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 $2.44\times {10}^4$ year and $6.58\times {10}^3$ year respectively. Calculate the isotopic composition of this sample.

• A. ${}_{259}Pu=42\%,$ ${}_{240}Pu=58\%$
• B. ${}_{239}Pu=41\%,$ ${}_{240}Pu=59\%$
• C. ${}_{259}Pu=40\%,$ ${}_{240}Pu=60\%$
• D. ${}_{239}Pu=39\%,$ ${}_{240}Pu=61\%$

Answer 1

The correct option is D ${}_{239}Pu=39\%,$ ${}_{240}Pu=61\%$

The specific activity of each isotope is given by $A=\lambda N=\frac{0.693N}{t_{\frac{1}{2}}}$
For ${}^{239}Pu:A=\left(\frac{0.693}{2.44\times {10}^{4}y}\right)\left(\frac{6.02\times {10}^{23}}{239g}\right)=7.15\times {10}^{16}/y.g$
For ${}^{240}Pu:A=\left(\frac{0.693}{6.58\times {10}^{3}y}\right)\left(\frac{6.02\times {10}^{23}}{239g}\right)=2.64\times {10}^{17}/y.g$
The number of seconds in a year is $\left(365d\right)\left(\frac{24h}{d}\right)\left(\frac{60min}{h}\right)\left(\frac{60s}{min}\right)=3.15\times {10}^{7}s$
For ${}^{239}Pu:A=2.27\times {10}^9/s.g$
For ${}^{240}Pu:A=8.37\times {10}^9/s.g$
Let the fraction of ${}^{239}Pu=x$ then the fraction of ${}^{240}Pu=1-x$
$(2.27\times {10}^9)x+(1-x)(8.37\times {10}^9)=6.0\times {10}^9\Rightarrow (8.37\times {10}^9)-(6.10\times {10}^9)x=6.0\times {10}^9$
$\Rightarrow 2.37\times {10}^9=(6.10\times {10}^9)x\Rightarrow x=0.39\Rightarrow 39$%${}^{239}Pu$
$\therefore 61$% ${}^{240}Pu$