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Photoelectric emission takes place from a certain metal at threshold frequency ' $v$ '. If the radiation of frequency ' $2 v$ ' is incident on the metal plate, the maximum velocity of the emitted photoelectron will be ( $\mathrm{m}=$ mass of electron, $\mathrm{h}$ = Planck's constant)
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Verified Answer
The correct answer is:
$\sqrt{\frac{2 \mathrm{hv}}{\mathrm{m}}}$
Einstein's photo electric equation
Concept: The maximum kinetic energy of photoelectron is written:
$\frac{1}{2} \mathrm{mv}_{\mathrm{m}}^2 \leq \mathrm{hv}-\theta$
The threshold frequency is given by
$\mathrm{hv}_{\mathrm{c}}=\theta$
Given, $\mathrm{v}_{\mathrm{c}}=\mathrm{v}$, now for radiation with frequency $2 \mathrm{v}$
$\begin{aligned}
& \frac{1}{2} \mathrm{mv}_{\mathrm{m}}^2 \leq \mathrm{h}(2 \mathrm{v}-\mathrm{h}(\mathrm{v})) \\
& \Rightarrow \mathrm{v}_{\mathrm{m}} \sqrt{\frac{2 \mathrm{hv}}{\mathrm{m}}}
\end{aligned}$
Option (A) is correct.
Concept: The maximum kinetic energy of photoelectron is written:
$\frac{1}{2} \mathrm{mv}_{\mathrm{m}}^2 \leq \mathrm{hv}-\theta$
The threshold frequency is given by
$\mathrm{hv}_{\mathrm{c}}=\theta$
Given, $\mathrm{v}_{\mathrm{c}}=\mathrm{v}$, now for radiation with frequency $2 \mathrm{v}$
$\begin{aligned}
& \frac{1}{2} \mathrm{mv}_{\mathrm{m}}^2 \leq \mathrm{h}(2 \mathrm{v}-\mathrm{h}(\mathrm{v})) \\
& \Rightarrow \mathrm{v}_{\mathrm{m}} \sqrt{\frac{2 \mathrm{hv}}{\mathrm{m}}}
\end{aligned}$
Option (A) is correct.
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