In a semiconductor,the concentration of elect | vedclass

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(B) Given:Electron concentration $(n_e)$ = $8 	imes 10^{14} 	ext{ cm}^{-3}$Hole concentration $(n_h)$ = $5 	imes 10^{12} 	ext{ cm}^{-3}$Since $n_e

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$N$-type

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(B) Given:Electron concentration $(n_e)$ = $8 	imes 10^{14} 	ext{ cm}^{-3}$Hole concentration $(n_h)$ = $5 	imes 10^{12} 	ext{ cm}^{-3}$Since $n_e > n_h$,the number of free electrons is significantly greater than the number of holes.In a semiconductor,if the concentration of electrons is greater than the concentration of holes,it is classified as an $N$-type semiconductor.Therefore,the correct option is $B$.

In a semiconductor,the concentration of electrons is $8 \times 10^{14} \text{ cm}^{-3}$ and that of the holes is $5 \times 10^{12} \text{ cm}^{-3}$. The semiconductor is

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