$A$ source contains two phosphorus radionuclides $_{15}^{32} P \left(T_{1/2} = 14.3 \ d\right)$ and $_{15}^{33} P \left(T_{1/2} = 25.3 \ d\right)$. Initially,$10\%$ of the decays come from $_{15}^{33} P$. How long must one wait until $90\%$ of the decays come from $_{15}^{33} P$?

(D) Let $N_1$ and $N_2$ be the number of nuclei of $_{15}^{32} P$ and $_{15}^{33} P$ respectively. The activity $A$ is given by $A = \lambda N = \frac{\ln 2}{T_{1/2}} N$.
Initially,$A_2 / (A_1 + A_2) = 0.1$,which implies $A_2 = (1/9) A_1$.
Substituting $A = \frac{\ln 2}{T_{1/2}} N$,we get $\frac{N_2}{T_2} = \frac{1}{9} \frac{N_1}{T_1}$,so $N_2(0) = \frac{1}{9} \frac{T_2}{T_1} N_1(0) = \frac{1}{9} \frac{25.3}{14.3} N_1(0) \approx 0.1966 N_1(0)$.
After time $t$,the activities are $A_1(t) = A_1(0) 2^{-t/T_1}$ and $A_2(t) = A_2(0) 2^{-t/T_2}$.
We want $A_2(t) / (A_1(t) + A_2(t)) = 0.9$,which implies $A_2(t) = 9 A_1(t)$.
Substituting the expressions for activity: $A_2(0) 2^{-t/T_2} = 9 A_1(0) 2^{-t/T_1}$.
Since $A_2(0) = (1/9) A_1(0)$,we have $(1/9) A_1(0) 2^{-t/T_2} = 9 A_1(0) 2^{-t/T_1}$.
$2^{-t/T_2} / 2^{-t/T_1} = 81$,which means $2^{t(1/T_1 - 1/T_2)} = 81$.
Taking $\log_{10}$ on both sides: $t(1/14.3 - 1/25.3) \log_{10} 2 = \log_{10} 81$.
$t(0.06993 - 0.03953) \times 0.3010 = 1.9085$.
$t(0.0304) \times 0.3010 = 1.9085 \implies t \approx 208.5 \ d$.