In an npn transistor, 10^{10} electrons enter | vedclass

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(C) The number of electrons entering the emitter is $n_E = 10^{10}$.Since $4\%$ of the electrons are lost in the base, the number of electrons reachin

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$0.96$

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(C) The number of electrons entering the emitter is $n_E = 10^{10}$.Since $4\%$ of the electrons are lost in the base, the number of electrons reaching the collector is $n_C = n_E - (0.04 	imes n_E) = 0.96 	imes n_E$.The emitter current is given by $I_E = \frac{n_E 	imes e}{t}$, where $e$ is the electronic charge and $t$ is the time.The collector current is given by $I_C = \frac{n_C 	imes e}{t} = \frac{0.96 	imes n_E 	imes e}{t}$.The current transfer ratio $\alpha$ is defined as the ratio of collector current to emitter current:$\alpha = \frac{I_C}{I_E} = \frac{n_C}{n_E} = \frac{0.96 	imes n_E}{n_E} = 0.96$.

In an $npn$ transistor, $10^{10}$ electrons enter the emitter in $10^{-6} \; s$. If $4\%$ of the electrons are lost in the base, the current transfer ratio $(\alpha)$ will be:

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