In an $NPN$ transistor, the collector current is $28 \text{ mA}$. If $80\%$ of the electrons emitted by the emitter reach the collector, what is the base current in $\text{mA}$ (in $\text{ mA}$)?

In an $NPN$ transistor,$10^{10}$ electrons enter the emitter in $10^{-6} \ s$ and $2 \%$ of electrons recombine with holes in the base. The current ratios $\alpha$ and $\beta$ of the transistor are respectively (nearly):

In a transistor,the base is very lightly doped as compared to the emitter because by doing so:

The correct relation between $\alpha$ (ratio of collector current to emitter current) and $\beta$ (ratio of collector current to base current) of a transistor is :

According to a graph drawn between the input and output voltages of a transistor connected in common emitter configuration, the region in which the transistor acts as a switch is

In the case of a common emitter transistor amplifier,the ratio of collector current to emitter current is $0.96$. Then,the current gain $(\beta)$ of the amplifier is: