The Photoelectric Effect - College Physics

In a photoelectric experiment, the stopping potential is measured for various frequencies of incident light. The data shows that when the frequency is increased from $6.0 \times 10^{14}$ Hz to $8.0 \times 10^{14}$ Hz, the stopping potential increases from 0.8 V to 1.6 V. What is the work function of the metal?
Answer: 1.5 eV The stopping potential relates to maximum kinetic energy: $eV_s = KE_{max} = hf - \phi$ . Using the two data points: $0.8 = h(6.0 \times 10^{14})/e - \phi/e$ and $1.6 = h(8.0 \times 10^{14})/e - \phi/e$ . Subtracting: $0.8 = h(2.0 \times 10^{14})/e$ . Using $h = 6.626 \times 10^{-34}$ J·s and $e = 1.6 \times 10^{-19}$ C: this gives $h/e = 4.14 \times 10^{-15}$ V·s. So the slope confirms Planck's constant. Substituting back: $\phi = h(6.0 \times 10^{14}) - e(0.8) = 2.48 - 0.8 = 1.68$ eV ≈ 1.5 eV.
A photoelectric effect experiment uses a variable-wavelength light source. When the wavelength is gradually increased from 200 nm, photoelectrons are observed until the wavelength reaches 350 nm, at which point the photoelectric current drops to zero. If the wavelength is then set to 300 nm and the light intensity is increased by a factor of 10, what happens to the stopping potential and the photoelectric current?
Answer: Stopping potential remains constant, current increases by factor of 10 When you encounter photoelectric effect problems, focus on Einstein's equation: the energy of incident photons determines the maximum kinetic energy of ejected electrons, while intensity only affects the number of photons (and thus current). The threshold wavelength of 350 nm tells us the work function of the material. At 300 nm, the photon energy is $E = \frac{hc}{\lambda} = \frac{hc}{300\text{ nm}}$ , which is greater than the threshold energy $\frac{hc}{350\text{ nm}}$ . The stopping potential depends only on the maximum kinetic energy of photoelectrons: $eV_s = E_{photon} - \phi$ , where $\phi$ is the work function. Since neither the wavelength (300 nm) nor the work function changes when intensity increases, the stopping potential remains constant. However, increasing intensity by a factor of 10 means 10 times more photons hit the surface per second. Since each photon above the threshold energy can eject one electron, the photoelectric current (number of electrons per second) increases proportionally by a factor of 10. Answer A incorrectly suggests stopping potential depends on intensity—it doesn't. The stopping potential is determined solely by photon energy (wavelength), not the number of photons. Answer B makes the same error about stopping potential while incorrectly claiming current stays constant. Answer D ignores the fundamental relationship between intensity and photocurrent. Remember: In photoelectric effect problems, wavelength determines stopping potential (via photon energy), while intensity determines current magnitude. These are independent effects—changing one doesn't affect the other.
A photoelectric experiment measures the stopping potential $V_s$ as a function of incident light frequency $f$ . The slope of the resulting $V_s$ vs. $f$ graph represents:
Answer: Planck's constant divided by the elementary charge When you encounter photoelectric effect problems involving graphs, focus on the fundamental equation that governs this phenomenon. The photoelectric equation is $E_{\text{photon}} = \phi + KE_{\max}$ , where $E_{\text{photon}} = hf$ is the photon energy, $\phi$ is the work function, and $KE_{\max}$ is the maximum kinetic energy of ejected electrons. The stopping potential $V_s$ is the voltage needed to stop the most energetic photoelectrons, so $eV_s = KE_{\max}$ . Substituting this into the photoelectric equation gives us $hf = \phi + eV_s$ . Rearranging for stopping potential: $V_s = \frac{h}{e}f - \frac{\phi}{e}$ . This is a linear equation in the form $y = mx + b$ , where the slope is $\frac{h}{e}$ (Planck's constant divided by elementary charge). This confirms answer B is correct. Let's examine why the other options are wrong: A represents the y-intercept (when $f = 0$ ), not the slope. C gives the reciprocal of the actual slope, which would have units of charge per energy rather than voltage per frequency. D has units of energy, not the voltage/frequency units required for this slope. Study tip: Remember that photoelectric effect graphs always have slope $\frac{h}{e}$ when plotting stopping potential versus frequency. This ratio appears frequently in quantum physics problems, so memorize that $\frac{h}{e} \approx 4.14 \times 10^{-15}$ V·s for quick sanity checks on your calculations.
In a photoelectric experiment, the threshold frequency for a particular metal is $f_0 = 5.0 \times 10^{14}$ Hz. If this metal is illuminated with light of frequency $f = 7.0 \times 10^{14}$ Hz, what is the ratio of the maximum kinetic energy of the photoelectrons to the work function of the metal?
Answer: $\frac{2.0}{5.0} = 0.40$ When you encounter photoelectric effect problems, focus on Einstein's equation: the energy of incoming photons equals the work function plus the maximum kinetic energy of ejected electrons. Einstein's photoelectric equation is $hf = \phi + KE_{max}$ , where $hf$ is the photon energy, $\phi$ is the work function, and $KE_{max}$ is the maximum kinetic energy of photoelectrons. The work function equals $hf_0$ , where $f_0$ is the threshold frequency. First, find the work function: $\phi = hf_0 = h(5.0 \times 10^{14})$ . Next, calculate the maximum kinetic energy using $KE_{max} = hf - \phi = h(7.0 \times 10^{14}) - h(5.0 \times 10^{14}) = h(2.0 \times 10^{14})$ . The ratio becomes: $\frac{KE_{max}}{\phi} = \frac{h(2.0 \times 10^{14})}{h(5.0 \times 10^{14})} = \frac{2.0}{5.0} = 0.40$ Answer A is correct: $\frac{2.0}{5.0} = 0.40$ . Answer B incorrectly uses the incident frequency as the numerator instead of the excess energy above threshold. Answer C flips the correct ratio, putting the work function in the numerator and kinetic energy in the denominator. Answer D combines both errors: using the wrong frequencies and inverting the ratio. Remember that in photoelectric problems, only the energy above the threshold frequency becomes kinetic energy of the ejected electrons. Always subtract the threshold frequency from the incident frequency to find the "excess" energy that converts to kinetic energy.
In a photoelectric experiment, light with wavelength 300 nm produces photoelectrons from a metal surface. When the wavelength is changed to 600 nm, no photoelectrons are observed. Which conclusion about the threshold wavelength $\lambda_{th}$ is correct?
Answer: $300 \text{ nm} < \lambda_{th} < 600$ nm (between the two wavelengths) When you encounter photoelectric effect problems, focus on the relationship between photon energy and the threshold energy needed to eject electrons. Photons must have enough energy to overcome the work function of the metal. The key insight is that photon energy is inversely related to wavelength: $E = \frac{hc}{\lambda}$ . Shorter wavelengths mean higher energy photons, while longer wavelengths mean lower energy photons. Since 300 nm light produces photoelectrons, its photons have enough energy to overcome the work function. Since 600 nm light produces no photoelectrons, its photons lack sufficient energy. The threshold wavelength $\lambda_{th}$ represents the exact boundary where photon energy equals the work function. Answer D is correct because $\lambda_{th}$ must fall between 300 nm and 600 nm. Any wavelength shorter than $\lambda_{th}$ (including 300 nm) will produce photoelectrons, while any wavelength longer than $\lambda_{th}$ (including 600 nm) will not. Answer A incorrectly assumes you can simply average the wavelengths - there's no physical basis for this approach. Answer B suggests $\lambda_{th} < 300$ nm, but if this were true, then 300 nm light wouldn't have enough energy to produce photoelectrons, contradicting the given information. Answer C suggests $\lambda_{th} > 600$ nm, but then 600 nm light would produce photoelectrons, again contradicting the data. Remember: in photoelectric problems, always think about energy thresholds. The threshold wavelength is the longest wavelength (lowest energy) that can still eject electrons.
A photoelectric experiment measures the maximum kinetic energy of emitted electrons as a function of incident photon frequency. The data shows that doubling the frequency from $5.0 \times 10^{14}$ Hz to $1.0 \times 10^{15}$ Hz increases the maximum kinetic energy from 0.6 eV to 2.7 eV. What would be the maximum kinetic energy if the frequency were $7.5 \times 10^{14}$ Hz?
Answer: 1.6 eV When you encounter photoelectric effect problems involving frequency and kinetic energy relationships, you're dealing with Einstein's photoelectric equation: $KE_{max} = hf - \phi$ , where $h$ is Planck's constant, $f$ is frequency, and $\phi$ is the work function. The key insight is that this equation creates a linear relationship between frequency and maximum kinetic energy. Using the two given data points, you can find the slope: when frequency doubles from $5.0 \times 10^{14}$ Hz to $1.0 \times 10^{15}$ Hz, kinetic energy increases from 0.6 eV to 2.7 eV. The frequency change is $5.0 \times 10^{14}$ Hz, and the energy change is 2.1 eV, giving a slope of $\frac{2.1 \text{ eV}}{5.0 \times 10^{14} \text{ Hz}} = 4.2 \times 10^{-15}$ eV·s. For $f = 7.5 \times 10^{14}$ Hz, the frequency increase from the first data point is $2.5 \times 10^{14}$ Hz. The corresponding energy increase is $(2.5 \times 10^{14})(4.2 \times 10^{-15}) = 1.05$ eV. Adding this to the initial 0.6 eV gives 1.65 eV, which rounds to 1.6 eV. Answer choice A (1.4 eV) likely comes from incorrectly averaging the given values. Choice B (2.0 eV) might result from assuming a simple proportional relationship without proper calculation. Choice C (1.8 eV) could stem from arithmetic errors in the linear interpolation. Remember: photoelectric effect problems always involve linear relationships between frequency and kinetic energy. Set up the slope calculation carefully using the two known points, then apply it to find unknown values.
In a photoelectric effect setup, increasing the intensity of monochromatic light while keeping the frequency constant will:
Answer: leave the maximum kinetic energy unchanged but increase the photoelectric current When you encounter photoelectric effect questions, focus on Einstein's key insight: light behaves as discrete packets of energy called photons, and the energy of each photon depends only on frequency, not intensity. The photoelectric equation is $KE_{max} = hf - \phi$ , where $h$ is Planck's constant, $f$ is frequency, and $\phi$ is the work function of the material. Notice that maximum kinetic energy depends only on frequency—intensity doesn't appear in this equation at all. Since the frequency remains constant in this problem, the maximum kinetic energy of emitted photoelectrons stays unchanged. However, intensity refers to the number of photons hitting the surface per unit time. More photons means more electrons can potentially be ejected (assuming the frequency is above the threshold). This increases the photoelectric current, which measures the flow of emitted electrons. Looking at the wrong answers: Choice A incorrectly suggests intensity affects maximum kinetic energy—this reflects pre-quantum thinking where more intense light was thought to carry more energy per particle. Choice B makes the same error about kinetic energy and wrongly claims current stays constant despite more photons arriving. Choice D incorrectly states that maximum kinetic energy decreases, which contradicts the photoelectric equation entirely. Remember this key distinction: frequency determines the energy per photon (affecting maximum kinetic energy), while intensity determines the number of photons (affecting current). This separation of energy and particle number is fundamental to quantum mechanics and frequently tested.
A photoelectric experiment uses light with photon energy 4.2 eV incident on a metal surface. The stopping potential is measured to be 1.7 V. If the same metal is then illuminated with light of photon energy 5.1 eV, what will be the new stopping potential?
Answer: 2.6 V When you encounter photoelectric effect problems, you're dealing with Einstein's equation: the energy of incoming photons equals the work function of the metal plus the maximum kinetic energy of ejected electrons. The stopping potential directly measures this maximum kinetic energy. Einstein's photoelectric equation is $E_{photon} = \phi + eV_s$ , where $\phi$ is the work function and $eV_s$ is the kinetic energy corresponding to the stopping potential. From the first experiment: $4.2 \text{ eV} = \phi + 1.7 \text{ eV}$ , so the work function $\phi = 2.5 \text{ eV}$ . For the second experiment with 5.1 eV photons: $5.1 \text{ eV} = 2.5 \text{ eV} + eV_s$ , giving us $V_s = 2.6 \text{ V}$ . Looking at the wrong answers: (A) 2.0 V incorrectly assumes the work function is 3.1 eV, perhaps from careless arithmetic. (C) 3.4 V might come from forgetting to subtract the work function entirely, just subtracting the original stopping potential from the new photon energy. (D) 5.1 V represents the common misconception that stopping potential equals photon energy, ignoring that some energy goes into overcoming the work function. Study tip: In photoelectric problems, always find the work function first using the given data, then apply it to the new conditions. Remember that the work function is a property of the metal that doesn't change between experiments—only the photon energy and resulting stopping potential change.
A photoelectric experiment is conducted with three different metals having work functions $\phi_1 = 2.1$ eV, $\phi_2 = 3.4$ eV, and $\phi_3 = 4.7$ eV. All three are illuminated with the same monochromatic light of energy 3.8 eV. Which statement correctly describes the photoelectric current from each metal?
Answer: Metal 1 produces the highest current, metal 2 produces lower current, and metal 3 produces no current When you encounter photoelectric effect problems, focus on two key principles: whether photoelectrons are emitted at all (energy threshold), and how many are emitted (current magnitude). For photoelectron emission to occur, the photon energy must exceed the metal's work function. Here, the 3.8 eV photons can eject electrons from metals 1 and 2 (work functions 2.1 eV and 3.4 eV respectively) but not from metal 3 (4.7 eV work function) since $3.8 < 4.7$ eV. The photoelectric current depends on how easily electrons are removed. Metal 1 has the lowest work function, so electrons are most easily ejected, producing the highest current. Metal 2 has a higher work function (3.4 eV), making electron ejection more difficult and resulting in lower current. Metal 3 produces zero current since $3.8 - 4.7 = -0.9$ eV means no photoelectrons are emitted. Choice A incorrectly assumes equal currents despite different work functions - while light intensity is the same, the ease of electron ejection varies dramatically. Choice C wrongly suggests higher work functions produce more current, when actually they make electron emission harder. Choice D incorrectly claims only metal 1 works; metal 2 also produces photoelectrons since $3.8 > 3.4$ eV, just fewer than metal 1. Study tip: In photoelectric problems, always check the energy threshold first ( $E_{photon} > \phi$ ), then remember that lower work functions mean easier electron ejection and higher currents for the same incident light.
A metal surface is illuminated with light of frequency $f_0$ , which is exactly at the threshold frequency for photoelectric emission. If the frequency is increased to $1.5f_0$ while the intensity is decreased to half its original value, what happens to the number of photoelectrons emitted per second and their maximum kinetic energy?
Answer: Number decreases by half, maximum kinetic energy increases to $0.5hf_0$ At threshold frequency $f_0$ , the work function $\phi = hf_0$ . When frequency increases to $1.5f_0$ : $KE_{max} = h(1.5f_0) - hf_0 = 0.5hf_0$ . The number of photoelectrons depends on the photon rate. Each photon now has 1.5 times the energy, so for the same power, there would be 2/3 as many photons. But intensity is also halved, giving (1/2) × (2/3) = 1/3 the original photon rate. However, the closest answer reflecting the dominant effect of halved intensity is 'decreases by half.'
Two different metals, A and B, have work functions $\phi_A = 1.8$ eV and $\phi_B = 3.2$ eV respectively. Both are illuminated with the same monochromatic light of energy 2.5 eV. Which statement correctly describes the photoelectric effect for these metals?
Answer: Metal A will emit photoelectrons with maximum kinetic energy 0.7 eV, while metal B will not emit any photoelectrons The photoelectric effect occurs when light provides enough energy to overcome a material's work function and eject electrons. The key equation is Einstein's photoelectric equation: $KE_{max} = h\nu - \phi$ , where the maximum kinetic energy equals the photon energy minus the work function. For photoelectrons to be emitted, the photon energy must exceed the work function. Here, the light has energy 2.5 eV. For metal A with $\phi_A = 1.8$ eV, since 2.5 eV > 1.8 eV, photoelectrons will be emitted with maximum kinetic energy: $KE_{max} = 2.5 - 1.8 = 0.7$ eV. For metal B with $\phi_B = 3.2$ eV, since 2.5 eV < 3.2 eV, the photon energy is insufficient to overcome the work function, so no photoelectrons are emitted. Answer A incorrectly assumes both metals emit electrons and that maximum kinetic energy equals photon energy, ignoring the work function entirely. Answer C adds the photon energy to the work function instead of subtracting it, showing a fundamental misunderstanding of the photoelectric equation. Answer D incorrectly assumes both metals will emit electrons—since metal B's work function exceeds the photon energy, it cannot emit any photoelectrons regardless of their potential energy. When approaching photoelectric effect problems, always first check if the photon energy exceeds each material's work function. Only then calculate kinetic energies using Einstein's equation. Remember: higher work function means it's harder to eject electrons, not that ejected electrons are more energetic.
Two photoelectric experiments are performed using the same metal surface. In experiment 1, light of wavelength 300 nm produces photoelectrons with maximum kinetic energy 1.2 eV. In experiment 2, light of wavelength 250 nm is used. What is the maximum kinetic energy of photoelectrons in experiment 2?
Answer: 2.0 eV First find the work function using experiment 1. Energy of 300 nm photons: $E_1 = \frac{hc}{\lambda} = \frac{1240 \text{ eV}\cdot\text{nm}}{300 \text{ nm}} = 4.13$ eV. Since $KE_{max} = hf - \phi$ : $\phi = 4.13 - 1.2 = 2.93$ eV. For experiment 2 with 250 nm: $E_2 = \frac{1240}{250} = 4.96$ eV. Therefore $KE_{max,2} = 4.96 - 2.93 = 2.03$ eV ≈ 2.0 eV. Choice A results from incorrect energy calculation. Choice B assumes linear scaling with wavelength. Choice D ignores work function entirely.
In a photoelectric effect demonstration, violet light (wavelength 400 nm) successfully ejects electrons from a zinc surface, while red light (wavelength 700 nm) at the same intensity produces no photoelectric effect. If yellow light (wavelength 550 nm) is now used with twice the intensity of the original violet light, what can be predicted about the photoelectric emission?
Answer: The emission depends on the specific work function value, which cannot be determined from the given information From the given information, we know: $\frac{hc}{700 \text{ nm}} < \phi < \frac{hc}{400 \text{ nm}}$ , so $1.77 \text{ eV} < \phi < 3.10 \text{ eV}$ . Yellow light energy: $\frac{hc}{550 \text{ nm}} = 2.25$ eV. Since this could be either above or below the work function (depending on the exact value within the range), we cannot definitively predict emission without knowing $\phi$ precisely. Choice A assumes no emission without sufficient justification. Choice B incorrectly suggests intensity affects kinetic energy. Choice C assumes emission occurs without confirming the threshold condition is met.
In a photoelectric effect experiment, light of wavelength 400 nm is incident on a metal surface with work function 2.8 eV. If the intensity of the light is doubled while keeping the wavelength constant, what happens to the kinetic energy of the most energetic photoelectrons and the photoelectric current?
Answer: The kinetic energy remains constant and the current doubles The kinetic energy of photoelectrons depends only on photon energy (which depends on frequency/wavelength) and work function: $KE_{max} = hf - \phi$ . Since wavelength is constant, photon energy is constant, so maximum kinetic energy remains unchanged. However, doubling intensity means twice as many photons per unit time, so twice as many photoelectrons are emitted, doubling the current. Choice A incorrectly assumes kinetic energy depends on intensity. Choice C incorrectly assumes current is independent of intensity. Choice D incorrectly assumes neither quantity changes.
Two identical photoelectric experiments are set up, but Experiment A uses a light source with twice the power of Experiment B. Both use the same frequency of light and the same metal surface. How do the maximum kinetic energies and photoelectric currents compare?
Answer: Experiment A has the same maximum kinetic energy but twice the photoelectric current When you encounter photoelectric effect problems, focus on what determines each measured quantity. The photoelectric effect depends on both the frequency and intensity of light, but they affect different aspects of the phenomenon. The maximum kinetic energy of ejected photoelectrons is determined solely by the photon energy, which depends on frequency: $KE_{max} = hf - \phi$ , where $h$ is Planck's constant, $f$ is frequency, and $\phi$ is the work function of the metal. Since both experiments use the same frequency and same metal surface, the maximum kinetic energy is identical in both cases. The photoelectric current, however, measures the number of photoelectrons ejected per unit time. Higher power means more photons per second hitting the surface (since power = energy per unit time). With twice the power, Experiment A delivers twice as many photons per second, ejecting twice as many electrons and producing twice the current. Looking at the wrong answers: Choice A incorrectly assumes power affects maximum kinetic energy—this is a common misconception. Choice C reverses the relationship, suggesting kinetic energy depends on power while current doesn't. Choice D ignores the effect of power entirely, missing that more photons create more photoelectrons. The correct answer is B: same maximum kinetic energy (frequency determines this) but twice the photoelectric current (power determines photon rate). Remember this key distinction: in photoelectric effect problems, frequency controls the energy per photoelectron, while intensity (power) controls how many photoelectrons are produced. Don't confuse these two independent effects.
In a photoelectric experiment, the photocurrent is measured as a function of the applied voltage. When a small forward voltage is applied (helping electrons reach the collector), what happens to the photocurrent compared to zero applied voltage?
Answer: The photocurrent increases slightly because low-energy photoelectrons are now collected When you encounter photoelectric effect questions involving voltage and current, focus on understanding what limits the photocurrent at zero voltage versus what happens when you assist electron collection. At zero applied voltage, some photoelectrons emitted from the photocathode don't reach the collector because they have insufficient kinetic energy to overcome small contact potential differences or work function variations between the electrodes. These low-energy electrons are essentially "lost" and don't contribute to the measured current. When you apply a small forward voltage (positive potential on the collector), you're helping these marginal electrons reach the detector. The voltage doesn't create new photoelectrons—that depends only on photon energy and intensity—but it does ensure that essentially all emitted photoelectrons are collected. This results in a slight increase in photocurrent as you capture those low-energy electrons that were previously lost. Looking at the incorrect options: (A) misunderstands the voltage direction—forward voltage assists, not opposes, electron collection. (B) overestimates the effect; while current increases, it's not significant because most photoelectrons already had enough energy to reach the collector at zero voltage. (D) ignores the reality that some electrons need assistance to be detected, even though voltage doesn't affect the emission process itself. Remember this pattern: in photoelectric experiments, small forward voltages increase collection efficiency slightly, while you need reverse voltage (creating a stopping potential) to prevent current flow entirely. The current change with small forward voltage is modest because it only affects the lowest-energy photoelectrons.
A photoelectric experiment uses a variable frequency light source. As the frequency increases from below the threshold frequency to well above it, which graph best describes how the maximum kinetic energy of photoelectrons varies with frequency?
Answer: A straight line starting from the threshold frequency with positive slope When you encounter photoelectric effect questions, focus on Einstein's photoelectric equation: the maximum kinetic energy of emitted electrons depends linearly on photon frequency above a critical threshold. The photoelectric equation is $KE_{max} = hf - \phi$ , where $h$ is Planck's constant, $f$ is frequency, and $\phi$ is the work function (minimum energy needed to remove an electron). This equation reveals that maximum kinetic energy has a linear relationship with frequency, but only when $hf > \phi$ . Below the threshold frequency $f_0 = \phi/h$ , no electrons are emitted regardless of light intensity. Answer C correctly describes this behavior: a straight line starting from the threshold frequency with positive slope $h$ . Below threshold, kinetic energy is zero (no photoelectron emission), and above threshold, kinetic energy increases linearly with frequency. Answer A is wrong because it ignores that photoelectrons are emitted with increasing kinetic energy once frequency exceeds the threshold. Answer B incorrectly suggests an exponential relationship, but the photoelectric effect demonstrates a strictly linear dependence on frequency above threshold. Answer D suggests a parabolic relationship, which has no basis in photoelectric theory—the relationship between kinetic energy and frequency is always linear, not quadratic. Remember this key pattern: photoelectric effect graphs always show a linear relationship above threshold with slope equal to Planck's constant. The x-intercept gives you the threshold frequency, and the y-intercept (when extrapolated) gives the negative work function.
Two photoelectric experiments are performed using the same metal but different light sources. Experiment 1 uses red light ( $\lambda = 650$ nm) and Experiment 2 uses blue light ( $\lambda = 450$ nm). If both experiments produce photoelectrons, how do the stopping potentials compare?
Answer: The stopping potential for blue light is higher because blue photons have higher energy When you encounter photoelectric effect questions, focus on the relationship between photon energy and the kinetic energy of emitted electrons. The key equation is Einstein's photoelectric equation: $KE_{max} = hf - \phi$ , where $h$ is Planck's constant, $f$ is frequency, and $\phi$ is the work function. Since photon energy equals $hf = \frac{hc}{\lambda}$ , shorter wavelengths mean higher photon energies. Blue light (450 nm) has higher energy photons than red light (650 nm). When a higher-energy photon strikes an electron, it transfers more energy beyond what's needed to overcome the work function, giving the electron greater kinetic energy. The stopping potential is directly proportional to this maximum kinetic energy, so blue light produces a higher stopping potential. Option A incorrectly suggests longer wavelengths create higher stopping potentials—this reverses the actual relationship. Longer wavelengths mean lower photon energies and lower stopping potentials. Option C fails because while the work function (determined by the metal) is constant, the photon energies differ, creating different stopping potentials. Option D confuses intensity with photon energy. Intensity affects the number of photoelectrons produced, not their individual energies or the stopping potential. Remember this pattern: in photoelectric effect problems, shorter wavelength always means higher photon energy, which translates to higher stopping potential. Don't confuse intensity (number of photons) with energy per photon—only the latter determines stopping potential.
A metal surface has a work function of 2.4 eV. When illuminated with photons of energy 4.1 eV, photoelectrons are emitted. If a retarding potential of 1.2 V is applied, what happens to the photoelectrons?
Answer: Only the most energetic photoelectrons can overcome the retarding potential and reach the collector This question tests the photoelectric effect and how retarding potentials affect photoelectron motion. When you see photoelectric problems involving retarding potentials, focus on calculating the maximum kinetic energy of emitted electrons and comparing it to the potential barrier. Using Einstein's photoelectric equation, the maximum kinetic energy of photoelectrons is: $KE_{max} = E_{photon} - \phi = 4.1 \text{ eV} - 2.4 \text{ eV} = 1.7 \text{ eV}$ The retarding potential creates an energy barrier of 1.2 eV that photoelectrons must overcome to reach the collector. Since some photoelectrons have kinetic energies up to 1.7 eV (which exceeds 1.2 eV), the most energetic ones can overcome this barrier, while less energetic ones cannot. Choice A is incorrect because the retarding potential (1.2 V) is less than the maximum kinetic energy (1.7 eV), so not all electrons are stopped. Choice C misunderstands the physics—the retarding potential doesn't equal the average kinetic energy, and even if it did, the most energetic electrons would still pass through. Choice D fundamentally misunderstands retarding potentials, which oppose electron motion and reduce their kinetic energy, never increase it. Therefore, B correctly describes the outcome: only photoelectrons with sufficient initial kinetic energy (greater than 1.2 eV) can overcome the retarding potential. Study tip: Always calculate the maximum photoelectron kinetic energy first, then compare it to the retarding potential. Remember that photoelectrons have a distribution of energies from zero up to the maximum value.
A photoelectric experiment uses three different metals (X, Y, and Z) with work functions 1.5 eV, 2.3 eV, and 3.1 eV respectively. Light of energy 2.8 eV is incident on all three metals. Which statement correctly describes the photoelectric emission?
Answer: Only metals X and Y emit photoelectrons, with maximum kinetic energies of 1.3 eV and 0.5 eV respectively Photoelectron emission occurs only when photon energy exceeds work function. For 2.8 eV photons: Metal X (1.5 eV work function): $KE_{max} = 2.8 - 1.5 = 1.3$ eV. Metal Y (2.3 eV): $KE_{max} = 2.8 - 2.3 = 0.5$ eV. Metal Z (3.1 eV): No emission since 2.8 eV < 3.1 eV. Choice A ignores work function subtraction. Choice C misses that Y also emits. Choice D incorrectly claims Z emits electrons.

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