Pure $ ext{Si}$ at $500 ext{ K}$ has an equal number of electron ($n_e$) and hole ($n_h$) concentration of $

Question

Pure $	ext{Si}$ at $500	ext{ K}$ has an equal number of electron ($n_e$) and hole ($n_h$) concentration of $1.5 	imes 10^{16} 	ext{ m}^{-3}$. Doping by indium increases the hole concentration $n_h$ to $4.5 	imes 10^{22} 	ext{ m}^{-3}$. The doped semiconductor is of:

Accepted Answer

1. **Identify the Type of Semiconductor:** Indium is a Group 13 (trivalent) impurity. Doping an intrinsic silicon crystal with a trivalent impurity creates excess holes, making it a p-type semiconductor.
2. **Apply Mass Action Law:** For any semiconductor in thermal equilibrium, the product of majority and minority charge carrier concentrations is equal to the square of the intrinsic carrier concentration: $n_e n_h = n_i^2$.
3. **Calculate Electron Concentration ($n_e$):** Given the intrinsic concentration $n_i = 1.5 	imes 10^{16} 	ext{ m}^{-3}$ and the new hole concentration $n_h = 4.5 	imes 10^{22} 	ext{ m}^{-3}$, we can solve for $n_e$:
   $$n_e = \frac{n_i^2}{n_h}$$
   $$n_e = \frac{(1.5 	imes 10^{16})^2}{4.5 	imes 10^{22}}$$
   $$n_e = \frac{2.25 	imes 10^{32}}{4.5 	imes 10^{22}}$$
   $$n_e = 0.5 	imes 10^{10} 	ext{ m}^{-3} = 5 	imes 10^9 	ext{ m}^{-3}$$
4. **Conclusion:** The semiconductor is p-type with an electron concentration of $5 	imes 10^9 	ext{ m}^{-3}$.

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