A pure Si crystal has 5 times 10^{28} atoms m | vedclass

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(A) The concentration of $Si$ atoms is $N = 5 	imes 10^{28} \, m^{-3}$.Since it is doped with $1 \, ppm$ (part per million) of pentavalent $As$ atoms

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$4.5 	imes 10^9 \, m^{-3}$

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(A) The concentration of $Si$ atoms is $N = 5 	imes 10^{28} \, m^{-3}$.Since it is doped with $1 \, ppm$ (part per million) of pentavalent $As$ atoms,the number of donor electrons $n_e$ is approximately equal to the concentration of $As$ atoms.$n_e \approx N_D = 1 	imes 10^{-6} 	imes 5 	imes 10^{28} = 5 	imes 10^{22} \, m^{-3}$.In thermal equilibrium,the product of electron and hole concentration is $n_e n_h = n_i^2$.Therefore,$n_h = \frac{n_i^2}{n_e} = \frac{(1.5 	imes 10^{16})^2}{5 	imes 10^{22}}$.$n_h = \frac{2.25 	imes 10^{32}}{5 	imes 10^{22}} = 0.45 	imes 10^{10} = 4.5 	imes 10^9 \, m^{-3}$.

$A$ pure $Si$ crystal has $5 \times 10^{28}$ atoms $m^{-3}$. It is doped with $1 \, ppm$ concentration of pentavalent $As$. The number of holes is (Take $n_i = 1.5 \times 10^{16} \, m^{-3}$)

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