• # question_answer The threshold frequency for a certain photosensitive metal is ${{V}_{0}}.$When it is illuminated by light of frequency $v=2{{V}_{0}},$the maximum velocity of photoelectrons is ${{\upsilon }_{0}}.$What will be the maximum velocity of the photoelectrons when the same metal is illuminated by light of frequency$v=5{{v}_{0}}?$ A) $\sqrt{2}{{v}_{0}}$            B) $2{{v}_{0}}$C) $2\sqrt{2}{{v}_{0}}$                      D) $4{{v}_{0}}$

$\frac{1}{2}mu_{\max }^{2}=h(v-{{v}_{o}})$ For light of frequency $v=2{{v}_{0}},\frac{1}{2}mu_{o}^{2}=h(2{{v}_{o}}-{{v}_{o}})=h{{v}_{o}}$ For light of frequency, $v=5{{v}_{o}},$we have $\frac{1}{2}m{{u}^{2}}=h(5{{v}_{o}}-{{v}_{o}})=4\,h{{v}_{o}}$ $\therefore$${{u}^{2}}=4u_{o}^{2}$or $u=2{{u}_{o}}$ Hence, the correction option is [b].