The Photoelectric Effect - AP Physics 2

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?
Answer: Photon energy depends on frequency, so changing intensity changes the number of emitted electrons, not their maximum energy. 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 metal is illuminated with monochromatic light of adjustable frequency. At $f=6.0\times10^{14}\,\text{Hz}$ , electrons are emitted; at $f=5.5\times10^{14}\,\text{Hz}$ , none are emitted, even when intensity is increased. Which inference about the threshold frequency $f_0$ is supported?
Answer: $f_0$ lies between $5.5$ and $6.0\times10^{14}\,\text{Hz}$ . This question tests understanding of the photoelectric effect. The threshold frequency f₀ is the minimum frequency at which photons have just enough energy to overcome the work function: hf₀ = W. The observation that electrons are emitted at 6.0×10¹⁴ Hz but not at 5.5×10¹⁴ Hz (even with increased intensity) indicates the threshold frequency must lie between these values. If f₀ were above 6.0×10¹⁴ Hz, no electrons would be emitted at that frequency; if f₀ were below 5.5×10¹⁴ Hz, electrons would be emitted at that frequency. Choice D incorrectly claims threshold depends on intensity, when it actually depends only on the material's work function. The principle is: threshold frequency is a sharp boundary determined by the material's work function, independent of light intensity.
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?
Answer: Intensity changes the photon arrival rate, affecting the number of emitted electrons but not $K_{\max}$ . 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.
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?
Answer: Stopping potential depends on photon energy, which is set by frequency 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.
Monochromatic light shines on a metal cathode in a photoelectric tube. At $f=4.1\times10^{14}\,\text{Hz}$ and high intensity, no electrons are emitted. At $f=4.6\times10^{14}\,\text{Hz}$ and low intensity, electrons are emitted immediately, but the current is small. Which statement best accounts for the difference?
Answer: At higher frequency, each photon has more energy, allowing emission even at low intensity. The photoelectric effect. This phenomenon requires each photon to have energy hf exceeding the metal's work function W₀ for electron emission to occur. At 4.1×10¹⁴ Hz, the photon energy is below the work function, so no electrons can escape regardless of how many photons strike the surface (high intensity). At 4.6×10¹⁴ Hz, photon energy exceeds W₀, allowing immediate emission even with few photons (low intensity), though the small current reflects the low photon arrival rate. Choice B incorrectly invokes energy accumulation over time, which contradicts the observed immediate emission above threshold. The critical insight: individual photon energy (set by frequency) determines whether emission is possible; intensity only affects emission rate.
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?
Answer: Doubling intensity increases the number of photons per second, increasing current without changing $V_s$ . 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).
In a vacuum tube, monochromatic light shines on a clean sodium surface. At $f=4.8\times10^{14}\,\text{Hz}$ , electrons are emitted immediately; increasing intensity increases the current but the maximum electron kinetic energy is unchanged. At $f=3.9\times10^{14}\,\text{Hz}$ , no electrons are emitted even after 60 s at very high intensity. Which statement best explains these observations?
Answer: Electrons are emitted only if the photon energy $hf$ exceeds the work function, so there is a threshold frequency. 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.
A student shines monochromatic light on a metal and measures the photoelectron current. At a frequency below threshold, the current is zero for all tested intensities. At a frequency above threshold, the current increases with intensity and electrons are emitted immediately. Which statement is best supported?
Answer: There is a minimum frequency required for emission because each photon must supply at least the work function. The photoelectric effect. This phenomenon requires each photon to possess minimum energy hf₀ equal to the metal's work function W₀ for electron liberation, establishing a threshold frequency f₀ = W₀/h. Below threshold, photons lack sufficient energy to overcome the electron binding energy, resulting in zero emission regardless of intensity (photon quantity). Above threshold, each photon can eject one electron immediately, with higher intensity providing more photons and thus higher current. Choice A incorrectly suggests intensity can compensate for low frequency, contradicting the quantum nature where each photon-electron interaction is independent. The fundamental concept: emission depends on individual photon energy (frequency); current depends on photon flux (intensity).
A metal surface is illuminated with light of two different frequencies at the same intensity. With $f_1$ , no electrons are emitted. With $f_2>f_1$ , electrons are emitted and their maximum kinetic energy is measured. Which single change would most directly increase the maximum kinetic energy of the emitted electrons?
Answer: Increase the light frequency above $f_2$ while keeping intensity constant. This question tests understanding of the photoelectric effect. The maximum kinetic energy of photoelectrons is determined by the equation KE_max = hf - W, where h is Planck's constant, f is the light frequency, and W is the metal's work function. Since f₁ produces no electrons while f₂ > f₁ does, we know f₂ is above threshold and KE_max = hf₂ - W. To increase maximum kinetic energy, we must increase the photon energy hf, which requires increasing frequency above f₂. Changing intensity (choices A and B) only affects the number of photons and thus emission rate, not individual electron energies. Choice D incorrectly assumes energy accumulation over time, violating the photon model's principle of discrete, instantaneous interactions. The strategy is: to increase maximum electron energy, increase photon frequency; intensity changes only affect emission rate.
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?
Answer: There is a minimum frequency needed for emission, set by the metal's work function. 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.
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?
Answer: Red photons have too little energy to overcome the work function, so emission requires higher frequency. 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.
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?
Answer: The photon energy increased enough to exceed the work function at $4.4\times10^{14}\,\text{Hz}$ . 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?
Answer: Higher intensity increases the number of emitted electrons per second but not their maximum kinetic energy. 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 metal is illuminated with two different monochromatic lights. With blue light, electrons are emitted and a stopping potential of $1.5\,\text{V}$ is measured. With red light at the same intensity, no electrons are emitted. Which statement is most consistent with the photon model?
Answer: Red light has lower frequency, so its photons may have energy less than the work function. The photoelectric effect. In this quantum phenomenon, photon energy depends solely on frequency through E = hf, with blue light having higher frequency than red light. The blue light ejects electrons with 1.5 V stopping potential, indicating its photon energy exceeds the metal's work function with excess energy becoming electron kinetic energy. The red light's lower frequency means its photons may have energy below the work function, preventing electron emission entirely regardless of intensity. Choice D incorrectly claims electron energy depends on intensity, contradicting both the photon model and the experimental observation of frequency-dependent emission. The critical principle: color determines frequency, frequency determines photon energy, and emission requires hf ≥ W₀.
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?
Answer: The metal has a threshold frequency between $4.8$ and $5.2\times10^{14}\,\text{Hz}$ . 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.
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?
Answer: The metal has a threshold frequency near $f_0$ , and electron maximum kinetic energy increases with frequency. 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 fixed frequency $f=6.0\times10^{14}\,\text{Hz}$ strikes a zinc plate. As intensity is increased, the photoelectron current increases, but the stopping potential remains $V_s=1.2\,\text{V}$ . Electrons are emitted with no measurable time delay. Which observation supports the photon model of light?
Answer: The stopping potential stays constant when intensity changes at fixed frequency. 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.
A metal surface is illuminated with monochromatic light. At $f=5.2\times10^{14}\,\text{Hz}$ , electrons are emitted and a stopping potential of $0.80\,\text{V}$ is measured. When the frequency is increased to $6.2\times10^{14}\,\text{Hz}$ at the same intensity, electrons are still emitted and the stopping potential increases to $1.20\,\text{V}$ . Which conclusion is best supported?
Answer: The maximum kinetic energy increases with frequency because photon energy is proportional to $f$ . The photoelectric effect. This phenomenon demonstrates that light consists of photons with energy E = hf, where higher frequency means higher energy per photon. When frequency increases from 5.2×10¹⁴ Hz to 6.2×10¹⁴ Hz, each photon carries more energy, resulting in ejected electrons with greater maximum kinetic energy K_max = hf - W₀. The stopping potential $V_s$ measures this maximum kinetic energy through eV_s = K_max, explaining why $V_s$ increases from 0.80 V to 1.20 V. Choice D incorrectly claims electron energy depends on intensity, when actually intensity only affects the number of photons and thus the current. The fundamental principle: frequency determines photon energy and thus maximum electron energy; intensity determines photon count.
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?
Answer: Electrons are emitted only when $f$ exceeds a threshold, and intensity mainly changes how many electrons are emitted. 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 photoelectric experiment uses the same metal surface and the same light frequency above threshold. When intensity is doubled, the measured photoelectron current doubles, but the stopping potential does not change. Which interpretation is most consistent with the results?
Answer: Doubling intensity increases the number of photons per second, increasing emitted electrons per second. 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).

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