The Photoelectric Effect
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AP Physics 2 › The Photoelectric Effect
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 frequency because photon energy increases with $f$.
Electrons are emitted only after accumulating energy over time, so higher $f$ reduces the delay.
Electrons can be emitted at any frequency if the stopping potential is adjusted appropriately.
The maximum electron kinetic energy increases with intensity because more photons hit the surface.
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 surface is illuminated by monochromatic light. For $f=4.2\times10^{14},\text{Hz}$, no electrons are emitted even at high intensity. For $f=4.4\times10^{14},\text{Hz}$, electrons are emitted immediately and a stopping potential can be measured. Which statement best identifies what changed between the two cases?
The photon energy increased enough to exceed the work function at $4.4\times10^{14},\text{Hz}$.
The metal emitted electrons at both frequencies, but detection failed at the lower frequency.
The intensity must have increased, causing electrons to gain more energy and escape.
The electrons had more time to accumulate energy from the wave at the higher frequency.
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.
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 allows emission at any frequency because the wave energy builds up in electrons.
Higher intensity decreases the work function by heating the metal, so stopping potential stays fixed.
Higher intensity increases the energy per photon, so the stopping potential must increase.
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).
A student measures photoelectrons from a zinc surface using a variable stopping potential. With violet light, electrons are emitted and the stopping potential is $V_s=1.2,\text{V}$. Switching to red light at the same intensity produces no emitted electrons. Increasing the red-light intensity by a factor of 5 still produces no electrons. Which statement best explains why no electrons are emitted with red light?
Red light has fewer photons, so electrons cannot be emitted regardless of frequency.
Higher intensity red light should increase electron kinetic energy until emission occurs.
Red photons have too little energy to overcome the work function, so emission requires higher frequency.
Electrons need more time to absorb red light energy, so emission would occur after a long delay.
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.
Monochromatic light of frequency $f$ shines on a metal surface. At $f=f_0$ no electrons are emitted, even when intensity is increased by a factor of 10. At $f=1.2f_0$, electrons are emitted immediately, and the maximum kinetic energy increases when $f$ is increased further. Which conclusion is most consistent with the photoelectric effect?
The electron maximum kinetic energy increases with intensity at fixed frequency.
Electrons gradually accumulate wave energy, so higher intensity should eventually cause emission at $f_0$.
The metal has a threshold frequency near $f_0$, and electron maximum kinetic energy increases with frequency.
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.
Light of frequency slightly above a metal’s threshold shines on the surface. Electrons are emitted immediately. When the frequency is held constant and the intensity is reduced, the current decreases but emitted electrons still have the same maximum kinetic energy. Which claim is supported by these observations?
Photon energy depends on intensity, so reducing intensity reduces the maximum kinetic energy.
Electrons store energy from many waves, so reducing intensity should eventually stop emission after a delay.
Electrons are emitted for any frequency, but low intensity makes them too slow to detect.
Photon energy depends on frequency, so changing intensity changes the number of emitted electrons, not their maximum energy.
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.
A student shines light on a metal and observes: below a certain frequency, no photoelectrons are emitted; above it, electrons are emitted immediately. At a fixed frequency above threshold, increasing intensity increases the current but does not change the stopping potential. Which option best summarizes the key threshold behavior?
There is a minimum exposure time needed for emission, set by the metal’s work function.
There is a minimum intensity needed for emission, set by the metal’s work function.
There is a minimum frequency needed for emission, set by the metal’s work function.
There is no minimum frequency because any light will eject electrons if intensity is large enough.
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.
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 raises the threshold frequency, so emission becomes harder at the same $f$.
Doubling intensity increases the energy of each photon, so $V_s$ should increase.
Doubling intensity lets electrons absorb energy continuously, so $V_s$ stays fixed only after a delay.
Doubling intensity increases the number of photons per second, increasing current without changing $V_s$.
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).
Light shines on a clean sodium surface in vacuum. With frequency $f_1$ (below a certain value), no electrons are detected even after 60 s at high intensity. When the frequency is raised slightly to $f_2$, electrons are emitted immediately; increasing intensity at $f_2$ increases the emission rate but the maximum kinetic energy of the electrons is unchanged. Which statement best explains these observations?
Increasing intensity at fixed frequency increases the maximum kinetic energy of emitted electrons.
Electrons require time to accumulate energy from the wave before being emitted at higher frequency.
Electrons are emitted only when $f$ exceeds a threshold, and intensity mainly changes how many electrons are emitted.
Electrons are emitted at any frequency if the light is intense enough and shines long enough.
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.
A student varies the frequency of light incident on a metal while keeping intensity constant. No electrons are emitted for $f<5.0\times10^{14},\text{Hz}$. For $f>5.0\times10^{14},\text{Hz}$, electrons are emitted immediately and the stopping potential increases linearly with $f$. Which claim is directly supported by these observations?
The intensity determines the maximum electron kinetic energy, explaining the linear trend.
The metal has a threshold frequency near $5.0\times10^{14},\text{Hz}$ set by its work function.
The stopping potential is independent of frequency because all photons eject identical electrons.
Electrons are emitted at any frequency if the illumination lasts long enough.
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.