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(C) Given: Collector current $I_C = 8 \ mA$. Since $80 \%$ of the electrons from the emitter reach the collector,the current gain $\alpha$ is given by

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$\alpha=0.8, \beta=4.0$

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(C) Given: Collector current $I_C = 8 \ mA$. Since $80 \%$ of the electrons from the emitter reach the collector,the current gain $\alpha$ is given by the ratio of collector current to emitter current,which is $0.8$. Thus,$\alpha = \frac{I_C}{I_E} = 0.8$. We can find the emitter current $I_E$ as $I_E = \frac{I_C}{\alpha} = \frac{8 \ mA}{0.8} = 10 \ mA$. The base current $I_B$ is given by $I_E - I_C = 10 \ mA - 8 \ mA = 2 \ mA$. The current gain $\beta$ is calculated as $\beta = \frac{\alpha}{1 - \alpha} = \frac{0.8}{1 - 0.8} = \frac{0.8}{0.2} = 4.0$. Therefore,$\alpha = 0.8$ and $\beta = 4.0$.

In an $n-p-n$ transistor amplifier,the collector current is $8 \ mA$. If $80 \%$ of the electrons from the emitter reach the collector,then ($\alpha$ and $\beta$ are current ratios):

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