Physics 9702 · Paper 4

The Photoelectric Effect

Shine light on a metal and electrons fly off — but only if the light's frequency is high enough, no matter how bright it is. That stubborn fact is what proved light comes in packets.

Einstein's equation

One photon, one electron: the photon's energy hf splits into the work function Φ (the minimum energy to escape) plus the electron's kinetic energy. Below the threshold frequency no single photon carries enough energy, so no amount of intensity helps.

The is the emission of electrons from a metal surface when electromagnetic radiation is incident on it (9702/42/M/J/25 Q9(a)). One photon interacts with one electron and hands over all its energy.

hf = \Phi + \tfrac{1}{2}mv_{max}^2

Symbols

$hf$ = energy of the incoming photon (J)
$\Phi$ = work function — minimum energy to escape the metal (J)
$\tfrac{1}{2}mv_{max}^2$ = maximum kinetic energy of the freed electron (J)

The photon's energy pays the escape fee $\Phi$ first; whatever is left becomes the electron's kinetic energy. The freest electrons keep the most, so the equation uses maximum KE — not an average.
At the $f_0$ the KE is zero, so $hf_0 = \Phi$ . Below it, no single photon has enough energy — and piling on intensity just sends more too-weak photons, so nothing escapes. This is the evidence for the photon model: energy comes in whole packets set by frequency, not spread out as a wave (9702/42/O/N/23 Q8(b)(ii)).

Einstein's equation as a graph: E_max = hf − Φ. Above the threshold frequency it is a straight line of gradient h (the Planck constant); extended back, its intercept on the energy axis is −Φ. The gradient must be read as ~6.6×10⁻³⁴, not quoted from the data sheet.

Rearranged, $E_{max} = hf - \Phi$ is a straight line: plot maximum KE against frequency and the gradient is the Planck constant, the (extrapolated) intercept on the energy axis is $-\Phi$ — mind the minus sign (9702/42/O/N/23 Q8(c)). Below $f_0$ the line sits flat on zero (9702/41/O/N/24 Q8(c)).

Worked example

Threshold to work function to speed (2024)

A metal has threshold frequency $8.8\times10^{14}$ Hz. Find its work function, then the maximum electron speed for light of $11\times10^{14}$ Hz (9702/41/O/N/24 Q8(b)).

$\Phi = hf_0 = 6.63\times10^{-34} \times 8.8\times10^{14} = 5.8\times10^{-19}\ \text{J}$ .
$\tfrac{1}{2}mv_{max}^2 = hf - \Phi = 6.63\times10^{-34}\times11\times10^{14} - 5.8\times10^{-19} = 1.46\times10^{-19}\ \text{J}$ .
$v_{max} = \sqrt{2 \times 1.46\times10^{-19}/9.11\times10^{-31}} = 5.7\times10^{5}\ \text{m s}^{-1}$ .

Answer

\Phi = 5.8\times10^{-19}\ \text{J};\quad v_{max} = 5.7\times10^{5}\ \text{m s}^{-1}

Common mistake

The and threshold move together but inversely on wavelength: the metal with the smallest work function has the longest threshold wavelength

\lambda_0 = hc/\Phi

(9702/42/F/M/24 Q7(c)(ii)). Reaching for the biggest number in the table gets this backwards.

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