Q129 / 200 · NEET 2009 (Screening) · Physics
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 $\mathrm{Vm}^{-1}$
- A$8 \times 10^7$
- B$5 \times 10^{-11}$
- C$8 \times 10^{-11}$
- Correct answerD$5 \times 10^7$
Solution
$$
\begin{aligned}
\text { Energy } & =2 \mathrm{eV} \\
\mathrm{eV}_0 & =2 \mathrm{eV} \\
\Rightarrow \quad \mathrm{V}_0 & =2
\end{aligned}
$$
Now, electric field
$$
\begin{aligned}
E & =\frac{2}{4 \times 10^{-8}} \\
& =0.5 \times 10^8 \\
& =5 \times 10^7 \mathrm{Vm}^{-1}
\end{aligned}
$$
Hence, the correct answer is option: 'D'.