Rankers Physics

Relation between Current and Drift Velocity: Practice Problem & Solution

The mean free path of electrons in a metal is $4 \times 10^{-8} \text{ m}$. The electric field which can give on an average $2 \text{ eV}$ energy to an electron in the metal will be in units $\text{V/m}$. (2009)
$5 \times 10^{-11}$
$8 \times 10^{-11}$
$5 \times 10^7$
$8 \times 10^7$

Solution Explained:

To solve this problem, we apply the core principles of Relation between Current and Drift Velocity. Understanding the underlying formula is key to arriving at the correct answer below:

The energy gained by an electron is $E_k = e E \lambda$. Therefore, the electric field is $E = \frac{E_k}{e \lambda} = \frac{2 \text{ eV}}{e \times 4 \times 10^{-8} \text{ m}} = \frac{2}{4 \times 10^{-8}} \text{ V/m} = 0.5 \times 10^8 \text{ V/m} = 5 \times 10^7 \text{ V/m}$.

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