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Concepts Of Physics MCQ Edition [Volume 2]
Exam
Concepts Of Physics MCQ Edition [Volume 2]
Subject
Physics
Chapter
Semiconductors and Semiconductor Devices

Concepts Of Physics MCQ Edition [Volume 2] · Physics · Semiconductors and Semiconductor Devices

The product of the conduction electron concentration and the hole...

The product of the conduction electron concentration and the hole concentration remains independent of the quantity of any doped impurity. In pure germanium, the concentration of conduction electrons is 6×10196 \times 10^{19} per cubic metre. After a germanium sample is doped with phosphorus impurity, the conduction electron concentration rises to 2×10232 \times 10^{23} per cubic metre. Determine the hole concentration in the doped germanium.

Options

  1. A3.6×10163.6 \times 10^{16} per cubic metre
  2. B1.8×10161.8 \times 10^{16} per cubic metre
  3. C9.0×10159.0 \times 10^{15} per cubic metre
  4. D1.2×10161.2 \times 10^{16} per cubic metre

Correct answer

B. 1.8×10161.8 \times 10^{16} per cubic metre

Step-by-step solution

In an intrinsic semiconductor, the concentration of conduction electrons is equal to the concentration of holes, which is the intrinsic carrier concentration nin_i.

Given ni=6×1019 m−3n_i = 6 \times 10^{19} \text{ m}^{-3}.

According to the mass action law, the product of electron and hole concentrations remains constant at a given temperature:

nenh=ni2n_e n_h = n_i^2

After doping, the new electron concentration is ne=2×1023 m−3n_e = 2 \times 10^{23} \text{ m}^{-3}.

Substituting the values into the mass action law:

nh=ni2ne=(6×1019)22×1023n_h = \dfrac{n_i^2}{n_e} = \dfrac{(6 \times 10^{19})^2}{2 \times 10^{23}}

nh=36×10382×1023n_h = \dfrac{36 \times 10^{38}}{2 \times 10^{23}}

nh=18×1015=1.8×1016 m−3n_h = 18 \times 10^{15} = 1.8 \times 10^{16} \text{ m}^{-3}
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