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Q.
In NPN transistor, $10^{10}$ electrons enter in emitter region in $10^{-6} s$. If $2 \%$ electrons are lost in base region then collector current and current amplification factor $(\beta)$ respectively are :-
Solution:
$i _{ e }=\frac{ N _{ e }}{ t }=\frac{10^{10} \times 1.6 \times 10^{-19}}{10^{-6}}=1.6\, mA$
$i _{ b }=2 \%$ of $i _{ e }$ so $i _{ c }=98 \%$ of $i _{ e }$
$i _{ c }=\frac{98}{100}=1.6\, mA \simeq 1.57\, mA$
$\beta=\frac{ i _{ c }}{ i _{ b }}$
or $\frac{\alpha}{1-\alpha}$