$10^{10}$ electrons enter the emitter of a junction transistor in a time of $0.4 \mu s$. If $5 \%$ of the electrons are lost in the base, then the collector current is (in $\text{ mA}$)

Consider an $NPN$ transistor amplifier in common-emitter configuration. The current gain of the transistor is $100$. If the collector current changes by $1\, mA$,what will be the change in emitter current in $mA$?

Given below are two statements:
Statement $I:$ In a typical transistor,all three regions emitter,base and collector have same doping level.
Statement $II:$ In a transistor,collector is the thickest and base is the thinnest segment.
In the light of the above statements,choose the most appropriate answer from the options given below.

For a transistor,$\alpha_{dc}$ and $\beta_{dc}$ are the current ratios. Then the value of $\frac{\beta_{dc}-\alpha_{dc}}{\alpha_{dc} \cdot \beta_{dc}}$ is

In a $CE$ circuit,the collector of a transistor is kept at a constant voltage $V_C = 1.5 \ V$. When the base current changes from $100 \ \mu A$ to $150 \ \mu A$,the collector current changes from $5 \ mA$ to $10 \ mA$. The current gain $\beta$ for this transistor is:

If $\alpha$ and $\beta$ are the current gains in the $CB$ and $CE$ configurations respectively of the transistor circuit,then $\frac{\beta - \alpha}{\alpha \beta} = $