Question

A source contains two phosphorus radionuclides ${}^{\mathbf{32}}\mathbf{P}\left(T_{1/2}=14.3d\right)$and${}^{\mathbf{33}}\mathbf{P}\mathbf{(}{\mathbf{T}}_{\mathbf{1}\mathbf{/}\mathbf{2}}\mathbf{=}\mathbf{25}\mathbf{.}\mathbf{3}\mathbf{d}\mathbf{)}$. Initially, 10.0%of the decays come from${}^{\mathbf{33}}\mathbf{P}$. How long must one wait until 90.0%do so?

Short answer

One must wait 209 days long until 90.0% is done.

Step-by-step solution

The given data

a) Half-life of${}^{32}\mathrm{P}$,$T_{1/2_1}=14.3\mathrm{d}$,

b) Half-life of role="math" localid="1661598432502" ${}^{33}P$,$T_{1/2_2}=25.3d$

c) Initial percentage of decay of ${}^{33}\mathrm{P}$is 10.0%.

d) Final percentage of decay of is ${}^{33}\mathrm{P}$,$90.0\%$.

Understanding the concept of decay  

The two phosphorus isotopes are present in a given composition together, where the second isotope has already decayed 10%. This implies that its initial amount present differs nine-fold from the initial amount of the first isotope. From the concept of decay rate being exponential proportional to the time value, we can get the waiting time considering the undecayed amount present for both the nuclides at the given time.

Formula:

The rate of decay,$\mathbf{R}\mathbf{=}\mathbf{\lambda N}$ (i)

Where,$\mathbf{\lambda }$is the disintegration constant

N Is the number of undecayed nuclei

The undecayed rate of the sample remaining after a given time,$\mathbf{R}\mathbf{=}{\mathbf{R}}_{\mathbf{0}}{\mathbf{e}}^{\mathbf{-}\mathbf{\lambda t}}$ (ii)

The disintegration constant,$\mathbf{\lambda }\mathbf{=}\frac{\mathbf{In}\mathbf{2}}{{\mathbf{T}}_{\mathbf{1}\mathbf{/}\mathbf{2}}}$ (iii)

Where,${\mathbf{T}}_{\mathbf{1}\mathbf{/}\mathbf{2}}$is the half-life of the substance

Calculation of the time of 90% of decay of the phosphorus nuclide

Let us label the two isotopes with subscripts 1 (for ${}^{32}\mathrm{P}$) and 2 (for ${}^{33}\mathrm{P}$). Initially, 10% of the decays come from ${}^{33}P$, which implies that the initial rate $R_0=9R_{01}$. Now, using this relation in equation (i), we can get the following condition as follows:

$$\begin{gathered} R_{01}=\frac{1}{9}{R}_{02} \\ {\lambda }_1{N}_{01}=\frac{1}{9}{\lambda }_2{N}_{02}..................\left(a\right) \end{gathered}$$

Now, for the condition of the undecayed sample at time, we have the condition$R_1=9R_2$

This implies the following condition considering equation (ii):

$\left(\frac{R_{01}}{R_{02}}\right)e^{-\left({\lambda }_1-{\lambda }_2\right)t}=9$

Further solving the above equation for getting the time value, we get that

$$\begin{gathered} t=\frac{1}{{\lambda }_1-{\lambda }_2}In\left(\frac{R_{01}}{9R_{02}}\right) \\ =\frac{\mathrm{In}\left({\displaystyle \frac{{\mathrm{R}}_{01}}{9{\mathrm{R}}_{02}}}\right)}{\mathrm{In}2/{\mathrm{T}}_{1/2_1}-\mathrm{In}2/{\mathrm{T}}_{1/2_1}}\left(\mathrm{from}\mathrm{equation}\left(\mathrm{iii}\right)\right) \\ =\frac{\mathrm{In}{\left({\displaystyle \frac{1}{9}}\right)}^2}{\mathrm{In}2\left[{\left(14.3\mathrm{d}\right)}^{-1}-{\left(25.3\mathrm{d}\right)}^{-1}\right]}\left(\mathrm{from}\mathrm{condition}\left(\mathrm{a}\right)\right) \\ =209\mathrm{d} \end{gathered}$$

Hence, the value of the waiting time is 209 d.