Question

A mixture of ${}^{239}Pu$ and ${}^{240}Pu$ has a specific activity of $6\times {{10}^{9}}\,dis/s/g$ . The half-lives of the isotopes are $2.44\times {{10}^{4}}y$ and $6.08\times {{10}^{3}}y$respectively. Calculate the isotopic composition of this sample.
(A)- ${}_{239}Pu=45.1$
(B)- ${}_{239}Pu=54.9$
(C)- ${}_{239}Pu=55.1$
(D)- None of these

Answer 1

The two isotopes of polonium are radioactive, and undergo disintegration. Then, the number of atoms of polonium undergoing disintegration per unit time are related to its half-life or decay constant.

Complete step by step answer:
The radioactivity or the decay rate is the change in the number (dN) or disintegration of the radioactive atoms per unit of time. It is given as:
$A=-\dfrac{dN}{dt}=\lambda N$ -------- (a)
where, $\lambda$ is the decay constant, which is further related to the half-life as:
${{t}_{1/2}}\,\,(in\,years)=\,\dfrac{\ln 2}{\lambda }=\dfrac{0.693}{\lambda }$ -------- (b)
Therefore, from equation (a) and (b), we get,
$A=\lambda N=\dfrac{0.693N}{{{t}_{1/2}}}$ -------- (c)

Now, using equation (c), we get, the activities of the two isotopes of Pu, as follows:
For ${}^{239}Pu$, with mass = 239 g and half-life = $2.44\times {{10}^{4}}y$, then the total number of atoms for disintegration is $N=\dfrac{6.022\times {{10}^{23}}}{239}$ , then we have,

$\Rightarrow A=\dfrac{0.693}{2.44\times {{10}^{4}}}\times \left( \dfrac{6.022\times {{10}^{23}}}{239} \right)\times \dfrac{1}{365\times 24\times 60\times 60}=2.27\times {{10}^{9\,}}\,dis{{\sec }^{-1}}{{g}^{-1}}$

Similarly, for ${}^{240}Pu$, with mass = 240 g and half-life = $6.08\times {{10}^{3}}y$, we have,
$\Rightarrow A=\dfrac{0.693}{6.08\times {{10}^{3}}}\times \left( \dfrac{6.022\times {{10}^{23}}}{240} \right)\times \dfrac{1}{365\times 24\times 60\times 60}=9.068\times {{10}^{9\,}}\,dis{{\sec }^{-1}}{{g}^{-1}}$

In the mixture of the two isotopes, let the fraction of ${}^{239}Pu$ be x, then the fraction of ${}^{240}Pu$ will be (1 – x).
So, as given the total specific activity of the mixture is $6\times {{10}^{9}}\,dis/s/g$, then we have,
$\Rightarrow (2.27\times {{10}^{9\,}}\,\times \,x\,)+(9.068\times {{10}^{9}}\times (1-x))=6\times {{10}^{9}}$
$\Rightarrow 9.068-6.798\,x\,=6$
$\Rightarrow 3.068=6.798\,x\,$
Then, $x=0.451$ and $1-x=0.549$
Thus, the percentage composition of ${}^{239}Pu = 45.1$ and ${}^{240}Pu = 54.9$.
Therefore, the isotopic composition of the mixture is - ${}_{239}Pu=45.1$.

So, the correct answer is “Option A”.

Note: In the decay rate of the radioactive polonium, it is related to its decay constant and the half-life, where the decay constant is inversely proportional to the half-life of the polonium.
The half-life is the time during which half of the number of atoms of the polonium have reduced or disintegrated.