For a transistor,$\frac{1}{\alpha_{DC}} - \frac{1}{\beta_{DC}}$ is equal to [where $\alpha_{DC}$ and $\beta_{DC}$ are current amplification factors].

In an $n-p-n$ transistor,$10^{8}$ electrons enter the emitter in $10^{-8} \,s$. If $1\%$ of electrons are lost in the base,the fraction of current that enters the collector and the current amplification factor $\beta$ are respectively:

$A$ transistor is used in common emitter mode as an amplifier. Then:

In an $NPN$ transistor,the collector current is $10\, mA$. If $90\%$ of the electrons emitted reach the collector,then:

In a $CE$ transistor amplifier,the signal voltage across the collector resistance is $2.5 \ V$. The input signal voltage is $0.02 \ V$. If the base and collector resistances are $1.5 \ k\Omega$ and $2.5 \ k\Omega$,then the current amplification factor is

For $I_b = 20 \ \mu A$ and $\beta = 100$,the value of the emitter current is ........ $mA$.