Question

In a given semiconductor, the ratio of the number density of electron to number density of hole is 2 : 1. If $ \frac{1}{7} $th of the total current is due to the hole and the remaining is due to the electrons, the ratio of the drift velocity of holes to the drift velocity of electrons is :

• A. \( \frac{2}{3} \)
• B. \( \frac{1}{3} \)
• C. \( \frac{3}{2} \)
• D. \( \frac{1}{7} \)

Hint:

When solving questions involving drift velocities, remember that the current due to holes and electrons depends on their respective charge densities and drift velocities. The total current is the sum of both contributions.

Answer 1

The Correct Option is D

We are given the following information: - The ratio of the number density of electrons to holes is \( \frac{n_e}{n_h} = 2 : 1 \), i.e., \( n_e = 2n_h \). - The current due to holes is \( \frac{1}{7} \)th of the total current, and the remaining current is due to electrons. The total current \( I \) can be expressed as: \[ I = I_e + I_h \] Where: \( I_e \) is the current due to electrons, and \( I_h \) is the current due to holes. Now, using the relationship for current \( I = nqAv_d \), where \( n \) is the number density, \( q \) is the charge, \( A \) is the cross-sectional area, and \( v_d \) is the drift velocity, we have: \[ I_e = n_e q A v_{d_e} \quad \text{and} \quad I_h = n_h q A v_{d_h} \] Since the ratio of the current is given as \( \frac{I_h}{I_e} = \frac{1}{7} \), we can write: \[ \frac{n_h v_{d_h}}{n_e v_{d_e}} = \frac{1}{7} \] Substituting \( n_e = 2n_h \), we get: \[ \frac{v_{d_h}}{v_{d_e}} = \frac{1}{7} \times \frac{n_e}{n_h} = \frac{1}{7} \times 2 = \frac{2}{7} \] Thus, the ratio of the drift velocity of holes to the drift velocity of electrons is \( \frac{v_{d_h}}{v_{d_e}} = \frac{1}{7} \).