• # question_answer A photosensitive metallic surface has work function $h{{v}_{0}}.$ If photons of energy $2\,h{{v}_{0}}$ fall on this surface, the electrons come out with a maximum velocity of $4\times {{10}^{6}}m/s.$ When the photon of energy is increased to $5\,h{{v}_{0}},$ what will be the maximum velocity of photo-electron?

$8\times {{10}^{6}}m/s$ $K{{E}_{\max }}=h\text{v}-{{\phi }_{0}}$ Where, ${{\phi }_{0}}$ is the work function. Given, ${{\phi }_{0}}=h{{\text{v}}_{\text{0}}}$             $\frac{1}{2}m\,{{(4\times {{10}^{6}})}^{2}}=2h{{\text{v}}_{\text{0}}}-h{{\text{v}}_{\text{0}}}$ i.e.        $\frac{1}{2}m\,{{(4\times {{10}^{6}})}^{2}}=h{{\text{v}}_{\text{0}}}$                   ? (i)             $\frac{1}{2}m\,{{({{V}_{max}})}^{2}}=5h{{\text{v}}_{\text{0}}}-h{{\text{v}}_{\text{0}}}=4h{{\text{v}}_{\text{0}}}$ i.e.        $\frac{1}{2}m\,{{({{V}_{max}})}^{2}}=4\times \frac{1}{2}m\,{{(4\times {{10}^{6}})}^{2}}$ [from Eq. (i)]                         $V_{\max }^{2}=4\times {{(4\times {{10}^{6}})}^{2}}$                         ${{V}_{\max }}=\sqrt{4\times {{(4\times {{10}^{6}})}^{2}}}$                                     $=2\times 4\times {{10}^{6}}$                                     $=8\times {{10}^{6}}\text{m}{{\text{s}}^{-1}}$