The Photoelectric Effect — Question 3

Question 3

Changing intensity changes current but leaves the stopping energy fixed

A monochromatic source illuminates a cathode above threshold. Its saturated photocurrent is $I_0$ and stopping potential is $V_0$. Intensity is multiplied by 3 while photon energy, illuminated area and quantum yield remain fixed. Neglect space charge and heating.

1. Derive the new saturation current from photon flux.

2. Find the new stopping potential and contrast the two observables.

Question 3 - Solution

Changing intensity changes current but leaves the stopping energy fixed

(a) Method and calculation

$$\dot N_\gamma=P/(hf),\qquad I_{\mathrm{sat}}=e\eta P/(hf).$$$$P\to3P\quad\Rightarrow\quad\boxed{I_{\mathrm{sat}}\to3I_0}.$$

(b) Interpretation and further calculation

$$eV_s=hf-\phi,\qquad f,\phi\text{ unchanged}.$$$$\boxed{V_s\to V_0},\qquad\boxed{K_{\max}\to eV_0}.$$

Physical check

This conclusion assumes the single-photon, fixed-yield regime; high-intensity nonlinear effects were excluded.