The mean free path of electrons in a metal is $4 \times 10^{-8} \mathrm{~m}$. The electric field which can…

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 $\mathrm{Vm}^{-1}$
  1. $8 \times 10^7$
  2. $5 \times 10^{-11}$
  3. $8 \times 10^{-11}$
  4. $5 \times 10^7$

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 $\begin{aligned} E & =\frac{2}{4 \times 10^{-8}} \\ & =0.5 \times 10^8 \\ & =5 \times 10^7 \mathrm{Vm}^{-1} \end{aligned}$

Asked in: NEET 2009 (Screening)

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