13.25 A source contains two phosphorous radio nuclides \({ }_{15}^{32} \mathrm{P}\) (T1/2 = 14.3 days) and \({ }_{15}^{33} \mathrm{P}\) (T1/2 = 25.3 days). Initially, 10% of the decays come from \({ }_{15}^{33} \mathrm{P}\). How long one must wait until 90% do so?

Hint: \(N=N_oe^{-\lambda t}\)
Step 1: Find the initial and final number of nuclie in both elements.

Half-life ofP1532, T1/2=14.3 days
Half-life ofP1533, T'1/2=25.3 days and P1533 nucleus decay is 10% of the total amount of decay.
The source has initially 10% ofP1533 nucleus and 90% of P1532 nucleus.
Suppose after t days, the source has 10% of P1532 nucleus and 90% ofP1533 nucleus.
Initially:
Number of P1533 nucleus=N
Number of P1532 nucleus=9 N
Finally:
Number of P1533 nucleus = 9 N'
Number of P1532 nucleus =N'

Step 2: Write the equations for both elements.
For \({ }_{15}^{33} \mathrm{P}\) nucleus, we can write the number ratio as:

N'9N=121T1/2
N'=9N(2)-125.3...........................(1)
For \({ }_{15}^{32} \mathrm{P}\) nucleus, we can write the number ratio as:
\(\begin{aligned} \frac{9 N^{\prime}}{N} &=\left(\frac{1}{2}\right)^{\frac{1}{T_{1 / 2}}} \\ 9 N^{\prime} &=N(2)^{\frac{-1}{14. 3}} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots .(2) \end{aligned}\)


Step 3: Find the required time.
On dividing the equation (1) by the equation (2), we get:
\(\frac{1}{9}=9 \times 2\left(\frac{-t}{25.3} +\frac{t}{14.3}\right)\)


181=211 t25.3×14.3
log 1- log81=-11t25.3×14.3log 2
-11t25.3×14.3=0-1.9080.301
t=25.3×14.3×1.90811×0.301208.5 days
Hence, it will take about 208.5 days for 90% decay ofP1533.