The graph between the stopping potential (V₀) and $(1/\mathrm{\lambda })$ is shown in the figure. ${\mathrm{\varphi }}_1, {\mathrm{\varphi }}_2\text{ and }{\mathrm{\varphi }}_3$ are work functions. Then, which of the following is/are correct?

${\mathrm{\varphi }}_1 : {\mathrm{\varphi }}_2 : {\mathrm{\varphi }}_3={\mathrm{eV}}_{0_1} : {\mathrm{eV}}_{0_2} : {\mathrm{eV}}_{0_3}$

${\mathrm{V}}_{0_1} : {\mathrm{V}}_{0_2} : {\mathrm{V}}_{0_3}=0.001 : 0.002 : 0.004=1 : 2 : 4$

Therefore, option (1) is correct.

By Einstein's photoelectric equation, $\frac{\mathrm{hc}}{\mathrm{\lambda }}-\mathrm{\varphi }={\mathrm{eV}}_0$

$\Rightarrow {\mathrm{V}}_0=\frac{\mathrm{hc}}{\mathrm{e\lambda }}-\frac{\mathrm{\varphi }}{\mathrm{e}}$                      …(i)

Comparing Eq. (i) by $\mathrm{y}=\mathrm{mx}+\mathrm{c},$ we get the slope of the line 

            $\mathrm{m}=\frac{\mathrm{hc}}{\mathrm{e}}=\tan \mathrm{\theta }$

$\Rightarrow$   Option (2) is correct.

From the graph it is clear that, $\frac{1}{{\mathrm{\lambda }}_{0_1}}=0.001{\mathrm{nm}}^{-1}$

$\Rightarrow {\mathrm{\lambda }}_{0_1}=\frac{1}{0.001}=1000\mathrm{nm}$

Also, $\frac{1}{{\mathrm{\lambda }}_{0_2}}=0.002{\mathrm{nm}}^{-1}\Rightarrow {\mathrm{\lambda }}_{0_2}=500\mathrm{nm}$

and ${\mathrm{\lambda }}_{0_3}=250\mathrm{nm}$

Ultraiolet colour light will have wavelength less than 400 nm. Therefore, this light will be able to show photoelectric effect on plates 1 and 2 and if the wave length is less than 250 nm , it will show photoelectric effect on plate 3 also .