Answer
half life is equal to $3$ $days$
Work Step by Step
Suppose number of radioactive nuclei present initially is $N_{0}$
it is given that after $9days$ number of nuclei decreases to one eighth of its initial number.
so after time $t= 9$ $days$ number of nuclei $N=\frac{N_{0}}{8}$
From the equation 31.5
$N=N_{0}e^{-\lambda t}$
here $\lambda$ is radioactive decay constant
so putting the values of
$t=9days$, $N=\frac{N_{0}}{8}$ in above equation
$\frac{N_{0}}{8}=N_{0}e^{-\lambda 9 days}$
$\frac{1}{8}=e^{-\lambda 9 days}$
$e^{\lambda 9 days}=8$
taking natural logarithm of both the side we will get
$ln{e^{\lambda 9 days}}=ln8$
$9days\lambda=ln2^3$
so $\lambda=\frac{3\times ln2}{9days}$
or $\lambda=\frac{ln2}{3days}$.........equation(1)
from equation 31.6
$T_{\frac{1}{2}}=\frac{ln2}{\lambda}$
putting the value of $\lambda$ from equation(1)
$T_{\frac{1}{2}}=\frac{ln2}{\frac{ln2}{3days}}$
$T_{\frac{1}{2}}=3days$
