A source contains two phosphorous radio nuclides

Solution:

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

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

${ }_{15}^{33} \mathrm{P}$ nucleus decay is $10 \%$ of the total amount of decay.

The source has initially $10 \%$ of ${ }_{15}^{33} \mathrm{P}$ nucleus and $90 \%$ of ${ }_{15}^{32} \mathrm{P}$ nucleus.

Suppose after $t$ days, the source has $10 \%$ of ${ }_{15}^{32} \mathrm{P}$ nucleus and $90 \%$ of ${ }_{15}^{33} \mathrm{P}$ nucleus.

Initially:

Number of ${ }_{15}^{33} \mathrm{P}$ nucleus $=N$

Number of ${ }_{15}^{32} \mathrm{P}$ nucleus $=9 \mathrm{~N}$

Finally:

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

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

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

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

$N^{\prime}=9 N(2)^{\frac{-f}{143}}$     ...(1)

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

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

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

On dividing equation (1) by equation (2), we get:

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

$\frac{1}{81}=2^{-\left(\frac{111}{253 \times 14.3}\right)}$

$\log 1-\log 81=\frac{-11 t}{25.3 \times 14.3} \log 2$

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

$t=\frac{25.3 \times 14.3 \times 1.908}{11 \times 0.301} \approx 208.5$ days

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