Question

A source contains two phosphorous radio nuclides ${}_{15}^{32}P(T_{1/2}=14.3d)$ and ${}_{15}^{33}P(T_{1/2}=25.3d)$. Initially, 10% of the decays come from ${}_{15}^{32}P$. How long one must wait until 90% to do so?

Answer 1

Let the number of atoms of ${}_{15}^{32}P$ be $N_1$ and atoms of ${}_{15}^{33}P$ be $N_2$.

The decay constants of two atoms be ${\lambda }_1$ and ${\lambda }_2$ respectively.

The initial activity of ${}_{15}^{33}P$ is $\frac{1}{9}$ times that of ${}_{15}^{32}P$.

Therefore, we can write the decay constants as:

$N_1{\lambda }_1=\frac{N_2{\lambda }_2}{9}$ ___(i)

Let after time t the activity of ${}_{15}^{33}P$ be $9$ times that of ${}_{15}^{32}P$

$N_1{\lambda }_1{e}^{-{\lambda }_{1}t}=9N_2{\lambda }_2{e}^{-{\lambda }_{2}t}$ ___(ii)

Now. Divide equation (ii) by (i) and taking the natural log of both sides we get

$-{\lambda }_{1}t=ln81-{\lambda }_{2}t$

$t=\frac{ln81}{{\lambda }_2-{\lambda }_1}$

where ${\lambda }_2=0.048/day$ and ${\lambda }_1=0.027/day$

$t=\frac{4.394}{0.048-0.027}=209.2days$

t comes out to be $209.2$ days