Explanation

(d)
$t_{99.9\%} = \frac{2.303}{k}\log\frac{a}{a - 99.9a}$
$t_{50\%} = \frac{0.693}{k}$
(b) $k = \frac{1}{t}\ln\frac{100}{100\ –\ 99.9} = \frac{1}{t}\ln\frac{100}{0.1}$
Or $\ \frac{\ln 2}{t_{1/2}} = \frac{1}{t}\ln 10^{3}$
Or $\frac{\log 2}{t_{1/2}} = \frac{1}{t}\log{10^{3} = 3/t}$
$t_{1/2} = \frac{\log 2}{3} \times t$
$\therefore t = 10t_{1/2}$