Question

Derive the expression for the stopping potential in terms of frequency of incident radiation and hence draw the graph showing the variation of the two quantities with respect to each other.

Answer 1

We know that, $hv=K{E}_{max}+W_0$
Where, $W_0$ = work function
$K{E}_{max}=hv-W_{0}also{W}_0=h{v}_0$
$K{E}_{max}=hv-h{v}_0\Rightarrow K{E}_{max}=h(v-v_0)$
${}_e{V}_0=h(v-v_0)V_0=\frac{h}{e}(v-v_0)[\because K{E}_{max}{=}_e{V}_0]$
$v=\frac{c}{\lambda }and{v}_0=\frac{c}{{\lambda }_0}$
$V_0=\frac{h}{e}[\frac{c}{\lambda }-\frac{c}{{\lambda }_0}]\Rightarrow V_0=\left(\frac{hc}{e}\right)[\frac{1}{\lambda }-\frac{1}{{\lambda }_0}]$
For photoelectric emission $\lambda <{\lambda }_{0}andv>v_0$
Graph between frequency (v) and stopping potential $V_0$, we know that,
$e{V}_0=hv-{\varphi }_0\Rightarrow V_0=\frac{h}{e}v-\frac{{\varphi }_0}{e}$
$So,V_0\propto v$