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The potential difference that must be applied to stop the fastest photoelectrons emitted by a nickel surface, having work function $5.01 \mathrm{eV}$, when ultraviolet light of $200 \mathrm{~nm}$ falls on it, must be
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The correct answer is:
$1.2 \mathrm{~V}$
Energy of incident light $\mathrm{E}(\mathrm{eV})=\frac{12375}{2000}$ $=6.2 \mathrm{eV}(200 \mathrm{~nm}=2000 Å)$
According to the relation $\mathrm{E}=\mathrm{W}_0+\mathrm{eV}_0$
$$
\begin{aligned}
\Rightarrow \quad \mathrm{V}_0 & =\frac{\mathrm{E}-\mathrm{W}_0}{\mathrm{e}} \\
& =\frac{(6.2-5.01) \mathrm{e}}{\mathrm{e}} \\
& =1.2 \mathrm{~V}
\end{aligned}
$$
According to the relation $\mathrm{E}=\mathrm{W}_0+\mathrm{eV}_0$
$$
\begin{aligned}
\Rightarrow \quad \mathrm{V}_0 & =\frac{\mathrm{E}-\mathrm{W}_0}{\mathrm{e}} \\
& =\frac{(6.2-5.01) \mathrm{e}}{\mathrm{e}} \\
& =1.2 \mathrm{~V}
\end{aligned}
$$
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