The Photoelectric Effect — Question 5

Question 5

A retarding curve can reveal a distribution of normal kinetic energies

Assume emitted electrons have normal kinetic energies $K_x$ uniformly distributed from 0 to $K_m$. At retarding voltage V, only electrons with $K_x>eV$ are collected. At $V=.300\,\mathrm V$, the current is $.750I_0$, where $I_0$ is the zero-retardation current.

1. Derive I(V) and infer $K_m$.

2. Find the stopping voltage and the voltage that gives half the initial current.

Question 5 - Solution

A retarding curve can reveal a distribution of normal kinetic energies

(a) Method and calculation

$$\frac{I(V)}{I_0}=\int_{eV}^{K_m}\frac{dK_x}{K_m}=1-\frac{eV}{K_m}\quad(0\le eV\le K_m).$$$$.750=1-\frac{.300\,\mathrm{eV}}{K_m}\quad\Rightarrow\quad\boxed{K_m=1.20\,\mathrm{eV}}.$$

(b) Interpretation and further calculation

$$\boxed{V_s=K_m/e=1.20\,\mathrm V}.$$$$I/I_0=1/2\quad\Rightarrow\quad\boxed{V=.600\,\mathrm V}.$$

Physical check

The specified distribution is for energy normal to the collector, not automatically the distribution of total electron energy.