The work function of metal ' \(A\) ' and ' \(B\) ' are in the ratio \(1: 2\). If light of frequency ' \(f\) ' and ' \(2 f\) ' is incident on surface ' \(A\) ' and ' \(B\) ' respectively, then the ratio of kinetic energies of emitted photo electrons is

\(\begin{aligned} & \text { For } \mathrm{A}, \mathrm{E}_{\mathrm{A}_{\max }}=\mathrm{h} v-\phi_{\mathrm{A}} \\ & \text { For } \mathrm{B}, \mathrm{E}_{\mathrm{B}_{\max }}=\mathrm{h}(2 v)-\phi_{\mathrm{B}} \\ & \frac{\mathrm{E}_{\mathrm{A}_{\max }}}{\mathrm{E}_{\mathrm{B}_{\max }}}=\frac{\mathrm{h} v-\phi_{\mathrm{A}}}{2 \mathrm{~h} v-\phi_{\mathrm{B}}} \\ & \text { As } \frac{\phi_{\mathrm{A}}}{\phi_{\mathrm{B}}}=\frac{1}{2} \Rightarrow \phi_{\mathrm{B}}=2 \phi_{\mathrm{A}}\end{aligned}\)
\(\begin{aligned} \therefore \quad \frac{\mathrm{E}_{\mathrm{A}_{\max }}}{\mathrm{E}_{\mathrm{B}_{\max }}} & =\frac{\frac{\mathrm{h} v-\phi_{\mathrm{A}}}{\phi_{\mathrm{A}}}}{\frac{2 \mathrm{~h} v-\phi_{\mathrm{B}}}{\phi_{\mathrm{A}}}}=\frac{\frac{\mathrm{h} v}{\phi_{\dot{A}}}-1}{\frac{2 h v}{\phi_{\mathrm{A}}}-2} \\ & =\frac{\mathrm{hv}-\phi_{\mathrm{A}}}{\phi_{\mathrm{A}}} \times \frac{\phi_{\mathrm{A}}}{2\left(\mathrm{~h} v-\phi_{\mathrm{A}}\right)}=\frac{1}{2}\end{aligned}\)