In an NPN transistor $10^{10}$ electrons enter the emitter in $10^{-6} \mathrm{~s}$ and $2 \%$ electrons…
In an NPN transistor $10^{10}$ electrons enter the emitter in $10^{-6} \mathrm{~s}$ and $2 \%$ electrons recombine with holes in base. The current ratios ' $\alpha$ ' and ' $\beta$ ' of a transistor are respectively (nearly)
$0 \cdot 98,49$
$49,0.98$
$0 \cdot 49,98$
$98,0.49$
Solution
$\mathrm{I}_{\mathrm{e}}=\frac{\mathrm{n}_{\mathrm{e}} \times \mathrm{e}}{\mathrm{t}} \text { and } \mathrm{I}_{\mathrm{c}}=\frac{\mathrm{n}_{\mathrm{c}} \times \mathrm{e}}{\mathrm{t}}$
From the data given,
$\begin{aligned}
I_c & =\frac{(98 / 100) n_e \times e}{t}=\frac{98}{100} I_c \\
\alpha & =\frac{I_c}{I_e}=\frac{98 I_e}{100 I_e}=0.98
\end{aligned}$
$\therefore \quad \beta=\frac{\alpha}{1-\alpha}=\frac{0.98}{1-0.98}=\frac{0.98}{0.02}=49$