Photoelectric Effect - Physics 2

1. Which of the following correctly compares the maximum kinetic energy of photoelectrons $KE_{max}$ in the original experiment versus the separate experiment with the second metal at frequency $2f$?
   Answer: $KE_{max}$ 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 $\phi = 0$. Whenever you see a photoelectric effect question, anchor yourself to Einstein's equation: $KE_{max} = hf - \phi$, where $hf$ is the photon energy and $\phi$ is the work function. Intensity never appears here — it only affects how many electrons are ejected, not their energy. For the original experiment, the maximum kinetic energy is $KE_1 = hf - \phi$. Note that doubling the intensity (mentioned in the passage) is a red herring — it changes nothing about $KE_{max}$. For the second experiment, the frequency doubles to $2f$ and the work function doubles to $2\phi$, giving $KE_2 = h(2f) - 2\phi = 2hf - 2\phi$. Now compare the two: $KE_2 - KE_1 = (2hf - 2\phi) - (hf - \phi) = hf - \phi$. So the second experiment yields a $KE_{max}$ that is greater by exactly $hf - \phi$, confirming C is correct. A is wrong because doubling both frequency and work function does not produce canceling effects — the photon energy gain ($hf$) and the work function increase ($\phi$) are generally unequal, so the effects cancel only if $\phi = hf$. B claims the difference is always $hf$, but this ignores that the work function also doubles. The correct difference is $hf - \phi$, which depends on $\phi$. D inverts the relationship, incorrectly claiming the second experiment yields less kinetic energy. Since $KE_2 - KE_1 = hf - \phi > 0$ when $hf > \phi$ (a requirement for photoemission to occur at all), this is backwards. Your strategy: always write out $KE_{max} = hf - \phi$ for each scenario explicitly, then subtract — don't try to reason about "effects canceling" in the abstract.
2. What is the minimum photon frequency $f_{min}$ required so that an electron ejected from the cesium cathode can reach the anode, and what is the maximum kinetic energy of electrons at the anode when $f = 2f_{min}$? (Use $h = 4.14 \times 10^{-15} \text{ eV}\cdot\text{s}$, $e = 1 \text{ (in eV units)}$.)
   Answer: $f_{min} = \phi_{Cs}/h \approx 5.07 \times 10^{14} \text{ Hz}$; 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 $1.5\text{ eV}$. Photoelectric effect questions become much more manageable when you track energy at two separate locations: the cathode (where emission happens) and the anode (where electrons arrive). At the cathode, Einstein's photoelectric equation tells you the kinetic energy an electron has just after leaving the surface: $KE_{cathode} = hf - \phi$. For emission to occur at all, $hf \geq \phi$. Here, the anode's positive potential doesn't lower the barrier for emission — it only helps electrons travel after they've already escaped. So the minimum frequency for emission is simply $f_{min} = \phi_{Cs}/h = 2.1\text{ eV} / (4.14 \times 10^{-15}\text{ eV}\cdot\text{s}) \approx 5.07 \times 10^{14}\text{ Hz}$. At $f = 2f_{min}$, the photon energy is $hf = 2hf_{min} = 2\phi_{Cs} = 4.2\text{ eV}$. After overcoming the work function, the electron leaves the cathode with $KE_{cathode} = 4.2 - 2.1 = 2.1\text{ eV} = \phi_{Cs}$. As it accelerates through the $+1.5\text{ V}$ potential toward the anode, it gains $1.5\text{ eV}$ of additional kinetic energy, giving $KE_{anode} = 2.1 + 1.5 = 3.6\text{ eV}$. That's answer B. A is wrong because the positive anode voltage doesn't reduce the work function — emission and transit are separate physical processes. C incorrectly skips subtracting the work function entirely, as if the electron keeps all the photon's energy. D makes the critical error of treating the positive anode as decelerating electrons — but electrons (negative charge) accelerate toward the positive anode, gaining energy. Your key strategy: always apply $KE = hf - \phi$ at the surface first, then separately apply the work done by the electric field during transit: $\Delta KE = q\Delta V = (+e)(+1.5\text{ V}) = +1.5\text{ eV}$.
3. If the student repeats the experiment using a different metal whose work function is larger by $\Delta\phi$, which of the following correctly describes how the new graph differs from the original?
   Answer: 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. Whenever you see a photoelectric effect question involving a $V_0$ vs. $f$ graph, anchor yourself to the Einstein photoelectric equation: $eV_0 = hf - \phi$, which rearranges to $V_0 = \frac{h}{e}f - \frac{\phi}{e}$. This is a linear equation where the slope is $h/e$ — a universal constant involving Planck's constant and the electron charge — while the y-intercept is $-\phi/e$ and the x-intercept (threshold frequency) is $f_0 = \phi/h$. Since $h/e$ depends only on fundamental constants, the slope never changes regardless of which metal you use. When the work function increases by $\Delta\phi$, the new y-intercept becomes $-(\phi + \Delta\phi)/e$, shifting downward by $\Delta\phi/e$. The new threshold frequency becomes $(\phi + \Delta\phi)/h$, which is larger — a higher frequency is needed before any electrons are released. The two lines are therefore parallel, with the new line shifted down and to the right. This confirms D as correct. A is wrong on two counts: the x-intercept shift formula $\Delta\phi/(e \cdot m^2)$ is dimensionally incorrect and physically fabricated, and the slope never changes. B is wrong because a larger work function has no effect on the slope $h/e$ — the rate of change of stopping potential with frequency is the same for every metal. C gets the slope right but reverses the direction of the x-intercept shift; a larger work function demands a higher threshold frequency, not lower. A reliable study tip: in photoelectric problems, always separate what's universal (the slope $h/e$) from what's material-specific (the work function $\phi$). Any question manipulating the metal only ever shifts the line vertically — never tilts it.
4. In a photoelectric experiment, light of frequency $f$ is directed at a metal surface with work function $\phi$, where $hf > \phi$. The experimenter then performs two independent modifications: (I) the light frequency is halved while intensity is quadrupled, and (II) the light frequency is doubled while intensity is halved. Which of the following correctly ranks the maximum kinetic energies $KE_I$, $KE_{II}$, and $KE_{original}$ after each modification?
   Answer: $KE_{II} > KE_{original}$, 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. Whenever you see a photoelectric effect question, anchor yourself to Einstein's equation: $KE_{max} = hf - \phi$. This tells you that maximum kinetic energy depends only on photon frequency and the work function — intensity is irrelevant to energy, it only affects how many electrons are ejected per second. For the original setup, $KE_{original} = hf - \phi > 0$ since $hf > \phi$. In modification II, frequency doubles, giving $KE_{II} = 2hf - \phi$. Since $hf > \phi$, we know $2hf - \phi > hf - \phi$, confirming $KE_{II} > KE_{original}$. In modification I, frequency becomes $f/2$, so the photon energy is $hf/2$. Here's the critical trap: we don't know whether $hf/2 > \phi$. If $hf/2 \leq \phi$, no electrons are emitted at all — no matter how much you crank up the intensity. So $KE_I$ may be undefined (no emission), making B correct. Answer A assumes both modified frequencies remain above threshold, which isn't guaranteed. The problem only tells you $hf > \phi$, not $hf/2 > \phi$, so you can't safely conclude $KE_I$ is a valid positive number. Answer C is doubly wrong: it claims quadrupled intensity "partially compensates" for lower photon energy. Intensity never compensates for energy — each photon still carries only $hf/2$. Answer D invents a fictitious mechanism — "inter-electron collisions reducing kinetic energy" — that has no basis in photoelectric theory. Doubling frequency always increases $KE_{max}$. Study tip: On any photoelectric question, immediately ask two things: (1) Is the frequency above threshold? (2) What is $hf - \phi$? Intensity is a distractor — treat it as irrelevant to energy every time.
5. A metal surface is illuminated with light whose frequency is exactly at the threshold frequency $f_0$ of the metal. The experimenter then measures the stopping potential and the photoelectric current. Which of the following correctly describes both measurements, and for the correct reason?
   Answer: 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. Whenever you see a photoelectric effect question, anchor your thinking to the Einstein photoelectric equation: $KE_{max} = hf - \phi$, where $\phi = hf_0$ is the work function. The stopping potential is defined by $eV_s = KE_{max}$, so it directly reflects the ejected electrons' kinetic energy — nothing more, nothing less. At exactly the threshold frequency, $hf_0 - hf_0 = 0$, meaning ejected electrons emerge with zero kinetic energy. Since $V_s = KE_{max}/e = 0$, no retarding voltage is needed to stop them — the stopping potential is zero. Now, what about current? These electrons have zero kinetic energy, so they cannot travel across the gap to reach the collector on their own. With no electrons arriving at the collector, the photoelectric current is also zero. This makes A the correct answer: stopping potential is zero because there's no kinetic energy to counteract, and current is zero because the electrons can't traverse the gap. B is wrong because it invokes a "built-in contact potential" to carry zero-kinetic-energy electrons to the collector — this misapplies how contact potentials work and doesn't reflect standard photoelectric experimental conditions. C incorrectly sets the stopping potential equal to $hf_0/e$, which would mean stopping electrons carrying the full photon energy — but the work function consumes that energy entirely, leaving nothing to stop. D fabricates a "resonance condition" that doesn't exist in photoelectric physics; threshold frequency is simply the minimum needed, not a special resonance point. Your key study tip: always compute $KE_{max}$ first — stopping potential and current behavior both follow directly from it.
6. An engineer claims that increasing the intensity of the light source by a factor of 10 will cause metal Q to emit photoelectrons even if the light source's maximum frequency is reduced to $1.2 f_P$. Which of the following best evaluates this claim?
   Answer: 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. Whenever you see a question involving the photoelectric effect and intensity, your first instinct should be to ask: does any individual photon have enough energy to exceed the work function? That single question determines everything. The photoelectric effect is governed by photon energy, not total light energy delivered to a surface. A single photon must have energy $E = hf \geq hf_{threshold}$ to liberate one electron. For metal Q, the threshold frequency is $f_Q = 1.5f_P$, meaning each photon must carry at least $1.5hf_P$ of energy. If the light source's maximum frequency is reduced to $1.2f_P$, every photon in the beam carries at most $1.2hf_P$, which falls short of Q's work function. No electron is emitted — period. This is why B is correct: intensity is irrelevant when no single photon clears the energy threshold. A commits the classic photoelectric misconception — that energy can accumulate across multiple photons to eventually free an electron. Electrons are not liberated by pooled energy; emission is an all-or-nothing, single-photon interaction. C compounds the same error by implying the intensity ratio can somehow "bridge the energy gap." The gap is in frequency (energy per photon), not total power. No intensity multiplier fixes a per-photon energy deficit. D correctly identifies that metal P does emit (since $1.2f_P > f_P$), but then suggests the reasoning "incorrectly extends" to Q as if it's a logical error rather than a physics constraint. The issue isn't faulty logic — it's that Q's threshold simply isn't met. Study tip: On any photoelectric effect question, immediately compare photon frequency to threshold frequency. Intensity controls how many electrons are emitted, never whether they are.
7. A physicist uses the photoelectric effect to measure Planck's constant by plotting stopping potential $V_0$ vs. frequency $f$ for five different frequencies, all well above the threshold. She obtains a best-fit slope of $m = 4.0 \times 10^{-15} \text{ eV}\cdot\text{s}$ and a y-intercept of $-2.3 \text{ V}$, from which she infers the work function $\phi = 2.3 \text{ eV}$. A colleague points out that the voltmeter used to measure stopping potential reads $0.12 \text{ V}$ too low due to a systematic offset. How does this systematic error affect the physicist's extracted values of $h$ and $\phi$?
   Answer: 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. Whenever you see a question about systematic measurement errors, your first instinct should be to ask: does this error shift every data point by the same constant, or does it scale with the measurement? A constant offset has a very specific effect on a best-fit line. The photoelectric equation relates stopping potential to frequency as $V_0 = \frac{h}{e}f - \frac{\phi}{e}$, so a plot of $V_0$ vs. $f$ has slope $h/e$ and y-intercept $-\phi/e$. If the voltmeter reads 0.12 V too low, every measured $V_0$ is shifted down by the same constant $\delta = 0.12\text{ V}$. Because the offset is identical at every frequency, the rise between any two points is unchanged — the 0.12 V subtracts out of both. The slope $\Delta V_0 / \Delta f$ is therefore unaffected, meaning $h$ is correctly extracted. However, the entire line is shifted downward by 0.12 V, making the y-intercept 0.12 V more negative than the true value. Since $\phi = e \times |y\text{-intercept}|$, a more negative intercept yields a larger inferred work function — overestimated by 0.12 eV. That's exactly what answer A describes, making it correct. Answer B is wrong because a uniform vertical shift does not change the slope; $h$ is not underestimated. Answer C gets the slope reasoning right but reverses the intercept logic — a more negative intercept produces a larger inferred $\phi$, not smaller. Answer D is wrong because while the slope is unaffected, the intercept is genuinely distorted; the offset does not cancel in the intercept extraction. Study tip: For any systematic offset error, ask separately: does it affect the slope (differences between points)? And does it affect the intercept (absolute vertical position)? A constant additive error only corrupts the intercept — never the slope.

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