A metal is illuminated at several frequencies. Measured (frequency in 10¹⁴ Hz, stopping potential in V) pairs are (6.0, 0.482), (7.0, 0.895), (8.0, 1.309), and (9.0, 1.722). Treat the best-fit line through the rounded data as having gradient 0.4135 V per 10¹⁴ Hz and intercept −2.00 V. Determine Planck's constant, work function and threshold frequency. Predict the effects of doubling intensity above threshold, and explain why classical wave theory fails here. Use e = 1.602 × 10⁻¹⁹ C.

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Accepted Answer

h = 6.62 × 10⁻³⁴ J s, work function = 2.00 eV, threshold frequency = 4.84 × 10¹⁴ Hz. Doubling intensity doubles photocurrent under unsaturated collection but does not change stopping voltage. Einstein's equation hf = Φ + K_max and eV_s = K_max combine to V_s = (h/e)f − Φ/e. Because the horizontal axis unit is 10¹⁴ Hz, h/e = 0.4135/(10¹⁴) V s. Therefore h = e(4.135 × 10⁻¹⁵) = 6.62 × 10⁻³⁴ J s. The vertical intercept is −Φ/e = −2.00 V. Hence Φ = 2.00 eV = 3.204 × 10⁻¹⁹ J. At threshold V_s = 0, so f₀ = Φ/h, equivalently 2.00/(4.135 × 10⁻¹⁵) = 4.84 × 10¹⁴ Hz. Greater intensity means more photons per second, so unsaturated photocurrent doubles, but each photon retains energy hf and the stopping potential is unchanged. Classical intensity-based energy transfer cannot explain an immediate threshold frequency or intensity-independent maximum electron energy.

Determine Planck's constant from stopping voltage

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Stopping-potential observations

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Stopping-potential observations — supplied values
Frequency / 10¹⁴ Hz	Stopping potential / V
6	0.482
7	0.895
8	1.309
9	1.722