Explanation

(d) Given that, change in emitter current, ${\mathrm{\Delta}I}_{E} =$ 8 mA
and change in collector current, ${\mathrm{\Delta}I}_{C} = 7.9$ mA
We know that,
$\alpha = \frac{{\mathrm{\Delta}I}_{C}}{{\mathrm{\Delta}I}_{E}}$
$\Rightarrow = \frac{7.9}{8} \Rightarrow$ $\alpha \approx$0.99
Also, we know that
$\beta = \frac{\alpha}{1 - \alpha}$
$\Rightarrow = \frac{\frac{7.9}{8}}{1 - \frac{7.9}{8}} = \frac{7.9}{8 - 7.9}$
Or $\beta = \frac{7.9}{0.1}$ = 79
Hence, the required answer is $\alpha =$0.99 and $\beta = 79.$