AP Physics 2 Quiz: The Photoelectric Effect

What this quiz covers

This quiz focuses on The Photoelectric Effect, giving you a quick way to practice the rules, question types, and explanations that matter most for AP 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.

Question 1

1. Blue light has higher frequency, so each photon can exceed the work function (correct answer)
2. Blue light is dimmer, so electrons are less likely to be trapped
3. Red light needs more time for electrons to store energy from the wave
4. Any light will eject electrons if the intensity is high enough

Explanation: The photoelectric effect. Blue light has higher frequency than red light, meaning each blue photon carries more energy according to E = hf. Even though the red light has very high intensity (many photons per second), each red photon lacks sufficient energy to overcome the metal's work function. The dim blue light has fewer photons per second, but each blue photon has enough energy to eject an electron immediately upon absorption. Choice C incorrectly invokes the classical wave model where electrons could accumulate energy over time. The key principle is that photoelectric emission depends on individual photon energy (frequency), not the total energy delivered (intensity).

Question 2

1. Increasing intensity increases photon flux but not the energy of each photon. (correct answer)
2. Increasing intensity increases photon energy, so electrons should have higher maximum kinetic energy.
3. Increasing intensity lets electrons accumulate energy and eventually exceed the work function.
4. Increasing intensity lowers the threshold frequency by heating the metal.

Explanation: This question tests understanding of the photoelectric effect. In the photoelectric effect, intensity represents the number of photons per unit time, while frequency determines the energy of each photon (E = hf). When intensity is tripled at constant frequency, three times as many photons strike the surface per second, but each photon still has the same energy. Since maximum kinetic energy depends only on photon energy minus work function (KE_max = hf - W), it remains unchanged when only intensity changes. Choice B incorrectly assumes intensity affects individual photon energy, confusing photon flux with photon energy. The fundamental principle is: intensity controls the rate of photoelectron emission (more photons means more electron-photon interactions), while frequency controls the energy of each emitted electron.

Question 3

1. Below threshold, photons have insufficient energy to liberate electrons regardless of intensity. (correct answer)
2. Below threshold, electrons require more time to accumulate energy from the wave before emission.
3. Below threshold, fewer photons strike the surface, so emission stops due to low photon count.
4. Below threshold, intensity determines electron energy, so emission should still occur but with lower $K_{\max}$.

Explanation: The photoelectric effect. This phenomenon demonstrates that electrons require a minimum photon energy hf₀ = W₀ to escape the metal surface, where f₀ is the threshold frequency and W₀ is the work function. Below this threshold, individual photons lack sufficient energy to liberate electrons, making emission impossible regardless of how many photons strike the surface (intensity). Above threshold, photon energy exceeds W₀, enabling immediate electron emission with excess energy appearing as kinetic energy, measured by the stopping potential. Choice B incorrectly invokes gradual energy accumulation, which would predict delayed emission rather than the observed complete absence of emission. The key insight: photon energy is quantized by frequency; below threshold, no amount of intensity can compensate for insufficient photon energy.

Question 4

1. Doubling intensity doubles photon energy, so $V_s$ should increase
2. Stopping potential depends on photon energy, which is set by frequency (correct answer)
3. Stopping potential depends on exposure time, not on frequency
4. Stopping potential is fixed because electrons always leave with the same speed

Explanation: The photoelectric effect. The stopping potential directly measures the maximum kinetic energy of ejected electrons, which depends on the energy of individual photons through Kmax = hf - φ. Since photon energy E = hf depends only on frequency, and frequency remains constant in this experiment, each ejected electron has the same maximum kinetic energy regardless of intensity. Doubling intensity doubles the number of photons per second, explaining the doubled photocurrent, but doesn't change the energy per photon. Choice A represents the misconception that intensity affects photon energy, contradicting the quantum nature of light. The strategy is to recognize that stopping potential reveals individual photon energy, which depends solely on frequency.

Question 5

1. The metal has a threshold frequency between $4.8$ and $5.2\times10^{14}\,\text{Hz}$. (correct answer)
2. The metal emits electrons at all frequencies, but the detector missed them at $4.8\times10^{14}\,\text{Hz}$.
3. The metal requires higher intensity, not higher frequency, to emit electrons.
4. The metal will emit electrons at $4.8\times10^{14}\,\text{Hz}$ if illuminated long enough.

Explanation: This question tests understanding of the photoelectric effect. The photoelectric effect demonstrates that electrons are emitted from a metal surface only when the incident photon frequency exceeds a threshold frequency f₀, where the photon energy hf₀ equals the work function. The observation that no electrons are emitted at 4.8×10¹⁴ Hz but electrons are emitted at 5.2×10¹⁴ Hz indicates the threshold frequency lies between these values. The small but nonzero kinetic energy at 5.2×10¹⁴ Hz confirms this frequency is just above threshold, as KE_max = hf - hf₀ is small when f is slightly greater than f₀. Choice D incorrectly assumes that longer illumination time allows energy accumulation, which violates the photon model where each photon-electron interaction is independent and instantaneous. Remember: the existence of a sharp threshold frequency that doesn't depend on intensity or exposure time is key evidence for the photon model.

Question 6

1. Intensity changes the work function, so electron energies stay fixed even as emission rate changes.
2. Intensity changes photon energy, but the stopping potential masks the increase in electron energy.
3. Intensity changes the photon arrival rate, affecting the number of emitted electrons but not $K_{\max}$. (correct answer)
4. Intensity provides energy continuously, so $K_{\max}$ should rise slowly as exposure time increases.

Explanation: The photoelectric effect. In this quantum process, each photon transfers its entire energy hf to a single electron, and if hf exceeds the work function, the electron escapes with maximum kinetic energy K_max = hf - W₀. When intensity increases fivefold at constant frequency, five times more photons arrive per second, each still carrying the same energy hf. This increases the number of ejected electrons per second (higher current) but doesn't change the maximum energy any single electron can have, since that depends only on individual photon energy. Choice D incorrectly assumes continuous energy transfer over time, contradicting the instantaneous, discrete nature of photon-electron interactions. The fundamental rule: intensity affects how many electrons escape; frequency affects how fast they escape.

Question 7

1. The maximum electron kinetic energy increases with frequency because photon energy increases with $f$. (correct answer)
2. The maximum electron kinetic energy increases with intensity because more photons hit the surface.
3. Electrons can be emitted at any frequency if the stopping potential is adjusted appropriately.
4. Electrons are emitted only after accumulating energy over time, so higher $f$ reduces the delay.

Explanation: This question tests understanding of the photoelectric effect. In the photoelectric effect, the maximum kinetic energy of emitted electrons depends on the photon energy (E = hf) minus the work function: KEₘₐₓ = hf - W₀. Since stopping potential directly measures this maximum kinetic energy (eVₛ = KEₘₐₓ), an increase in frequency causes a proportional increase in stopping potential. The data shows that increasing frequency from 7.5×10¹⁴ Hz to 8.0×10¹⁴ Hz increases the stopping potential from 0.80 V to 1.00 V, confirming that photon energy increases with frequency. Choice B incorrectly claims that intensity affects photon energy, confusing the number of photons (intensity) with energy per photon (frequency-dependent). The key insight is that stopping potential depends only on frequency because it reflects individual photon-electron interactions, not the total number of interactions.

Question 8

1. Higher intensity increases the energy per photon, so the stopping potential must increase.
2. Higher intensity allows emission at any frequency because the wave energy builds up in electrons.
3. Higher intensity decreases the work function by heating the metal, so stopping potential stays fixed.
4. Higher intensity increases the number of emitted electrons per second but not their maximum kinetic energy. (correct answer)

Explanation: This question tests understanding of the photoelectric effect. In the photoelectric effect, light intensity determines the number of photons per second hitting the surface, while frequency determines the energy of each photon (E = hf). When frequency is fixed above threshold, increasing intensity increases the number of photon-electron interactions per second, thus increasing the photoelectric current proportionally. However, the maximum kinetic energy of individual electrons (and thus the stopping potential) remains unchanged because it depends only on the photon energy minus the work function: KEₘₐₓ = hf - W₀. Choice B incorrectly claims that intensity affects energy per photon, reflecting the misconception that light intensity and photon energy are related. The strategy is to remember that intensity affects quantity (current) while frequency affects quality (maximum electron energy).

Question 9

1. Electrons are emitted at any frequency if the light is intense enough
2. Light below a threshold frequency cannot eject electrons regardless of intensity (correct answer)
3. Higher intensity increases the maximum kinetic energy of emitted electrons
4. Electrons absorb energy gradually from the wave until they escape

Explanation: The photoelectric effect. In this phenomenon, light energy comes in discrete packets called photons, where each photon's energy depends only on its frequency through E = hf. When a photon's energy is below the work function (threshold energy) of the metal, no electrons can be ejected regardless of how many photons hit the surface or how long we wait. Above the threshold frequency, photons have enough energy to overcome the work function and eject electrons immediately. Choice A represents the classical wave misconception that sufficient intensity can compensate for low frequency, but this contradicts the quantum nature where each photon-electron interaction is independent. The key strategy is to remember that frequency determines whether emission is possible at all, while intensity only affects how many electrons are emitted per second.

Question 10

1. Doubling intensity increases the number of photons per second, increasing current without changing $V_s$. (correct answer)
2. Doubling intensity lets electrons absorb energy continuously, so $V_s$ stays fixed only after a delay.
3. Doubling intensity increases the energy of each photon, so $V_s$ should increase.
4. Doubling intensity raises the threshold frequency, so emission becomes harder at the same $f$.

Explanation: This question tests understanding of the photoelectric effect. In the photon model of light, each photon's energy depends only on frequency (E = hf), not on the light's intensity. Doubling intensity at fixed frequency means doubling the number of photons per second hitting the surface, which doubles the number of electron emissions and thus the current. However, since each photon still has the same energy (same frequency), the maximum kinetic energy of emitted electrons remains unchanged, keeping the stopping potential constant. Choice A incorrectly claims that intensity affects individual photon energy, reflecting the misconception that brighter light means more energetic photons. The strategy to remember is that intensity is about photon quantity (affecting current), while frequency is about photon quality (affecting stopping potential).

Question 11

1. Electrons are emitted at any frequency if the light is intense enough and shines long enough.
2. Electrons require time to accumulate energy from the wave before being emitted at higher frequency.
3. Electrons are emitted only when $f$ exceeds a threshold, and intensity mainly changes how many electrons are emitted. (correct answer)
4. Increasing intensity at fixed frequency increases the maximum kinetic energy of emitted electrons.

Explanation: This question tests understanding of the photoelectric effect. In the photoelectric effect, light energy comes in discrete packets called photons, where each photon's energy depends only on its frequency (E = hf), not on the light's intensity. For electrons to be emitted from a metal surface, the photon energy must exceed a minimum value called the work function (W₀), which corresponds to a threshold frequency (f₀ = W₀/h). Below this threshold frequency, no electrons are emitted regardless of intensity or exposure time, while above it, emission occurs immediately. Choice B incorrectly suggests that high intensity can compensate for low frequency, reflecting the classical wave misconception that energy accumulates continuously. The key strategy is to remember that frequency determines whether emission is possible (photon energy vs. work function), while intensity only affects how many electrons are emitted per second.

Question 12

1. There is a minimum exposure time needed for emission, set by the metal's work function.
2. There is a minimum intensity needed for emission, set by the metal's work function.
3. There is no minimum frequency because any light will eject electrons if intensity is large enough.
4. There is a minimum frequency needed for emission, set by the metal's work function. (correct answer)

Explanation: This question tests understanding of the photoelectric effect. The photoelectric effect exhibits a clear threshold behavior: there exists a minimum frequency (threshold frequency f₀) below which no electrons are emitted, determined by the metal's work function through f₀ = W₀/h. Below this frequency, no amount of intensity or exposure time produces emission because individual photons lack sufficient energy. Above threshold, emission occurs immediately, with intensity affecting only the emission rate (current) while the stopping potential depends on frequency alone. Choice B incorrectly identifies intensity as the threshold parameter, confusing the number of photons with the energy requirement per photon. The key principle is that photoelectric emission is governed by individual photon-electron interactions, where frequency determines if emission is possible and intensity determines how often it occurs.

Question 13

1. The intensity must have increased, causing electrons to gain more energy and escape.
2. The metal emitted electrons at both frequencies, but detection failed at the lower frequency.
3. The electrons had more time to accumulate energy from the wave at the higher frequency.
4. The photon energy increased enough to exceed the work function at $4.4\times10^{14}\,\text{Hz}$. (correct answer)

Explanation: This question tests understanding of the photoelectric effect. The photoelectric effect requires that each photon have sufficient energy (E = hf) to overcome the metal's work function for electron emission to occur. At 4.2×10¹⁴ Hz, the photon energy (h × 4.2×10¹⁴ Hz) was below the work function, so no electrons were emitted regardless of intensity. At 4.4×10¹⁴ Hz, the photon energy exceeded the work function, enabling immediate electron emission. This demonstrates the existence of a threshold frequency (f₀ = W₀/h) below which no emission occurs. Choice C incorrectly suggests that electrons accumulate energy over time, reflecting the classical wave theory that cannot explain the immediate emission above threshold. The fundamental concept is that each photon-electron interaction is an all-or-nothing event: the photon either has enough energy to eject an electron or it doesn't.

Question 14

1. Higher intensity red light should increase electron kinetic energy until emission occurs.
2. Red photons have too little energy to overcome the work function, so emission requires higher frequency. (correct answer)
3. Electrons need more time to absorb red light energy, so emission would occur after a long delay.
4. Red light has fewer photons, so electrons cannot be emitted regardless of frequency.

Explanation: This question tests understanding of the photoelectric effect. The photoelectric effect shows that photon energy depends only on frequency (E = hf), and electrons are emitted only when this energy exceeds the metal's work function. Red light has lower frequency than violet light, so red photons carry less energy than violet photons. Since violet light produces photoelectrons but red light does not (even at 5× intensity), the red photon energy must be below zinc's work function. Choice C incorrectly suggests that higher intensity can increase electron kinetic energy, confusing intensity (number of photons) with individual photon energy (which depends only on frequency). The fundamental principle is that no amount of low-energy photons can combine to eject an electron—each photon-electron interaction is independent, and emission requires a single photon with E > W₀.

Question 15

1. The metal has a threshold frequency near $f_0$, and electron maximum kinetic energy increases with frequency. (correct answer)
2. The electron maximum kinetic energy increases with intensity at fixed frequency.
3. Electrons gradually accumulate wave energy, so higher intensity should eventually cause emission at $f_0$.
4. The metal emits electrons at any frequency if the light is intense enough.

Explanation: This question tests understanding of the photoelectric effect. In the photoelectric effect, electrons are emitted from a metal surface only when the incident photon energy (E = hf) exceeds the metal's work function, establishing a threshold frequency below which no emission occurs. The observation that no electrons are emitted at frequency f₀ even with 10× intensity confirms that emission depends on photon frequency, not intensity. Above the threshold frequency, the maximum kinetic energy of emitted electrons increases linearly with frequency according to KEₘₐₓ = hf - W₀. Choice D incorrectly suggests that electrons gradually accumulate wave energy, reflecting the classical wave theory that fails to explain the instantaneous emission above threshold. The key principle is that each photon-electron interaction is independent: frequency determines if emission is possible, intensity only affects how many such interactions occur.

Question 16

1. Electrons are emitted for any frequency, but low intensity makes them too slow to detect.
2. Photon energy depends on intensity, so reducing intensity reduces the maximum kinetic energy.
3. Photon energy depends on frequency, so changing intensity changes the number of emitted electrons, not their maximum energy. (correct answer)
4. Electrons store energy from many waves, so reducing intensity should eventually stop emission after a delay.

Explanation: This question tests understanding of the photoelectric effect. The photoelectric effect demonstrates that photon energy depends solely on frequency (E = hf), not on light intensity. When frequency is held constant above threshold, each photon still carries the same energy, so the maximum kinetic energy of emitted electrons (KEₘₐₓ = hf - W₀) remains unchanged. Reducing intensity simply means fewer photons per second hit the surface, resulting in fewer electron emissions and thus lower current, but each emitted electron still has the same maximum energy. Choice B incorrectly claims that photon energy depends on intensity, confusing the number of photons with energy per photon. The key principle is that intensity controls how many identical photon-electron interactions occur, while frequency controls the outcome of each individual interaction.

Question 17

1. The metal has a threshold frequency near $5.0\times10^{14}\,\text{Hz}$ set by its work function. (correct answer)
2. The intensity determines the maximum electron kinetic energy, explaining the linear trend.
3. Electrons are emitted at any frequency if the illumination lasts long enough.
4. The stopping potential is independent of frequency because all photons eject identical electrons.

Explanation: The photoelectric effect. This phenomenon exhibits a threshold frequency f₀ = W₀/h, where W₀ is the metal's work function, below which no electrons are emitted because photon energy hf is insufficient to overcome the binding energy. The observations show no emission below 5.0×10¹⁴ Hz and immediate emission above it, directly indicating this frequency corresponds to the threshold. Above threshold, the linear increase in stopping potential with frequency follows from K_max = hf - W₀, where the excess energy becomes electron kinetic energy. Choice B incorrectly attributes electron energy to intensity rather than frequency, contradicting the observed frequency dependence. Remember: threshold frequency reveals work function; linear V_s vs. f confirms E = hf.

Question 18

1. Doubling intensity doubles photon energy, so the stopping potential should increase.
2. Doubling intensity increases the number of photons per second, increasing emitted electrons per second. (correct answer)
3. Doubling intensity lowers the work function by heating the metal, keeping stopping potential constant.
4. Doubling intensity allows electrons to accumulate energy over time, so emission becomes delayed.

Explanation: The photoelectric effect. In this quantum phenomenon, light consists of discrete photons, each carrying energy hf determined solely by frequency. When intensity doubles at fixed frequency, twice as many photons strike the surface per second, but each photon still has the same energy hf. Since each photon can eject at most one electron, doubling the photon arrival rate doubles the photoelectron current while the maximum kinetic energy K_max = hf - W₀ remains unchanged, keeping the stopping potential constant. Choice A incorrectly assumes intensity affects individual photon energy, contradicting the fundamental quantum nature of light. The key principle: intensity controls photon quantity (affecting current); frequency controls photon quality (affecting electron energy).

Question 19

1. The stopping potential stays constant when intensity changes at fixed frequency. (correct answer)
2. The current increases when intensity increases at fixed frequency.
3. The metal plate remains electrically neutral overall during illumination.
4. The emitted electrons travel toward the anode in the vacuum tube.

Explanation: The photoelectric effect. In this phenomenon, individual photons transfer their energy hf to electrons, and if hf exceeds the work function, electrons are emitted with maximum kinetic energy K_max = hf - W₀. The stopping potential V_s directly measures this maximum kinetic energy through eV_s = K_max, making it dependent only on photon frequency, not intensity. When intensity increases at fixed frequency, more photons arrive per second, increasing the photoelectron current, but each photon still has the same energy hf. Choice B describes a consequence of higher intensity but doesn't specifically support the photon model like the constant stopping potential does. Remember: stopping potential reveals individual photon energy; current reveals photon quantity.

Question 20

1. Electrons absorb energy continuously, so waiting longer eventually ejects them at any frequency.
2. Increasing intensity increases the energy per photon, so $K_{\max}$ should increase at fixed frequency.
3. Electrons are emitted only if the photon energy $hf$ exceeds the work function, so there is a threshold frequency. (correct answer)
4. Any frequency can eject electrons if the intensity is high enough because more photons push electrons out.

Explanation: The photoelectric effect. When light shines on a metal surface, electrons are emitted only if the photon energy hf exceeds the metal's work function W₀, establishing a threshold frequency f₀ = W₀/h. Above this threshold (4.8×10¹⁴ Hz for sodium), photons have enough energy to eject electrons immediately, while below it (3.9×10¹⁴ Hz), no emission occurs regardless of intensity or duration. The maximum kinetic energy of emitted electrons equals hf - W₀, which depends only on frequency, not intensity. Choice A incorrectly assumes continuous energy absorption over time, contradicting the instantaneous emission observed. The key strategy: photon energy depends on frequency alone; if hf < W₀, no electrons escape regardless of intensity or time.