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Let the energy of an emitted photoelectron be $\mathrm{E}$ and thewave-length of incident light be $\lambda$. What will be the change in $\mathrm{E}$ if $\lambda$ is doubled?
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The correct answer is:
$\mathrm{E} / 2$
We have $\mathrm{h} v=\mathrm{W}_{\mathrm{o}}+\mathrm{E}$, where $\mathrm{E}$ is the energy of emitted photoelectron
$\Rightarrow \frac{\mathrm{hc}}{\lambda}=\mathrm{W}_{\mathrm{o}}+\mathrm{E}$
As he and $\mathrm{W}_{\mathrm{o}}$ are constant,
$\mathrm{E} \propto \frac{1}{\lambda}$
Therefore, as $\lambda$ is doubled, $E$ will become half.
$\Rightarrow \frac{\mathrm{hc}}{\lambda}=\mathrm{W}_{\mathrm{o}}+\mathrm{E}$
As he and $\mathrm{W}_{\mathrm{o}}$ are constant,
$\mathrm{E} \propto \frac{1}{\lambda}$
Therefore, as $\lambda$ is doubled, $E$ will become half.
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