The work function of metal ' $A$ ' and ' $B$ ' are in the ratio $1: 2$. If light of frequency ' $f$ ' and '…

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
  1. $1: 1$
  2. $1: 2$
  3. $1: 3$
  4. $1: 4$

Solution

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

Asked in: MHT CET 2024 (02 May Shift 1)

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