Topic: Electronic Devices and Circuits
Two initially identical samples A & B of pure germanium are doped with donors to concentrations of $1\times10^{20}$ and $3\times10^{20}$ respectively. If the hole concentration in A is $9\times10^{12}$ then the hole concentration in B at the same temperature will be
- A. $3\times10^{12}m^{-3}$
- B. $7\times10^{12}m^{-3}$
- C. $11\times10^{12}m^{-3}$
- D. $27\times10^{12}m^{-3}$
Answer: A. $3\times10^{12}m^{-3}$
Explanation: Using the mass-action law, $n\cdot p = n_i^2$. For sample A: $N_{DA} \times P_A = n_i^2 \Rightarrow 1\times10^{20} \times 9\times10^{12} = n_i^2 = 9\times10^{32}$. For sample B at the same temperature, $n_i^2$ remains constant. So, $N_{DB} \times P_B = n_i^2 \Rightarrow 3\times10^{20} \times P_B = 9\times10^{32} \Rightarrow P_B = \frac{9\times10^{32}}{3\times10^{20}} = 3\times10^{12}m^{-3}$.