Question

The concentration of hole-electron pairs in pure silicon at 300 K is \(7 \times 10^{15} \, \text{m}^{-3}\). Antimony is doped into silicon in proportions of \(1 \, \text{atom per} \, 10^7 \, \text{atoms}\), and half of the impurity atoms contribute electrons. The factor by which the number of charge carriers increases due to doping is:

• A. 1
• B. 1
• C. 1
• D. \(8 \times 10^2\)

Hint:

When calculating carrier concentration in doped semiconductors, always consider the proportion of dopants and their ionization rates.

Answer 1

The Correct Option is C

The number of silicon atoms per cubic meter is $5 \times 10^{28}$. The number of antimony atoms is:
\[N_{Sb} = \frac{5 \times 10^{28}}{10^7} = 5 \times 10^{21} \text{ m}^{-3}\]
Since half of the Sb atoms contribute, the number of free electrons due to doping is:
\[n_e = \frac{5 \times 10^{21}}{2} = 2.5 \times 10^{21} \text{ m}^{-3}\]
The factor of increase in charge carriers is:
\[\text{Factor} = \frac{n_e}{n_i} = \frac{2.5 \times 10^{21}}{7 \times 10^{15}} \approx 8 \times 10^5\]