Question

A mixture of ${}^{239}Pu$ and ${}^{240}Pu$ has a specific activity of $6\times {{10}^{9}}$ dps per g sample. The half-lives of isotopes are $2.44\times {{10}^{4}}$ year and $6.58\times {{10}^{3}}$ year respectively. Calculate the isotopic composition of mixture.
A. $59.05$
B. $51.05$
C. $61.05$
D. $69.05$

Answer 1

To calculate the answer to this question, first we need to calculate half-life of atoms. radioactive substances have a specific rate of decay. There is a relation between half life and decay constant of radioactive substances.

Formula used:
$A=-\dfrac{dN}{dt}$
$\Rightarrow A=\lambda N$
where,$A$ is activity, $\lambda$ is the decay constant and $dN$ is the change in disintegration of atoms.
${{t}_{1/2}}=\dfrac{0.693}{\lambda }$
where, ${{t}_{1/2}}$ is the half-life.

Complete step by step answer:
Here, it is given that the specific activity of sample is $6\times {{10}^{9}}$ dps per g.
Let $a\,g$ be ${}^{239}Pu$ and $b\,g$ be ${}^{240}Pu$ present in mixture, therefore,
$a+b=1$
This is equation $(i)$
The rate of decay is the change in the disintegration of number of atoms per unit time. The formula can be written as:
$A=-\dfrac{dN}{dt}$
$\Rightarrow A=\lambda N$
This is equation $(1)$
where, $\lambda$ is the decay constant, $dN$ is the change in disintegration of atoms.
${{t}_{1/2}}=\dfrac{0.693}{\lambda }$
This is equation $(2)$ where, ${{t}_{1/2}}$ is the half-life.
If the value of $\lambda$ in equation $(2)$ is substituted in equation $(1)$,
$A=\dfrac{0.693}{{{t}_{1/2}}}$
This is equation $(3)$
Now, using this equation $(3)$ , we get the activities of isotopes of $pu$ .
For ${}^{239}Pu$ having mass which is equal to $239g$ and half-life $2.44\times {{10}^{4}}$ year.
We can calculate the total number of atoms for disintegration as follows:
$N=\dfrac{6.022\times {{10}^{23}}}{239}$
$A=\dfrac{0.693N}{{{t}_{1/2}}}$
Now, substituting the value of $N$ and ${{t}_{1/2}}$ in equation $(3)$ and converting half – life given in years into seconds, we get
$\Rightarrow A=\dfrac{0.693}{2.44\times {{10}^{4}}}\left( \dfrac{6.022\times {{10}^{23}}}{239} \right)\times \dfrac{a}{365\times 24\times 60\times 60}$
On further solving, we get,
$\Rightarrow A=2.27\times {{10}^{9}}a\,dps\,per\,g$
Similarly, we will calculate the activity of ${}^{240}Pu$, having mass equal to $240g$ and half-life of $6.58\times {{10}^{3}}$ year.
$A=\dfrac{0.693N}{{{t}_{1/2}}}$
Substituting all the values and converting half – life from years into seconds, we get,
$\Rightarrow A=\dfrac{0.693}{6.58\times {{10}^{3}}}\left( \dfrac{6.022\times {{10}^{23}}}{240} \right)\times \dfrac{b}{365\times 24\times 60\times 60}$
On further solving, we get,
$\Rightarrow A=8.39\times {{10}^{9}}b\,dps\,per\,g$
Now,
$2.27\times {{10}^{9}}a+8.39\times {{10}^{9}}b=6\times {{10}^{9}}$
$\Rightarrow 2.27a+8.39b=6$
This is equation $(ii)$
By solving equation $(i)$ and $(ii)$, we get
$a=0.3895g;\,\,b=0.6105g$
The percentage composition of ${}^{239}Pu$ is $38.95$
The percentage composition of ${}^{240}Pu$ is $61.05$

Hence correct answer is optionC.

Note:
Half-life is defined as the time that is taken by the concentration of a given reactant to reach $50$ to its initial concentration. It is expressed in seconds and denoted with the symbol ${{t}_{1/2}}$ .
Decay constant is the proportionality factor between the size of radioactive atoms and rate at which the population decreases. It is denoted with a symbol, $\lambda$ .