किसी स्त्रोत में फॉस्फोरस के दो रेडियो न्यूक्लाइड निहित हैं ${ }_{15}^{32} P \left(T_{1 / 2}=14.3 d \right)$ एवं ${ }_{15}^{33} P$ $\left(T_{1 / 2}=25.3 d \right) ।$ । प्रारंभ में ${ }_{15}^{33} P$ से $10\, \%$ क्षय प्राप्त होता है। इससे $90 \%$ क्षय प्राप्त करने के लिए कितने समय प्रतीक्षा करनी होगी?

Half life of $_{15}^{32} P, T_{1 / 2}=14.3$ days

Half life of $_{15}^{33} P, T_{1 / 2}^{\prime}=25.3$ days

nucleus decay is $10 \%$ of the total amount of decay. The source has initially $10 \%$ of $_{15}^{32} P$ nucleus and $90 \%$ of $_{15}^{32} P$ nucleus. Suppose after $t$ days, the source has $10 \%$ of $_{15}^{32} P$ nucleus and $90 \%$ of $_{15}^{33} P$ nucleus.

Initially:

Number of $_{15}^{33} P$ nucleus $= N$ Number of $_{15}^{32} P$ nucleus $=9 N$

Finally:

Number of $_{15}^{33} P$ nucleus $=9 N^{\prime}$ Number of $_{15}^{32} P$ nucleus $=N^{\prime}$

For $_{15}^{32} P$ nucleus, we can write the number ratio as:

$\frac{N^{\prime}}{9 N}=\left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}}$

$N^{\prime}=9 N(2)^{\frac{-t}{443}}\dots(i)$

For $_{15}^{33} P,$ we can write the number ratio as:

$\frac{9 N^{\prime}}{N}=\left(\frac{1}{2}\right)^{\frac{1}{T_{1/2}}}$

$9 N^{\prime}=N(2)^{\frac{-t}{25.3}}\dots(ii)$

On dividing equation (i) by equation (ii), we get

$\frac{1}{9}=9 \times 2^{\left(\frac{t}{25.3}-\frac{t}{14.3}\right)}$

$\frac{1}{81}=2^{\left(-\frac{11 t}{253 \times 443}\right)}$

$\log 1-\log 81=\frac{-11 t}{25.3 \times 14.3} \log 1$

$\frac{-11 t}{25.3 \times 14.3}=\frac{0-1.908}{0.301}$

$t=\frac{25.3 \times 14.3 \times 1.908}{11 \times 0.301} \approx 208.5 d a y s$

Hence, it will take about $208.5$ days for $90 \%$ decay of $_{15} P^{33}$