Question

The mean free path of conduction electrons in copper is about $4\times {10}^{-8}\text{m}$. For a copper block, find the electric field which can give, on an average, $1\text{eV}$ energy to conduction electrons.

• A. $2.5\times {10}^7{\text{Vm}}^{-1}$
• B. $2.5\times {10}^{-7}{\text{Vm}}^{-1}$
• C. $5\times {10}^{-7}{\text{Vm}}^{-1}$
• D. $5\times {10}^7{\text{Vm}}^{-1}$

Answer 1

The correct option is A $2.5\times {10}^7{\text{Vm}}^{-1}$

Let the electric field Be $E$.

Then the force on an electron is $F=eE$.

As the electron moves through a distance $d$, the work done on it is,

$W=eEd$.

This is equal to the energy transferred to the electron.

As the electron travels an average distance of $4\times {10}^{-8}\text{m}$ before a collision, and to get $1\text{eV}$ energy from the electric field,

$W=eE(4\times {10}^{-8}\text{m})=1\text{eV}$

$\Rightarrow E=2.5\times {10}^7{\text{Vm}}^{-1}$

Hence, option $\left(A\right)$ is correct.