• # question_answer A photosensitive metallic surface has work function, $h\,{{v}_{0}}.$If photons of energy$2h{{v}_{0}}$fall on this surface, the electrons come out with a maximum velocity of$4\times {{10}^{6}}\,m/s.$When the photon energy is increased to$5h{{v}_{0}},$then maximum velocity of photoelectrons will be A) $2\times {{10}^{6}}\,m/s$B) $2\times {{10}^{7}}\,m/s$C) $8\times {{10}^{5}}\,m/s$D) $8\times {{10}^{6}}\,m/s$

Einstein's photoelectric equation can be written as $\frac{1}{2}m{{v}^{2}}=hv-\phi$ $\Rightarrow$$\frac{1}{2}m\times {{(4\times {{10}^{6}})}^{2}}=2h{{v}_{0}}-h{{v}_{0}}$                     (i) and $\frac{1}{2}m\times {{v}^{2}}=5h{{v}_{0}}-h{{v}_{0}}$                       (ii) Dividing Eq. (ii) by (i), we get $\frac{{{v}^{2}}}{{{(4\times {{10}^{6}})}^{2}}}=\frac{4h{{v}_{0}}}{h{{v}_{0}}}$ $\Rightarrow$${{v}^{2}}=4\times 16\times {{10}^{12}}\Rightarrow \,\,{{v}^{2}}=64\times {{10}^{12}}$ $\Rightarrow$$v=8\times {{10}^{6}}\,m/s$