In pure silicon,the electron-hole concentrati | vedclass

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(A) In an intrinsic semiconductor,the initial electron and hole concentration is $n_i = 7 	imes 10^{15} \ m^{-3}$.The total initial charge carrier co

Vedclass · Accepted Answer

$1.8 	imes 10^5$

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(A) In an intrinsic semiconductor,the initial electron and hole concentration is $n_i = 7 	imes 10^{15} \ m^{-3}$.The total initial charge carrier concentration is $n_{initial} = n_e + n_h = 2n_i = 14 	imes 10^{15} \ m^{-3}$.The number density of donor impurity atoms $(N_D)$ is given by the ratio of silicon atoms to impurity atoms:$N_D = \frac{5 	imes 10^{28}}{10^7} = 5 	imes 10^{21} \ m^{-3}$.Since only half of the impurity atoms contribute electrons to the conduction band,the increase in electron concentration is:$n_e = \frac{N_D}{2} = 2.5 	imes 10^{21} \ m^{-3}$.After doping,the final charge carrier concentration is $n_{final} = n_e + n_h \approx n_e = 2.5 	imes 10^{21} \ m^{-3}$ (since $n_e \gg n_h$).The factor by which the charge carriers increase is given by:$	ext{Factor} = \frac{n_{final} - n_{initial}}{n_{initial}} \approx \frac{2.5 	imes 10^{21}}{14 	imes 10^{15}} \approx 1.78 	imes 10^5 \approx 1.8 	imes 10^5$.

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