Q10 / 120 · VITEEE 2015 · Chemistry

A certain metal when irradiated by light $(r=3.2 \times$ $10^{16} \mathrm{~Hz}$ ) emits photoelectrons with twice kinetic energy as did photoelectrons when the same metal is irradiated by light $\left(r=2.0 \times 10^{16} \mathrm{~Hz}\right)$. The $v_{0}$ of metal is

  • A
    $1.2 \times 10^{14} \mathrm{~Hz}$
  • Correct answer
    B
    $8 \times 10^{15} \mathrm{~Hz}$
  • C
    $1.2 \times 10^{16} \mathrm{~Hz}$
  • D
    $4 \times 10^{12} \mathrm{~Hz}$

Solution

$(\mathrm{KE})_{1}=\mathrm{hv}_{1}-\mathrm{hv}_{0}$ $(\mathrm{KE})_{2}=\mathrm{hv}_{2}-\mathrm{hv}_{0}$ $\begin{array}{ll}\text { As, } & (\mathrm{KE})_{1}^{2}=2 \times(\mathrm{KE})_{2} \\ \therefore & \mathrm{hv}_{1}-\mathrm{hv}_{0}=2\left(\mathrm{hv}_{2}-\mathrm{hv}_{0}\right) \\ \text { or, } & \mathrm{hv}_{0}=2 \mathrm{~h} \mathrm{v}_{2}-\mathrm{hv}_{1} \\ \text { or, } & \mathrm{v}_{0}=2 \mathrm{v}_{2}-\mathrm{v}_{1} \\ & =2 \times\left(2 \times 10^{16}\right)-\left(3.2 \times 10^{16}\right) \\ & =0.8 \times 10^{16} \mathrm{~Hz}=8 \times 10^{15} \mathrm{~Hz}\end{array}$

Hence, the correct answer is option: 'B'.

Relevant Questions