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Photoelectric emission is observed from a metallic surface for frequencies $v_1$ and $v_2$ of the incident light rays $\left(v_1>v_2\right)$. If the maximum values of kinetic energy of the photoelectrons emitted in the two cases are in the ratio of $1: k$, then the threshold frequency of the metallic surface is
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The correct answer is:
$\frac{k v_1-v_2}{k-1}$
Maximum kinetic energy of photoelectrons
$(\mathrm{KE})_{\max }=h v-h v_0$
$v=$ frequency of incident radiation
$v_0=$ threshold frequency
Ist Case $\quad\left(\mathrm{KE}_{\max }\right)_1=h v_1-h v_0$
IInd Case $\left(\mathrm{KF}_{\max }\right)_2=h v_2-h v_0$
$\begin{aligned} \frac{\left(\mathrm{KE}_{\max }\right)_1}{\left(\mathrm{KE}_{\max }\right)_2} & =\frac{h\left(v_1-v_0\right)}{h\left(v_2-v_0\right)} \\ \frac{1}{k} & =\frac{v_1-v_0}{v_2-v_0} \\ v_2-v_0 & =k v_1-k v_0 \\ k v_0-v_0 & =k v_1-v_2 \\ v_0 & =\frac{k v_1-v_2}{k-1}\end{aligned}$
$(\mathrm{KE})_{\max }=h v-h v_0$
$v=$ frequency of incident radiation
$v_0=$ threshold frequency
Ist Case $\quad\left(\mathrm{KE}_{\max }\right)_1=h v_1-h v_0$
IInd Case $\left(\mathrm{KF}_{\max }\right)_2=h v_2-h v_0$
$\begin{aligned} \frac{\left(\mathrm{KE}_{\max }\right)_1}{\left(\mathrm{KE}_{\max }\right)_2} & =\frac{h\left(v_1-v_0\right)}{h\left(v_2-v_0\right)} \\ \frac{1}{k} & =\frac{v_1-v_0}{v_2-v_0} \\ v_2-v_0 & =k v_1-k v_0 \\ k v_0-v_0 & =k v_1-v_2 \\ v_0 & =\frac{k v_1-v_2}{k-1}\end{aligned}$
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