15.5 The Photoelectric Effect

1. Example 1Calculator allowed

   Ejecting electrons from a metal

   Light of wavelength 400 nm shines on a metal with a work function of 2.30 eV. Find (a) the photon energy, (b) the maximum kinetic energy of ejected electrons, (c) the stopping potential and (d) the threshold frequency.

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   1. 1Step 1: (a) $E = \dfrac{1240\text{ eV·nm}}{400\text{ nm}} = 3.10$ eV.
   2. 2Step 2: (b) $K_{\max} = hf - \phi = 3.10 - 2.30 = 0.80$ eV.
   3. 3Step 3: (c) $eV_{\text{stop}} = K_{\max}$, so $V_{\text{stop}} = 0.80$ V.
   4. 4Step 4: (d) $f_0 = \dfrac{\phi}{h} = \dfrac{2.30\text{ eV}}{4.14 \times 10^{-15}\text{ eV·s}} = 5.56 \times 10^{14}$ Hz, which matches a threshold wavelength of 1240/2.30 = 539 nm.

   Answer: (a) 3.10 eV, (b) 0.80 eV, (c) 0.80 V, (d) 5.56 × 10¹⁴ Hz

2. Example 2Calculator allowed

   Planck's constant from data

   A student measures the stopping potential for light of four frequencies on the same metal and gets $K_{\max}$ = 0.38, 0.80, 1.21 and 1.63 eV at f = 6.0, 7.0, 8.0 and 9.0 × 10¹⁴ Hz. Find Planck's constant, the threshold frequency and the work function.

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   1. 1Step 1: Graph $K_{\max}$ against f. The points lie on a straight line, as $K_{\max} = hf - \phi$ predicts.
   2. 2Step 2: Slope: $\dfrac{1.63 - 0.38}{(9.0 - 6.0) \times 10^{14}} = \dfrac{1.25\text{ eV}}{3.0 \times 10^{14}\text{ Hz}} = 4.2 \times 10^{-15}$ eV·s, close to the accepted 4.14 × 10⁻¹⁵ eV·s.
   3. 3Step 3: Extend the line down to $K_{\max} = 0$: working back from 6.0 × 10¹⁴ Hz, you need to lose 0.38 eV, which takes $\dfrac{0.38}{4.2 \times 10^{-15}} \approx 0.9 \times 10^{14}$ Hz. So $f_0 \approx 5.1 \times 10^{14}$ Hz.
   4. 4Step 4: Work function: $\phi = hf_0$. Use your measured slope for h, so the answer matches your line: $\phi \approx (4.2 \times 10^{-15})(5.1 \times 10^{14}) \approx 2.1$ eV. (Check: the line's vertical intercept is $0.38 - (4.17 \times 10^{-15})(6.0 \times 10^{14}) \approx -2.1$ eV, which is $-\phi$.)

   Answer: h ≈ 4.2 × 10⁻¹⁵ eV·s; f₀ ≈ 5.1 × 10¹⁴ Hz; φ ≈ 2.1 eV

3. Example 3Calculator allowed

   Brighter light (classic trap)

   The metal from the first example (φ = 2.30 eV) is lit with 400 nm light. (a) The intensity is doubled. What happens to the current and to $K_{\max}$? (b) The light is switched to very bright 650 nm red light. What happens?

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   1. 1Step 1: (a) Doubling the intensity doubles the number of photons per second, so about twice as many electrons are freed and the current doubles. Each photon still has 3.10 eV, so $K_{\max}$ stays at 0.80 eV.
   2. 2Step 2: (b) A 650 nm photon has $\frac{1240}{650} = 1.91$ eV, which is less than 2.30 eV. No single photon can free an electron, so no electrons come out, however bright the light.
   3. 3Step 3: The trap is assuming more intensity means more energy per electron. That's the wave-model prediction the experiment disproved.

   Answer: (a) The current doubles; the maximum kinetic energy is unchanged. (b) No electrons are ejected.

Common mistakes

• !Thinking brighter light gives faster electrons. Intensity changes the number of electrons, not their maximum energy.
• !Forgetting the threshold: below $f_0$, no electrons come out at any intensity.
• !Mixing eV and J in $K_{\max} = hf - \phi$. Put every term in the same unit before subtracting.
• !Reading the slope of the $K_{\max}$-versus-f graph as the work function. The slope is h; the intercepts give $f_0$ and φ.

On the exam

• ✓Experimental questions are common: describe the stopping-potential setup, graph $K_{\max}$ against frequency, and get h from the slope and φ from an intercept.
• ✓Expect a question asking why the results support the photon model. Name a specific observation (threshold frequency, or $K_{\max}$ not depending on intensity) and explain it with E = hf.