Question

The figure is the plot of stopping potential versus the frequency of the light used in an experiment on photoelectric effect. Find (a) the ratio h/e and (b) the work function.

Answer 1

We have to take two cases.
$$\begin{gathered} Case\left(\mathrm{I}\right) \\ \mathrm{When}\mathrm{stopping}\mathrm{potential},V_0=1.656\mathrm{volts} \\ \mathrm{Frequency},v=5\times {10}^{14}\mathrm{Hz} \\ Case\left(\mathrm{II}\right)\mathrm{When}\mathrm{stopping}\mathrm{potential},V_0=0 \\ \mathrm{Frequency},v=1\times {10}^{14}\mathrm{Hz} \\ \left(b\right)\mathrm{From}\mathrm{Einstein}\text{'s}\mathrm{equation}, \\ e{V}_0=hv-W_0 \\ \mathrm{On}\mathrm{substituting}\mathrm{the}\mathrm{values}\mathrm{of}\mathrm{case}\left(1\right)\mathrm{and}\mathrm{case}\left(2\right),\mathrm{we}\text{get:} \\ 1.656\mathrm{e}=h\times 5\times {10}^{14}-W_0...\left(1\right) \\ 0=5\times h\times 1\times {10}^{14}-5\times W_0...\left(2\right) \\ \mathrm{Subtracting}\mathrm{equation}\left(2\right)\mathrm{from}\left(1\right),\mathrm{we}\mathrm{get}: \\ {\mathrm{W}}_0=\frac{1.656}{4}\mathrm{eV} \\ =0.414\mathrm{eV} \end{gathered}$$
(a) Putting the value of W₀ in equation (2), we get:
$$\begin{gathered} \Rightarrow 5{\mathrm{W}}_0=5h\times {10}^{14} \\ \Rightarrow 5\times 0.414=5\times h\times {10}^{14} \\ \Rightarrow h=4.414\times {10}^{-15}\mathrm{eVs} \\ \mathrm{Or} \\ \frac{h}{e}=4.414\times {10}^{-15}\mathrm{Vs} \end{gathered}$$