The Photoelectric Effect — Question 8

Question 8

Saturation current times stopping voltage is an energy bound rather than simultaneous output

A cathode at photon energy 4.00 eV has work function 2.00 eV and external yield $\eta=.250$. Incident optical power is P. The saturation current is $I_s$ and stopping voltage is $V_s$.

1. Find an upper bound on total emitted-electron kinetic power.

2. Express $I_sV_s/P$ and explain why this does not equal electrical power delivered at the stopping voltage.

Question 8 - Solution

Saturation current times stopping voltage is an energy bound rather than simultaneous output

(a) Method and calculation

$$\dot N_e=\eta P/E_\gamma,\qquad K\le K_{\max}=E_\gamma-\phi.$$$$\boxed{P_{K,e}\le\dot N_eK_{\max}=\eta P\left(1-\phi/E_\gamma\right)=P/8}.$$

(b) Interpretation and further calculation

$$\boxed{I_sV_s/P=\eta(E_\gamma-\phi)/E_\gamma=1/8}.$$$$I(V_s)=0\quad\Rightarrow\quad\boxed{P_{\mathrm{electrical}}(V_s)=V_sI(V_s)=0}.$$

Physical check

Saturation current and stopping voltage belong to different operating points. Multiplying them supplies a useful bound, not an actual operating power.