From a metallic surface photoelectric emission is observed for frequencies \(v_1\) and \(v_2\left(v_1>v_2\right)\) of the incident light. The maximum values of the kinetic energy of the photoelectrons emitted in the two cases are in the ratio \(1: \mathrm{x}\). Hence the threshold frequency of the metallic surface is

Using Einstein's photoelectric equation,
\(\begin{aligned}
& \mathrm{E}_{\mathrm{k}}=\mathrm{hv}-\phi_0 \\
& \mathrm{E}_{\mathrm{k}}=\mathrm{hv}-\mathrm{h} v_0 \\
& \therefore \quad \mathrm{E}_{\mathrm{K}_1}=\mathrm{h}\left(\mathrm{v}_1-\mathrm{v}_0\right) \text { and } \mathrm{E}_{\mathrm{K}_2}=\mathrm{h}\left(\mathrm{v}_2-\mathrm{v}_0\right) \\
& \text { Given } \frac{E_{K_1}}{E_{K_2}}=\frac{1}{x} \\
& \Rightarrow \frac{v_1-v_0}{v_2-v_0}=\frac{1}{x} \\
& \left(v_1-v_0\right) x=v_2-v_0 \\
& v_1 x-v_0 x=v_2-v_0 \\
& \therefore \quad v_1 x-v_2=v_0 x-v_0 \\
& v_1 x-v_2=v_0(x-1) \\
& \therefore \quad v_0=\frac{v_1 x-v_2}{x-1} \\
&
\end{aligned}\)