Question

A mixture of ${}^{239}\text{Pu}$ and ${}^{240}\text{Pu}$ 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. ${}_{239}Pu=45.1$%, ${}_{240}Pu=54.9$%
• B. ${}_{239}Pu=54.9$%, ${}_{240}Pu=45.1$%
• C. ${}_{239}Pu=55.1$%, ${}_{240}Pu=44.9$%
• D. None of these

Answer 1

The correct option is A ${}_{239}Pu=45.1$%, ${}_{240}Pu=54.9$%

Total activity of a sample is the sum of the individual activities of all its components.
Let the total mass of the sample be 1 g and the mass of Pu - 239 be x g. The mass of Pu - 240 will be 1-x g.
Total activity due to Pu-239 in the sample is
$\frac{\text{mass}}{\text{molar mass}}\times \text{Avogadro's number}\times \frac{0.693}{\text{half life period}}=\frac{x}{239}\times 6.023\times {10}^{23}\times \frac{0.693}{2.44\times {10}^4}$
Total activity due to Pu-240 in the sample is $\frac{\text{mass}}{\text{molar mass}}\times \text{Avogadro's number}\times \frac{0.693}{\text{half life period}}=\frac{1-x}{240}\times 6.023\times {10}^{23}\times \frac{0.693}{6.08\times {10}^3}$
Note: Here, the unit of specific activity is converted from dis/s/g to dis/y/g
Total activity of the sample is
$\frac{x}{239}\times 6.023\times {10}^{23}\times \frac{0.693}{2.44\times {10}^4}+\frac{1-x}{240}\times 6.023\times {10}^{23}\times \frac{0.693}{6.08\times {10}^3}=6\times {10}^9\times 365\times 24\times 60\times 60$
$7.16\times {10}^{16}x+2.86\times {10}^{17}(1-x)=1.89\times {10}^{17}$
$2.86\times {10}^{17}-2.14\times {10}^{17}x=1.89\times {10}^{17}$
$2.14\times {10}^{17}x=9.68\times {10}^{16}$
$x=0.451$ and $1-x=0.549$
Hence, the isotopic composition of the sample is ${}_{239}Pu=45.1$ % and ${}_{240}Pu=54.9$ %