Question

If $t_{3 / 4}$ and $t_{1 / 2}$ are time required for completion of $3 / 4$ decay and $1 / 4$ decay then $t_{3 / 4}=t_{1 / 2} \times n$, then $n$ is $\ldots \ldots$

Answer 1

$t_{3 / 4}=\frac{2.303}{\lambda} \log \frac{1}{1 / 4}$

$=\frac{2.303}{\lambda} \log 4=\frac{2.303}{\lambda} \log 2^{2}$

$=2 \times \frac{2.303}{\lambda} \log 2=2 \times t_{1 / 2} .\left(t_{1 / 2}=\frac{0.693}{\lambda}=\frac{\ln 2}{\lambda}\right)$