An ideal photoelectric experiment gives these pairs (frequency in units of 10¹⁴ Hz, stopping potential in V): (7, 0.798), (8, 1.212), (9, 1.626), (10, 2.04). Treat the points as exactly collinear. Use e = 1.602 × 10⁻¹⁹ C. (a) Determine h from the gradient and the work function in eV. (b) Calculate the threshold frequency. (c) Explain how doubling light intensity at a fixed frequency above threshold affects the stopping potential and, if collection is unsaturated and efficiency unchanged, the photocurrent.

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h = 6.632 × 10^-34 J s; work function 2.1 eV; threshold 5.072 × 10^14 Hz. Stopping potential stays fixed; photocurrent doubles under the stated conditions. Einstein's photon-energy balance is hf = Φ + K_max, and stopping the most energetic photoelectrons requires eV_s = K_max. Thus V_s = (h/e)f − Φ/e is linear. The gradient against the displayed frequency scale is 0.414 V per 10¹⁴ Hz, or 4.140 × 10^-15 V s. Multiplying by e gives h = 6.632 × 10^-34 J s. Extrapolating to f = 0 gives an intercept of −2.1 V. Therefore Φ = 2.1 eV, since one electronvolt is the energy gained by an electron across one volt. At threshold V_s = 0, so f₀ = Φ/h = 5.072 × 10^14 Hz when Φ and h are expressed in compatible units. All supplied frequencies lie above this threshold. Doubling intensity supplies twice as many photons per second but does not change each photon's energy. Maximum electron energy and stopping potential stay fixed; under the stated collection conditions the photocurrent doubles.

Threshold frequency from stopping voltages · set 6

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Stopping-potential observations

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Stopping-potential observations — supplied values
Frequency / 10¹⁴ Hz	Stopping potential / V
7	0.798
8	1.212
9	1.626
10	2.04