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Question: The number density of electrons and holes in pure silicon at \(27^{o}C\) are equal and its value is ...

The number density of electrons and holes in pure silicon at 27oC27^{o}C are equal and its value is 2.0×1016m−32.0 \times 10^{16}m^{- 3} on doping with indium the hole density increases to 4.5×1022m−34.5 \times 10^{22}m^{- 3} the electrons density in doped silicon is.

A

10×109m−310 \times 10^{9}m^{- 3}

B

8.86×109m−38.86 \times 10^{9}m^{- 3}

C

11×109m−311 \times 10^{9}m^{- 3}

D

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

Answer

8.86×109m−38.86 \times 10^{9}m^{- 3}

Explanation

Solution

: using, ne=ni2n_{e} = n_{i}^{2}

Here, ni=2×1016m−3n_{i} = 2 \times 10^{16}m^{- 3}

ρ=4.5×1022m−3\rho = 4.5 \times 10^{22}m^{- 3}

∴n=ni2ρ=(2×1016)24.5×1022=8.89×109m−3\therefore n = \frac{n_{i}^{2}}{\rho} = \frac{(2 \times 10^{16})^{2}}{4.5 \times 10^{22}} = 8.89 \times 10^{9}m^{- 3}