Topic: Electronic Devices
$1~cm^{3}$ of pure Germanium at $20^{\circ}C$ contains about $4.2\times10^{22}$ atoms, $2.5\times10^{13}$ free electrons and $2.5\times10^{13}$ holes. 0.0001% of Arsenic doping donates an extra $10^{17}$ free electrons in the same volume. The approximate number of holes in one $cm^{3}$ in the doped semiconductor under equilibrium condition is
- A. $6.25\times10^{9}$
- B. $2.5\times10^{9}$
- C. $10.5\times10^{9}$
- D. $10^{17}$
Answer: A. $6.25\times10^{9}$
Explanation: In pure Germanium, $n_{i}=n=p=2.5\times10^{13}$. When it is doped with arsenic, it donates an extra $10^{17}$ free electrons. So, number of electrons now are $n=(2.5\times10^{13}+10^{17})$. $n_{i}$ (intrinsic concentration) will remain same. Using mass action law, $n\cdot p=n_{i}^{2} \Rightarrow (2.5\times10^{13}+10^{17})\times p=(2.5\times10^{13})^{2} \Rightarrow p=6.25\times10^{9}$.