The mean free path of electrons in a metal is 4 × 10^ -8 ~m . The electric field which can give on an average 2 eV energy to an electron in the metal will be in units of V / m :

Correct answer: A. 5 × 10^7 aligned qV& =2 eV 1.6 ×10^ -19 V& =2 ×1.6 ×10^ -19 V V& =2 V E& = V d E& = 2 V 4 × 10^ -8 =5 ×10^7 aligned

NEET 2009

NEET 2009 · Chemistry · States of Matter

The mean free path of electrons in a metal is 4 × 10^ -8 ~m . The...

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 units of

\mathrm{V} / \mathrm{m}

A. $5 \times 10^7$

\begin{aligned}\mathrm{qV}& =2 \mathrm{eV}\\ \Rightarrow \quad 1.6 \times10^{-19} \mathrm{ V}& =2 \times1.6 \times10^{-19} \mathrm{ V}\\ \Rightarrow \quad \mathrm{V}& =2 \mathrm{ V}\\ \mathrm{E}& =\frac{\mathrm{V}}{\mathrm{d}}\\ \Rightarrow \quad \mathrm{E}& =\frac{2 \mathrm{ V}}{4 \times 10^{-8}}=5 \times10^7 \end{aligned}