The graph between the stopping potential $\left(V_0\right)$ and wave number $\left(\frac{1}{\lambda }\right)$ in nm^-1 for three metals is as shown in the figure. If $\varphi$ is the work function for three metals, then

 

According to Einstein photoelectric equation

$\frac{\mathrm{hc}}{\mathrm{\lambda }}=\mathrm{W}+{\mathrm{eV}}_0$

${\mathrm{V}}_0=\frac{\mathrm{hc}}{\mathrm{e\lambda }}-\frac{\mathrm{W}}{\mathrm{e}}=\frac{\mathrm{hc}}{\mathrm{e\lambda }}-\frac{\mathrm{hc}}{{\mathrm{\lambda }}_0\mathrm{e}}$

$\mathrm{where} {\mathrm{\lambda }}_0 \mathrm{is} \mathrm{the} \mathrm{threshold} \mathrm{wavelength}$

$\mathrm{slope} \mathrm{of} \mathrm{the} \mathrm{line} \frac{\mathrm{hc}}{\mathrm{e}}$

$\begin{array}{r}{\lambda }_{02}=1/0.002=500\mathrm{nm}\\ {\lambda }_{02}=1/0.004=250\mathrm{nm}\end{array}$

work function is proportional to $\frac{1}{{\mathrm{\lambda }}_0}$

$so work functions are in the ratio 1:2:4$