The Photoelectric Effect: Interpreting a Stopping-Voltage Graph
A student shines monochromatic light on a metal surface in a photocell and measures the stopping voltage for several frequencies. The data are: f = 5.0 × 10¹⁴ Hz, Vs = 0.20 V; f = 6.0 × 10¹⁴ Hz, Vs = 0.62 V; f = 7.0 × 10¹⁴ Hz, Vs = 1.03 V. The student plots Vs against f and obtains a straight line, then claims 'the gradient gives Planck's constant and the y-intercept gives the work function.' A classmate says the gradient should give h/e and the intercept should give −W/e. The teacher asks the class to decide, and to use the data to find h and W. Use e = 1.6 × 10⁻¹⁹ C.
1. Derive the relationship between Vs and f, and determine what the gradient and intercept of the graph represent. Then calculate h and W from the data.
Read the worked answer
Einstein's equation: hf = W + Kmax, and Kmax = eVs, so eVs = hf − W, giving Vs = (h/e)f − W/e.
So a graph of Vs against f has gradient h/e and y-intercept −W/e. The classmate is correct; the first student's claim is dimensionally wrong because Vs is in volts, not joules.
Gradient from the data: between first and last points, ΔVs = 1.03 − 0.20 = 0.83 V; Δf = 2.0 × 10¹⁴ Hz. Gradient = 0.83/(2.0 × 10¹⁴) = 4.15 × 10⁻¹⁵ V·s. Then h = gradient × e = 4.15 × 10⁻¹⁵ × 1.6 × 10⁻¹⁹ = 6.6 × 10⁻³⁴ J·s, close to the accepted 6.63 × 10⁻³⁴ J·s.
Intercept: using Vs = (h/e)f − W/e, W/e = (h/e)f − Vs = 4.15 × 10⁻¹⁵ × 5.0 × 10¹⁴ − 0.20 = 2.075 − 0.20 = 1.875 V. So W = 1.875 × 1.6 × 10⁻¹⁹ = 3.0 × 10⁻¹⁹ J (about 1.9 eV).
2. The student then doubles the intensity of the 6.0 × 10¹⁴ Hz light. Predict what happens to the stopping voltage and to the photocurrent, and explain why.
Read the worked answer
The stopping voltage is unchanged at 0.62 V. The stopping voltage depends only on the maximum kinetic energy of the emitted electrons, which is set by the photon energy (hf) and the work function (W): Kmax = hf − W. Doubling intensity does not change the frequency, so the photon energy is unchanged and Kmax is unchanged. The stopping voltage therefore remains 0.62 V.
The photocurrent, however, approximately doubles (assuming the photocell is not saturated). Doubling intensity at fixed frequency means twice as many photons arrive per second, so twice as many electrons are ejected per second, giving twice the current. This is the key distinction: intensity controls the number of photoelectrons, while frequency controls their maximum energy. A student who predicts a change in Vs has confused the two variables, which is exactly the judgement the photoelectric effect tests.