Physics 2 Quiz: Photoelectric Effect

What this quiz covers

This quiz focuses on Photoelectric Effect, 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.

Question 1

1. The slope (and thus $h$) is unaffected because the constant offset cancels in any rise-over-run calculation; however, $\phi$ is overestimated by $0.12 \text{ eV}$ because every measured $V_0$ is too low by $0.12 \text{ V}$, making the y-intercept more negative than the true value and causing the inferred work function to be too large. (correct answer)
2. Both $h$ and $\phi$ are underestimated: $h$ because the uniformly reduced $V_0$ values lower the apparent slope, and $\phi$ because correcting the readings upward would shift the y-intercept toward zero, implying a smaller inferred work function.
3. The slope (and thus $h$) is unaffected because the systematic offset is constant across all measurements; $\phi$ is underestimated by $0.12 \text{ eV}$ because the measured y-intercept is $0.12 \text{ V}$ more negative than the true value, and a more negative intercept corresponds to a smaller apparent work function.
4. Both $h$ and $\phi$ are unaffected, because a systematic voltmeter offset is equivalent to choosing a shifted reference potential, which displaces the entire $V_0$-axis uniformly and cancels out when both slope and intercept are extracted from the best-fit line.

Question 2

1. The stopping potential is zero and the current is zero, because at exactly the threshold frequency the photon energy equals the work function, leaving no kinetic energy for the ejected electrons; with zero kinetic energy, no retarding field is needed to stop them, and they cannot reach the collector to constitute a current. (correct answer)
2. The stopping potential is zero and the current is nonzero, because electrons are ejected with zero kinetic energy so no stopping potential is needed to halt them, but they can still drift to the collector under the circuit's built-in contact potential difference, producing a measurable current.
3. The stopping potential is $hf_0/e$ and the current is zero, because the stopping potential must account for the full photon energy delivered to the electron, and electrons produced with zero net kinetic energy cannot travel to the collector to form a current.
4. The stopping potential is zero and the current is nonzero, because photons at exactly the threshold frequency are absorbed with maximum efficiency due to a resonance condition, maximizing the ejection rate; even though individual electrons have zero kinetic energy, the collector's positive bias draws them across the gap, yielding a nonzero current.

Question 3

1. The new line has the same slope $m$ and its x-intercept shifts to a higher frequency by $\Delta\phi / (e \cdot m^2)$, and its y-intercept shifts downward by $\Delta\phi / e$, because the slope depends only on $h/e$ while the threshold frequency scales inversely with the square of the slope.
2. The new line has a steeper slope because a larger work function means photons must supply more energy per unit frequency to liberate electrons, effectively increasing the ratio of stopping potential gained per unit frequency above threshold.
3. The new line is parallel to the original (same slope $m = h/e$) and its x-intercept shifts to a lower frequency, while the y-intercept shifts downward by $\Delta\phi / e$, because a larger work function means the metal requires less photon energy to reach threshold.
4. The new line is parallel to the original (same slope $m = h/e$) and its x-intercept shifts to a higher frequency, while the y-intercept (extrapolated) shifts downward by $\Delta\phi / e$, because the slope $h/e$ is universal while both the threshold frequency and the extrapolated intercept reflect the larger work function. (correct answer)

Question 4

1. $KE_{max}$ is the same in both experiments, because doubling both the frequency and the work function produces exactly canceling effects, so the net kinetic energy is unchanged for any values of $f$ and $\phi$.
2. $KE_{max}$ in the second experiment is greater than in the first by exactly $hf$, because the photon energy increases by $2hf$ while the work function increases by $\phi$, and their difference always equals $hf$ regardless of $\phi$.
3. $KE_{max}$ in the second experiment exceeds that in the first by $hf - \phi$, where $\phi$ is the work function of the original metal, because the photon energy increases by $hf$ while the energy barrier increases by $\phi$, and these gains do not cancel unless $\phi = 0$. (correct answer)
4. $KE_{max}$ in the second experiment is less than in the first by $hf - \phi$, because the larger work function of the second metal removes more energy than is gained from the higher-frequency photons whenever $\phi > hf$.

Question 5

1. $f_{min} = (\phi_{Cs} - 1.5\text{ eV})/h \approx 1.45 \times 10^{14} \text{ Hz}$; at $f = 2f_{min}$, $KE_{anode} = 2hf_{min} - \phi_{Cs} + 1.5\text{ eV} = 1.5\text{ eV}$, because the anode's positive potential reduces the effective work function, lowering the minimum frequency needed for emission.
2. $f_{min} = \phi_{Cs}/h \approx 5.07 \times 10^{14} \text{ Hz}$; at $f = 2f_{min}$, $KE_{anode} = \phi_{Cs} + 1.5\text{ eV} = 3.6\text{ eV}$, because the photon energy at $2f_{min}$ equals $2\phi_{Cs}$, the work function removes $\phi_{Cs}$, leaving $\phi_{Cs} = 2.1\text{ eV}$ of kinetic energy at the cathode, and the anode potential adds $1.5\text{ eV}$. (correct answer)
3. $f_{min} = \phi_{Cs}/h \approx 5.07 \times 10^{14} \text{ Hz}$; at $f = 2f_{min}$, $KE_{anode} = 2\phi_{Cs} + 1.5\text{ eV} = 5.7\text{ eV}$, because the photon energy is $2hf_{min} = 2\phi_{Cs}$ and no work function is subtracted once the electron has left the surface, so the full photon energy plus the anode potential is available as kinetic energy.
4. $f_{min} = \phi_{Cs}/h \approx 5.07 \times 10^{14} \text{ Hz}$; at $f = 2f_{min}$, $KE_{anode} = 2\phi_{Cs} - 1.5\text{ eV} = 2.7\text{ eV}$, because the photon energy at $2f_{min}$ is $2\phi_{Cs}$, the work function removes $\phi_{Cs}$, leaving $\ phi_{Cs}$ at the cathode, and the anode potential decelerates rather than accelerates the electrons since they move toward a higher potential.

Question 6

1. The claim is correct, because a tenfold increase in intensity means ten times as many photons strike the surface per second, and the cumulative energy deposited eventually exceeds the work function of Q even at $1.2 f_P$.
2. The claim is incorrect, because at maximum frequency $1.2 f_P$, every incident photon has energy less than the work function of Q ($hf_Q = 1.5 h f_P$), so no individual photon can liberate an electron regardless of intensity or photon flux. (correct answer)
3. The claim is incorrect, but only because the intensity increase is insufficient; increasing intensity by a factor of at least $1.5/1.2 = 1.25$ times would be needed to bridge the energy gap and allow emission from Q.
4. The claim is correct for metal P but not for metal Q, because metal P's threshold is already exceeded at $1.2 f_P$ and higher intensity increases emission current, while the same logic incorrectly extends to Q where the threshold is not met.

Question 7

1. $KE_{II} > KE_{original} > KE_I$, because kinetic energy depends only on photon frequency and not on intensity; doubling frequency raises $KE$ and halving frequency lowers it, with the work function remaining constant and both modified frequencies still above threshold.
2. $KE_{II} > KE_{original}$, with $KE_I$ possibly undefined, because intensity never affects $KE_{max}$, doubling frequency always increases kinetic energy, but halving the frequency may bring it below the threshold $\phi/h$, in which case no electrons are emitted in case I. (correct answer)
3. $KE_{II} > KE_{original} > KE_I$, because intensity changes shift the photocurrent but not $KE_{max}$, and the quadrupled intensity in case I partially compensates for the lower photon energy, yielding a nonzero $KE_I$ regardless of whether $f/2$ is above the threshold.
4. $KE_{original} > KE_{II} > KE_I$, because doubling the frequency in case II also doubles the photon flux at fixed intensity, increasing the probability of energy loss through inter-electron collisions before emission and thereby reducing the net kinetic energy.