Question

A silicon crystal is doped with a group III element, the electron concentration falls below intrinsic concentration by a factor of \(10^6\), so the concentration of impurity present is:

• A. \( 1.5 \times 10^{16} \, \text{cm}^{-3} \)
• B. \( 1.5 \times 10^{10} \, \text{cm}^{-3} \)
• C. \( 1.5 \times 10^{14} \, \text{cm}^{-3} \)
• D. \( 2.25 \times 10^{14} \, \text{cm}^{-3} \)

Hint:

When doping a semiconductor, the impurity concentration is often related to the intrinsic concentration by a factor.

Answer 1

The Correct Option is C

Given an intrinsic concentration of \( n_i = 1.5 \times 10^{10} \, \text{cm}^{-3} \), the impurity concentration is calculated as follows:\[n_{\text{imp}} = \frac{n_i}{10^6} = 1.5 \times 10^{14} \, \text{cm}^{-3}\]
Final Answer: \[\boxed{1.5 \times 10^{14} \, \text{cm}^{-3}}\]