Answer
half lifr $T_{1/2}$=$8$ $days$
Work Step by Step
Given that number of nuclei present at the start of the experiment is $N_{0}=4.6\times10{^15}$
After time $t=20days $
number present is $N=8.14\times10^{14}$
Putting these values to decay equation $N=N_{0}e^{-\lambda t}$
$8.14\times10^{14}=4.6\times10^{15} e^{-\lambda 20days}$
$e^{\lambda 20days}=\frac{4.6\times10^{15}}{8.14\times10^{14}}$
$e^{\lambda 20days}=5.6511$
taking natural logarithm both the side
$ln(e^{\lambda 20days})=ln5.6511$
$\lambda20days=1.7318$
$\lambda=\frac{1.7318}{20days}$
$T_{1/2}=\frac{0.693}{\lambda}$
so $T_{1/2}=\frac{0.693}{\frac{1.7318}{20days}}$
$T_{1/2}=0.4\times20days$
$T_{1/2}=8days$
