Physics 9702 · Paper 4

Stopping Potential

How do you measure the maximum kinetic energy of photoelectrons without catching them? Push back with a voltage until the very fastest just fail to arrive.

eV_S = E_max

Apply a reverse voltage to the collecting plate. As it rises, slower electrons are turned back first; the $V_S$ is the voltage at which even the fastest electron is stopped and the current falls to zero. At that point its kinetic energy has all been spent climbing the potential:

eV_S = E_{max} = hf - \Phi

Symbols

The three-step reasoning behind it (9702/42/M/J/25 Q9(b)(i)): the current falls to zero when the applied voltage equals the energy per unit charge of the electrons; the photon's energy depends on frequency; so the maximum electron energy — and hence $V_S$ — depends on frequency.
A $V_S$ – $f$ graph gives everything: since $V_S = \dfrac{h}{e}f - \dfrac{\Phi}{e}$ , the gradient is $h/e$ and the intercepts give the threshold frequency and the work function. A 2025 “state three conclusions” part read off $f_0 = 1.5\times10^{15}$ Hz, $\Phi = 6.2$ eV, and $h = 6.6\times10^{-34}$ J s (9702/42/M/J/25 Q9(b)(ii)).

Common mistake

When the Planck constant comes from the graph, quote the value your gradient gives — around

6.6\times10^{-34}

— not the data-sheet

6.63\times10^{-34}

. Copying the book value defeats the point of the measurement and loses the mark (9702/42/M/J/25 Q9(b)(ii)).