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?

Answer:

\[8\times {{10}^{6}}m/s\] \[K{{E}_{\max }}=h\text{v}-{{\phi }_{0}}\] 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)]                         \[V_{\max }^{2}=4\times {{(4\times {{10}^{6}})}^{2}}\]                         \[{{V}_{\max }}=\sqrt{4\times {{(4\times {{10}^{6}})}^{2}}}\]                                     \[=2\times 4\times {{10}^{6}}\]                                     \[=8\times {{10}^{6}}\text{m}{{\text{s}}^{-1}}\]