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Physics Question on Semiconductor electronics: materials, devices and simple circuits

The number density of electrons and holes in pure silicon at 27∘C27 ^{\circ}C are equal and its value is 2.0×1016 m−32.0 \times 10^{16}\, m^{-3}. On doping with indium the hole density increases to 4.5×1022 m−34.5 \times 10^{22}\, m^{-3}, the electron density in doped silicon is

A

10×109 m−310 \times 10^9\, m^{-3}

B

8.89×109 m−38 .89 \times 10^9\,m^{-3}

C

11×109 m−311 \times 10^9 \,m^{-3}

D

16.78×109 m−316.78 \times 10^9\, m^{-3}

Answer

8.89×109 m−38 .89 \times 10^9\,m^{-3}

Explanation

Solution

Using, ne=ni2n_e = n_i^2 Here, ni=2×1016 m−3n_i = 2\times 10^{16}\,m^{-3}, ρ=4.5×1022 m−3\rho = 4.5 \times 10^{22}\,m^{-3} ∴n=ni2ρ\therefore n = \frac{n_i^2}{\rho} =(2×1016)24.5×1022 = \frac{(2\times 10^{16})^2}{4.5 \times 10^{22}} =8.89×109 m−3 = 8.89 \times 10^9 \,m^{-3}