The Photoelectric Effect — Question 4

Question 4

Photocurrent and optical power determine the external quantum yield

At wavelength 620 nm, incident power is $1.00\,\mathrm{mW}$ and saturated current is $0.100\,\mathrm{mA}$. The work function is $1.50\,\mathrm{eV}$. All emitted electrons are collected. Use $hc=1240\,\mathrm{eV\,nm}$, $e=1.60\times10^{-19}\,\mathrm C$, and $eV_s=K_{\max}=hf-\phi$ when emission is allowed.

1. Find the emitted-electron to incident-photon ratio.

2. Find the stopping voltage and explain why yield and stopping energy are independent pieces of information.

Question 4 - Solution

Photocurrent and optical power determine the external quantum yield

(a) Method and calculation

$$\dot N_\gamma=P/E_\gamma=10^{-3}/(3.20\times10^{-19})=3.125\times10^{15}\,\mathrm{s^{-1}}.$$$$\dot N_e=I/e=6.25\times10^{14}\,\mathrm{s^{-1}},\quad\boxed{\eta=.200}.$$

(b) Interpretation and further calculation

$$K_{\max}=2.00-1.50=.500\,\mathrm{eV}.$$$$\boxed{V_s=.500\,\mathrm V}.$$

Physical check

Yield counts how often a photon produces a collected electron; the work function determines the maximum energy of those electrons.