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