda_2)$ , which is approximately $3.6 \times 10^{-3} \text{ eV}$ — a splitting that is physically real but experimentally difficult to resolve. (correct answer)
Only one maximum kinetic energy value is produced because the two wavelengths are so close that their photons have effectively identical energies, and the photoelectric effect does not have sufficient energy resolution to distinguish emissions from the two lines.
Two distinct maximum kinetic energy values are produced, with each value equal to $hc/\lambda_i - \phi$ , but the higher-energy electrons from $\lambda_1$ suppress emission from $\lambda_2$ through space-charge effects, effectively producing only one observable peak.
One maximum kinetic energy value is produced because both wavelengths exceed the threshold, and the photoelectric effect selects only the highest-energy photons for emission while lower-energy photons are reflected without interaction.

Explanation: When a question asks about the photoelectric effect with multiple light sources, your first instinct should be to apply Einstein's photoelectric equation independently to each wavelength:

KE_{max} = \frac{hc}{\lambda} - \phi

. Each distinct photon energy produces its own distinct maximum kinetic energy — the work function subtracts the same amount from each, preserving any energy difference between the two photon beams. Here, the two wavelengths differ by only 0.6 nm, giving an energy splitting of

\Delta E = hc\left(\frac{1}{\lambda_1} - \frac{1}{\lambda_2}\right) \approx 3.6 \times 10^{-3} \text{ eV}

. Both wavelengths still exceed the threshold (since

hc/589\text{ nm} \approx 2.11\text{ eV} > \phi = 1.82\text{ eV}

), so both produce photoelectrons — at two genuinely distinct maximum kinetic energies. Answer A correctly captures this: two real, physically distinct values exist, separated by that small but nonzero splitting, even if resolving them experimentally is challenging. That's the right answer. Answer B is wrong because it conflates experimental difficulty with physical reality. The two energies are mathematically distinct regardless of whether a detector can resolve them — physics doesn't round off because instruments are imprecise. Answer C invents a "space-charge suppression" mechanism that simply doesn't apply here. Space-charge effects can limit total current, but they don't selectively erase one energy peak from the spectrum. Answer D describes a fictional selection rule. The photoelectric effect doesn't "choose" only the highest-energy photons — every photon above threshold can independently eject an electron. Study tip: Always treat each wavelength as an independent photon source. The photoelectric effect is a one-photon-one-electron process, so two wavelengths always mean two

KE_{max}

values — whether or not your detector can resolve the difference.

Question 8

A photon is absorbed by a molecule, promoting it from its ground electronic state to an excited electronic state. The excited molecule then undergoes rapid vibrational relaxation (losing energy to the surroundings as heat) before emitting a photon (fluorescence). The emitted fluorescence photon has a lower frequency than the absorbed photon. A student claims this violates conservation of energy because the emitted photon carries less energy than the absorbed photon. Which response most precisely refutes this claim while also identifying the correct relationship between the absorbed photon energy, the emitted photon energy, and the heat dissipated?

The claim is wrong because energy is conserved globally: $hf_{absorbed} = hf_{emitted} + Q$ , but this equality holds only when integrated over the entire ensemble of molecules, since individual molecules may violate energy conservation by amounts within the thermal fluctuation range $k_BT$ .
The claim is wrong because the Heisenberg uncertainty principle allows the emitted photon to have a different frequency without violating energy conservation, since the short lifetime of the excited state creates an energy uncertainty $\Delta E \approx \hbar/\tau$ that accounts for the frequency difference between absorbed and emitted photons.
The claim is wrong because photon energy is not conserved in molecular transitions — only the total number of photons is conserved. The molecule stores the energy difference as potential energy in its electronic configuration, which is not accessible to thermal measurements.
The claim is wrong because energy is conserved: $hf_{absorbed} = hf_{emitted} + Q$ , where $Q$ is the thermal energy released during vibrational relaxation. The absorbed photon energy is partitioned between the fluorescence photon and heat, so no violation occurs and the emitted photon frequency is lower by exactly the amount of energy lost to vibrational modes. (correct answer)

Explanation: When you see a question about fluorescence or photon emission, immediately think about energy conservation across the entire process — not just the photon-in versus photon-out comparison. Here's what actually happens: a photon is absorbed, exciting the molecule to a high-energy electronic state. Before fluorescence occurs, the molecule sheds some of that energy as heat through vibrational relaxation — collisions with neighboring molecules redistribute energy into thermal motion. Only then does the molecule emit a fluorescence photon from this lower-energy excited state. The complete energy balance is:

hf_{\text{absorbed}} = hf_{\text{emitted}} + Q

The absorbed photon's energy is partitioned between the emitted photon and the thermal energy

Q

released during vibrational relaxation. This is precisely what D states, making it correct. Energy is perfectly conserved — it simply changes form, with some becoming heat rather than light. This frequency downshift is known as the Stokes shift, a foundational concept in spectroscopy. A is wrong because energy conservation applies to every individual molecule, not just statistical ensembles. Individual molecules don't get a

k_BT

"free pass" to violate energy conservation — that's a misapplication of thermal fluctuation arguments. B is wrong because the Heisenberg uncertainty principle creates a small natural linewidth in the emission spectrum, but this effect is far too small to account for the entire Stokes shift. The frequency difference here comes from vibrational relaxation, not lifetime broadening. C is wrong on a fundamental level — photon number is not conserved in quantum optics; photon energy contributes to total energy conservation. The molecule does not store unexplained "inaccessible" potential energy. As a study tip: whenever a question claims energy isn't conserved, look for a hidden energy pathway — in molecular spectroscopy, that pathway is almost always heat.

Question 9

An LED emits light centered at $\lambda = 620 \text{ nm}$ (red) when forward-biased with a voltage $V = 2.1 \text{ V}$ . A second LED emits light centered at $\lambda = 450 \text{ nm}$ (blue) when forward-biased at $V = 2.8 \text{ V}$ . A student notes that the forward voltage of each LED approximates the photon energy divided by the electron charge (i.e., $eV \approx hc/\lambda$ ), and concludes that if the forward voltage is halved for each LED, the emitted photon wavelength will double.

Which of the following best evaluates the student's conclusion?

The student's conclusion is incorrect because the relationship $eV \approx hc/\lambda$ applies only to the peak emission wavelength at the rated current; at reduced voltage, the LED emits a continuous spectrum spanning all wavelengths from the bandgap edge to infinity, so the concept of a single emitted wavelength no longer applies.
The student's conclusion is correct because $eV = hc/\lambda$ establishes a direct inverse proportionality between voltage and wavelength, so halving V doubles $\lambda$ by the same algebraic relationship that defines the LED's operating point, provided current still flows.
The student's conclusion is incorrect because halving the forward voltage doubles the photon frequency rather than the wavelength, since $eV = hf$ gives $f \propto V$ , and doubling frequency corresponds to halving wavelength — the opposite of the student's prediction.
The student's conclusion is incorrect. While $eV \approx hc/\lambda$ holds near the nominal operating point, halving the forward voltage does not double the wavelength because a sufficiently reduced voltage may fall below the LED's turn-on threshold, dramatically reducing or eliminating light emission rather than shifting the emission wavelength, which is determined by the semiconductor bandgap — a material property independent of applied voltage. (correct answer)

Explanation: When analyzing LED behavior, you need to distinguish between two separate physical phenomena: the semiconductor's bandgap (which determines emission wavelength) and the applied voltage (which controls whether current flows at all). The relationship

eV \approx hc/\lambda

tells you that the forward voltage required to drive an LED approximately equals the photon energy — this is how the LED is designed, not a tunable dial. The emission wavelength is fixed by the semiconductor's bandgap energy, a material property baked into the crystal structure. You cannot shift it by simply changing the voltage. What voltage actually controls is whether charge carriers have enough energy to cross the junction. Halving the forward voltage may drop it below the LED's turn-on threshold (~1.5–2 V for red, ~2.5 V for blue), causing emission to collapse entirely rather than shift to a longer wavelength. Answer D correctly captures both points: the inverse proportionality argument doesn't transfer to real operation, and reduced voltage threatens to extinguish the LED rather than redshift it. Answer B is the trap most students fall into — it treats

eV = hc/\lambda

as a live algebraic relationship you can manipulate, ignoring that

\lambda

is fixed by the material, not freely adjustable. Answer A is partially true (emission does broaden at different currents) but wrong to claim a "continuous spectrum to infinity" — LEDs always emit near-bandgap energies. Answer C inverts the frequency-wavelength relationship nonsensically;

eV = hf

is correct, but voltage doesn't directly set the emission frequency in operation. Study tip: Whenever a question mixes a correct-looking equation with a physically wrong conclusion, ask yourself: which variables are truly free to change, and which are fixed by material properties?

Question 10

A laser operating at wavelength $\lambda = 500 \text{ nm}$ delivers an average power of $P = 2.0 \text{ mW}$ to a detector. The detector is a photovoltaic cell with a quantum efficiency of 40% (meaning 40% of incident photons generate one electron-hole pair each). Planck's constant is $h = 6.626 \times 10^{-34} \text{ J}\cdot\text{s}$ and $c = 3.00 \times 10^8 \text{ m/s}$ .

What is the approximate electric current generated by this detector?

$I \approx 3.2 \times 10^{-4} \text{ A}$ (correct answer)
$I \approx 8.0 \times 10^{-4} \text{ A}$
$I \approx 1.3 \times 10^{-3} \text{ A}$
$I \approx 6.4 \times 10^{-4} \text{ A}$

Explanation: When a question connects laser power, photon counting, and detector efficiency to electric current, you're working at the intersection of quantum optics and basic circuit concepts. The key chain of reasoning is: power → photon rate → electron rate → current. Start by finding the energy of one photon:

E_{photon} = \frac{hc}{\lambda} = \frac{(6.626 \times 10^{-34})(3.00 \times 10^8)}{500 \times 10^{-9}} \approx 3.98 \times 10^{-19} \text{ J}

The total photon arrival rate is:

\dot{N} = \frac{P}{E_{photon}} = \frac{2.0 \times 10^{-3}}{3.98 \times 10^{-19}} \approx 5.03 \times 10^{15} \text{ photons/s}

With 40% quantum efficiency, only 40% of those photons generate an electron-hole pair:

\dot{N}_{electrons} = 0.40 \times 5.03 \times 10^{15} \approx 2.01 \times 10^{15} \text{ electrons/s}

Converting to current using

I = \dot{N}_{electrons} \times e

I = (2.01 \times 10^{15})(1.6 \times 10^{-19}) \approx 3.2 \times 10^{-4} \text{ A}

This confirms A is correct. Choice B (

8.0 \times 10^{-4}

A) results from forgetting to apply quantum efficiency — using 100% instead of 40%. Choice D (

6.4 \times 10^{-4}

A) comes from applying efficiency incorrectly, doubling instead of halving the 40% factor. Choice C (

1.3 \times 10^{-3}

A) represents an error where the full photon rate is multiplied by an inflated efficiency or an incorrect electron charge value is used. A reliable strategy: always build the bridge step-by-step — power to photon rate, apply efficiency, then multiply by electron charge. Skipping the efficiency step is the most common trap here.