Answer
$f=\frac{1}{16}$
Work Step by Step
Suppose for sample $ A$
initial number of nuclei presents is $N_{0,A}$
radioactive decay constant is $\lambda_{A}$
half life is $ T_{1/2,A}$
Similarly for sample $ B$
initial number of nuclei presents is $N_{0, B}$
radioactive decay constant is $\lambda_{ B}$
half life is $ T_{1/2, B}$
Given that $ T_{1/2, B}=\frac{1}{2} T_{1/2, A}$........equation(1)
From equation 31.6 $T_{1/2}=\frac{ln2}{\lambda}$
$ T_{1/2,A}=\frac{ln2}{\lambda_{A}}$
$ T_{1/2,B}=\frac{ln2}{\lambda_{B}}$
putting these values to equation(1)
$ \frac{ln2}{\lambda_{B}}=\frac{1}{2} \frac{ln2}{\lambda_{A}}$.
$\lambda_{B}=2\lambda_{A}$..............................equation(2)
Suppose in time $t$ Sample $A$ decreases to one fourth of its initial number.
so after time $t$, $N_{t,A}=\frac{N_{0,A}}{4} $
from decay equation $N=N_{0}e^{-\lambda t}$ after time $t$
$N_{t,A}=\frac{N_{0,A}}{4}= N_{0,A}e^{-\lambda_{A} t}$
$\frac{1}{4}=e^{-\lambda_{A} t}$
$e^{\lambda_{A} t}=4$
taking natural log both side
$ln(e^{\lambda_{A} t})=ln4$
$\lambda_{A} t=2ln2$
$t=\frac{2ln2}{\lambda_{A}}$.................................equation(3)
in the same time Sample $B$ decrease to a fraction $f$ of the number present initially
so after time $t$ , $N_{t,B}=fN_{0,B} $
from decay equation $N=N_{0}e^{-\lambda t}$
after time $t=\frac{2ln2}{\lambda_{A}}$.
$N_{t,B}=fN_{0,B}=N_{0,B}e^{-\lambda_{B}\frac{2ln2}{\lambda_{A}}}$
$f=e^{-\lambda_{B}\frac{2ln2}{\lambda_{A}}}$
now putting $\lambda_{B}=2\lambda_{A}$ from equation (2)
we will get
$f=e^{-2\lambda_{A}\frac{2ln2}{\lambda_{A}}}$
$f=e^{-4ln2}$
$f=e^{ln(2)^{-4}}$
$f=(2)^{-4}$
$f=\frac{1}{16}$
