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 $\dots \dots$

Answer 1

Answer: 2

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

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

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