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The mean free path of electrons in a metal is $4 \times 10^{-8} \mathrm{~m}$. The electric field which can give on an average $2 \mathrm{eV}$ energy to an electron in the metal will be in unit of $\mathrm{Vm}^{-1}$
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Verified Answer
The correct answer is:
$5 \times 10^7$
$$
\begin{aligned}
\text { Energy } & =2 \mathrm{eV} \\
\mathrm{eV}_0 & =2 \mathrm{eV} \\
\Rightarrow \quad \mathrm{V}_0 & =2
\end{aligned}
$$
Now, electric field
$$
\begin{aligned}
E & =\frac{2}{4 \times 10^{-8}} \\
& =0.5 \times 10^8 \\
& =5 \times 10^7 \mathrm{Vm}^{-1}
\end{aligned}
$$
\begin{aligned}
\text { Energy } & =2 \mathrm{eV} \\
\mathrm{eV}_0 & =2 \mathrm{eV} \\
\Rightarrow \quad \mathrm{V}_0 & =2
\end{aligned}
$$
Now, electric field
$$
\begin{aligned}
E & =\frac{2}{4 \times 10^{-8}} \\
& =0.5 \times 10^8 \\
& =5 \times 10^7 \mathrm{Vm}^{-1}
\end{aligned}
$$
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