The Photoelectric Effect
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AP Physics 2 › The Photoelectric Effect
A student shines red light of very high intensity on a metal surface and observes no photoelectrons. Switching to dim blue light causes immediate emission of electrons. The student keeps the metal and setup unchanged. Which reasoning best accounts for the change?
Blue light is dimmer, so electrons are less likely to be trapped
Red light needs more time for electrons to store energy from the wave
Blue light has higher frequency, so each photon can exceed the work function
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).
Two trials use the same metal. Trial 1: light at frequency just above threshold produces electrons with low maximum kinetic energy. Trial 2: frequency is unchanged but intensity is tripled, producing three times as many electrons per second with the same maximum kinetic energy. Which statement best accounts for this result?
Increasing intensity lets electrons accumulate energy and eventually exceed the work function.
Increasing intensity increases photon energy, so electrons should have higher maximum kinetic energy.
Increasing intensity increases photon flux but not the energy of each photon.
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.
In a photoelectric tube, light of fixed intensity illuminates a metal cathode. As the frequency is increased, electrons are emitted and the stopping potential increases. When the frequency is decreased below a certain value, emission stops completely even though intensity is unchanged. Which explanation best matches the data?
Below threshold, fewer photons strike the surface, so emission stops due to low photon count.
Below threshold, electrons require more time to accumulate energy from the wave before emission.
Below threshold, photons have insufficient energy to liberate electrons regardless of intensity.
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.
A metal is illuminated by monochromatic light of frequency $f$ just above threshold. Electrons are emitted immediately. When intensity is doubled, the measured photocurrent doubles, but the stopping potential does not change. Which statement best explains why the stopping potential is unchanged?
Stopping potential depends on photon energy, which is set by frequency
Stopping potential depends on exposure time, not on frequency
Doubling intensity doubles photon energy, so $V_s$ should increase
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.
A metal plate is illuminated with monochromatic light of fixed intensity. At $f=4.8\times10^{14},\text{Hz}$, no electrons are emitted. At $f=5.2\times10^{14},\text{Hz}$, electrons are emitted immediately with small but nonzero maximum kinetic energy. Which conclusion about the metal is most consistent with the photon model?
The metal will emit electrons at $4.8\times10^{14},\text{Hz}$ if illuminated long enough.
The metal requires higher intensity, not higher frequency, to emit electrons.
The metal has a threshold frequency between $4.8$ and $5.2\times10^{14},\text{Hz}$.
The metal emits electrons at all frequencies, but the detector missed them at $4.8\times10^{14},\text{Hz}$.
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.
Light of frequency just above a metal’s threshold shines on the surface. Electrons are emitted, but their maximum kinetic energy is small. When the intensity is increased by a factor of 5 at the same frequency, more electrons are emitted per second, but the maximum kinetic energy is unchanged. What does this indicate about the role of intensity?
Intensity provides energy continuously, so $K_{\max}$ should rise slowly as exposure time increases.
Intensity changes the work function, so electron energies stay fixed even as emission rate changes.
Intensity changes the photon arrival rate, affecting the number of emitted electrons but not $K_{\max}$.
Intensity changes photon energy, but the stopping potential masks the increase in electron energy.
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.
In a photoelectric experiment, monochromatic light of frequency $f$ shines on a metal. At $f=7.5\times10^{14},\text{Hz}$, electrons are emitted and the stopping potential is $0.80,\text{V}$. When the frequency is increased to $8.0\times10^{14},\text{Hz}$ at the same intensity, the stopping potential increases to $1.00,\text{V}$. Which inference is best supported?
The maximum electron kinetic energy increases with intensity because more photons hit the surface.
The maximum electron kinetic energy increases with frequency because photon energy increases with $f$.
Electrons can be emitted at any frequency if the stopping potential is adjusted appropriately.
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.
A metal is illuminated with light of fixed frequency above threshold. At low intensity, a small photoelectric current is measured. When intensity is tripled (same frequency), the photoelectric current triples, but the stopping potential remains unchanged. Which statement best matches these results?
Higher intensity increases the energy per photon, so the stopping potential must increase.
Higher intensity decreases the work function by heating the metal, so stopping potential stays fixed.
Higher intensity allows emission at any frequency because the wave energy builds up in electrons.
Higher intensity increases the number of emitted electrons per second but not their maximum kinetic energy.
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).
Light shines on a clean zinc plate in vacuum. At $f=5.0\times10^{14},\text{Hz}$, no electrons are detected even after 60 s at high intensity. At $f=8.0\times10^{14},\text{Hz}$, electrons are emitted immediately; doubling intensity at this frequency increases the emission rate but the maximum kinetic energy is unchanged. Which statement best explains these observations?
Higher intensity increases the maximum kinetic energy of emitted electrons
Electrons absorb energy gradually from the wave until they escape
Electrons are emitted at any frequency if the light is intense enough
Light below a threshold frequency cannot eject electrons regardless of intensity
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.
Two experiments use the same metal. In Experiment 1, light of frequency $f$ produces photoelectrons with stopping potential $V_s$. In Experiment 2, the intensity is doubled but the frequency is unchanged, and the stopping potential remains $V_s$ while the current increases. Which statement is most consistent with the photon model?
Doubling intensity increases the energy of each photon, so $V_s$ should increase.
Doubling intensity increases the number of photons per second, increasing current without changing $V_s$.
Doubling intensity raises the threshold frequency, so emission becomes harder at the same $f$.
Doubling intensity lets electrons absorb energy continuously, so $V_s$ stays fixed only after a delay.
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).