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Pure Si\text{Si} at 500 K500\text{ K} has an equal number of electron (nen_e) and hole (nhn_h) concentration of 1.5×1016 m−31.5 \times 10^{16} \text{ m}^{-3}. Doping by indium increases the hole concentration nhn_h to 4.5×1022 m−34.5 \times 10^{22} \text{ m}^{-3}. The doped semiconductor is of:

A

n-type with electron concentration ne=5×1022 m−3n_e = 5 \times 10^{22} \text{ m}^{-3}

B

p-type with electron concentration ne=2.5×1023 m−3n_e = 2.5 \times 10^{23} \text{ m}^{-3}

C

n-type with electron concentration ne=2.5×1010 m−3n_e = 2.5 \times 10^{10} \text{ m}^{-3}

D

p-type with electron concentration ne=5×109 m−3n_e = 5 \times 10^9 \text{ m}^{-3}

Step-by-Step Solution

  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: nenh=ni2n_e n_h = n_i^2.
  3. Calculate Electron Concentration (nen_e): Given the intrinsic concentration ni=1.5×1016 m−3n_i = 1.5 \times 10^{16} \text{ m}^{-3} and the new hole concentration nh=4.5×1022 m−3n_h = 4.5 \times 10^{22} \text{ m}^{-3}, we can solve for nen_e: ne=ni2nhn_e = \frac{n_i^2}{n_h} ne=(1.5×1016)24.5×1022n_e = \frac{(1.5 \times 10^{16})^2}{4.5 \times 10^{22}} ne=2.25×10324.5×1022n_e = \frac{2.25 \times 10^{32}}{4.5 \times 10^{22}} ne=0.5×1010 m−3=5×109 m−3n_e = 0.5 \times 10^{10} \text{ m}^{-3} = 5 \times 10^9 \text{ m}^{-3}
  4. Conclusion: The semiconductor is p-type with an electron concentration of 5×109 m−35 \times 10^9 \text{ m}^{-3}.
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