In an intrinsic semiconductor at room temperature, the number of electrons and holes are equal. If the number of electrons is 10 to the power 16 per cubic meter, and we dope it with indium to create a p-type semiconductor such that the hole concentration becomes 10 to the power 20 per cubic meter, the new electron concentration is

A. 10^12 , m^-3 — According to the law of mass action, the product of electron and hole concentrations is equal to the square of the intrinsic carrier concentration, n_e n_h = n_i^2. Thus, n_e = (10^16)^2 / 10^20 = 10^32 / 10^20 = 10^12 , m^-3.

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$In an intrinsic semiconductor, n_{i} = 1 0^{16} m^{- 3} . After doping with indium, the hole concentration becomes n_{h} = 1 0^{20} m^{- 3} . The new electron concentration n_{e} is$

A. $1 0^{12} m^{- 3}$
B. $1 0^{10} m^{- 3}$
C. $1 0^{8} m^{- 3}$
D. $1 0^{14} m^{- 3}$

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Correct answer: A — $1 0^{12} m^{- 3}$

Solution

Using the mass action law n_{e} n_{h} = n_{i}^{2}, we find n_{e} = \frac{( 1 0 ^{16} ) ^{2}}{1 0 ^{20}} = \frac{1 0 ^{32}}{1 0 ^{20}} = 1 0^{12} m^{- 3} .

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In an intrinsic semiconductor, ni​=1016m−3. After doping with indium, the hole concentration becomes nh​=1020m−3. The new electron concentration ne​ is

Solution

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More Semiconductor Devices practice questions

$In an intrinsic semiconductor, n_{i} = 1 0^{16} m^{- 3} . After doping with indium, the hole concentration becomes n_{h} = 1 0^{20} m^{- 3} . The new electron concentration n_{e} is$