A crystal of intrinsic silicon at room temper | vedclass

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(D) Given:Intrinsic carrier concentration $n_i = 1.6 	imes 10^{16} / m^3$Donor concentration $N_D = 4.8 	imes 10^{20} / m^3$In an $n$-type semicondu

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$5.3 	imes 10^{11} / m^3$

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(D) Given:Intrinsic carrier concentration $n_i = 1.6 	imes 10^{16} / m^3$Donor concentration $N_D = 4.8 	imes 10^{20} / m^3$In an $n$-type semiconductor, the electron concentration $n_e \approx N_D = 4.8 	imes 10^{20} / m^3$.According to the law of mass action for semiconductors, $n_i^2 = n_e 	imes n_h$, where $n_h$ is the hole concentration.Substituting the values:$(1.6 	imes 10^{16})^2 = (4.8 	imes 10^{20}) 	imes n_h$$2.56 	imes 10^{32} = 4.8 	imes 10^{20} 	imes n_h$$n_h = \frac{2.56 	imes 10^{32}}{4.8 	imes 10^{20}}$$n_h = 0.533 	imes 10^{12} / m^3 = 5.33 	imes 10^{11} / m^3$.Thus, the concentration of holes is approximately $5.3 	imes 10^{11} / m^3$.

$A$ crystal of intrinsic silicon at room temperature has a carrier concentration of $1.6 \times 10^{16} / m^3$. If the donor concentration level is $4.8 \times 10^{20} / m^3$, then the concentration of holes in the semiconductor is

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