In a transistor,when the emitter current changes by $9.85 \,mA$,the collector current changes by $9.5 \,mA$. Then the base current is (in $\,mA$)

In the study of a transistor as an amplifier,if $\alpha = \frac{I_{C}}{I_{E}}$ and $\beta = \frac{I_{C}}{I_{B}}$,where $I_{C}$,$I_{B}$,and $I_{E}$ are the collector,base,and emitter currents respectively,then:

In a study of a transistor as an amplifier,the ratio of collector current to emitter current is $0.98$. If the collector current is $3 \text{ mA}$,then the base current will be approximately:

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):

If a transistor having $\alpha = 0.9$ is used in $CE$ configuration, then for a change of $0.4 \text{ mA}$ in base current, what will be the change in collector current (in $\text{ mA}$)?

$A$ transistor is operating in the active mode. $v_1$ is the potential barrier across the base-emitter junction and $v_2$ is the potential barrier across the collector-base junction. $b_1$ is the width of the depletion layer of the base-emitter junction and $b_2$ is the width of the collector-base junction.