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,$n_e \approx N_D = 4.8 	imes 10^{20} / m^3$According to

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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,$n_e \approx N_D = 4.8 	imes 10^{20} / m^3$According to the law of mass action for semiconductors,the product of electron and hole concentrations is equal to the square of the intrinsic carrier concentration:$n_e 	imes n_h = n_i^2$Substituting the given values:$4.8 	imes 10^{20} 	imes n_h = (1.6 	imes 10^{16})^2$$4.8 	imes 10^{20} 	imes n_h = 2.56 	imes 10^{32}$$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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