Q.A source contains two species of phosphorous nuclei, $_{15}^{32}$P ($T_{1/2}$ = 14.3 d) and $_{15}^{33}$P ($T_{1/2}$ = 25.3 d). At time t = 0, 90% of the decays are from $_{15}^{32}$P. How much time has to elapse for only 15% of the decays to be from $_{15}^{32}$P?

[!TLDR] $\lambda_1=0.04846$/d (P-32), $\lambda_2=0.02739$/d (P-33); ratio $A_1/A_2=9\,e^{-(\lambda_1-\lambda_2)t}$; setting this equal to $15 …

Decay constants: $\lambda_1=0.693/14.3=0.04846$ per day for $^{32}P$, $\lambda_2=0.693/25.3=0.02739$ per day for $^{33}P$. At t=0, 90% of decays are from $^{32}P$ and 10% from $^{33}P$, so $A_1(0)/A_2(0)=90/10=9$. The activity ratio at any later time evolves as $\frac{A_1(t)}{A_2(t)}=\frac{A_1(0)}{A_2(0)}\,e^{-(\lambda_1-\lambda_2)t}=9\,e^{-0.02107t}$ (since $\lambda_1-\lambda_2=0.04846-0.02739=0.02107$ per day). We want the time when only 15% of decays are from $^{32}P$ (and hence 85% from $^{33} …

Express each species' activity as $A_i(t)=A_i(0)e^{-\lambda_it}$, form the ratio $A_1(t)/A_2(t)$ (which simplifies since both activities share the same starting time …

Working with the raw NUMBER of nuclei $N_i(t)$ instead of the ACTIVITY $A_i(t)=\lambda_iN_i(t)$ -- since the problem states percentages of DECAYS (i.e. activity, not population), and the two species have different $\lambd …