Question

The concentration of electrons in an intrinsic semiconductor is \( 6 \times 10^{15} \, m^{-3} \). On doping with an impurity, the electron concentration increases to \( 4 \times 10^{22} \, m^{-3} \). In thermal equilibrium, the concentration of the holes in the doped semiconductor is:

• A. \( 18 \times 10^{-8} \, m^{-3} \)
• B. \( 1.5 \times 10^{-7} \, m^{-3} \)
• C. \( 9 \times 10^{8} \, m^{-3} \)
• D. \( 2 \times 10^{7} \, m^{-3} \)

Hint:

In semiconductors, doping increases the electron concentration, reducing the hole concentration while maintaining \( n_i^2 = n_e n_h \).

Answer 1

The Correct Option is C

For semiconductors, the intrinsic carrier concentration satisfies: \[ n_i^2 = n_e n_h \] where: - \( n_i = 6 \times 10^{15} \, m^{-3} \), - \( n_e = 4 \times 10^{22} \, m^{-3} \). Solving for \( n_h \): \[ n_h = \frac{n_i^2}{n_e} \] \[ = \frac{(6 \times 10^{15})^2}{4 \times 10^{22}} \] \[ = \frac{36 \times 10^{30}}{4 \times 10^{22}} \] \[ = 9 \times 10^8 \, m^{-3} \]