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A photon of energy $4 \mathrm{eV}$ is incident on a metal surface whose work function is $2 \mathrm{eV}$. The minimum reverse potential to be applied for stopping the emission of electrons is
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Verified Answer
The correct answer is:
$2 \mathrm{~V}$
When stopping potential is applied no electron will reach the cathode and the current will becomes zero.
$$
\therefore \quad e V_0=\frac{1}{2} m v_{\max }^2
$$
Also, $\frac{1}{2} m v_{\max }^2=h v-$ work function
$$
\begin{gathered}
=4 \mathrm{eV}-2 \mathrm{eV}=2 \mathrm{eV} \\
\Rightarrow \quad e V_0=2 \mathrm{eV} \Rightarrow \text { stopping potential }=2 \mathrm{~V}
\end{gathered}
$$
$$
\therefore \quad e V_0=\frac{1}{2} m v_{\max }^2
$$
Also, $\frac{1}{2} m v_{\max }^2=h v-$ work function
$$
\begin{gathered}
=4 \mathrm{eV}-2 \mathrm{eV}=2 \mathrm{eV} \\
\Rightarrow \quad e V_0=2 \mathrm{eV} \Rightarrow \text { stopping potential }=2 \mathrm{~V}
\end{gathered}
$$
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