Question

An intrinsic semiconductor has equal concentrations of hole and electron which is equal to \(4 \times 10^{8}\,\text{m}^{-3}\). During its conversion to extrinsic semiconductor, concentration of hole increases to \(8 \times 10^{10}\,\text{m}^{-3}\). The new electron concentration is:

• A. \(4 \times 10^{8}\,\text{m}^{-3}\)
• B. \(2 \times 10^{8}\,\text{m}^{-3}\)
• C. \(2 \times 10^{6}\,\text{m}^{-3}\)
• D. \(4 \times 10^{10}\,\text{m}^{-3}\)

Hint:

Always use \(np = n_i^2\) (mass action law) in semiconductor problems. If one carrier increases, the other decreases.

Answer 1

The Correct Option is C