Question

A mixture of ${}^{239}\text{Pu}$ and ${}^{240}\text{Pu}$ 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. $59.05$% ; $40.95$%
• B. $51.05$% ; $48.95$%
• C. $61.05$% ; $38.95$%
• D. $69.05$% ; $30.95$%

Answer 1

Answer: (C)

$61.05$% ; $38.95$%

Given, specific activity$=6\times {10}^9$ dps per g of mixture. Let a g ${}^{239}\text{Pu}$ and b g ${}^{240}\text{Pu}$ are present in mixture, Then
$a+b=1$ ... (i)
$\therefore rate=\lambda N$
For ${}^{239}\text{Pu}$,
$r_1=\frac{0.693\times 6.023\times {10}^{23}\times a}{2.44\times {10}^4\times 365\times 24\times 60\times 60\times 239}dse{c}^{-1}{g}^{-1}$
$=2.27\times {10}^9\times adse{c}^{-1}{g}^{-1}$
For ${}^{240}\text{Pu}$,
$r_2=\frac{0.693\times 6.023\times {10}^{23}\times b}{6.58\times {10}^3\times 365\times 24\times 60\times 60\times 240}dse{c}^{-1}{g}^{-1}$
$=8.39\times {10}^9\times bdse{c}^{-1}{g}^{-1}$
$\therefore 2.27\times {10}^{9}a+8.39\times {10}^{9}b=6\times {10}^9$
or, $2.27a+8.39b=6$ ... (ii)
By eqs. (i) and (ii), $b=0.6105g;a=0.3895g$
or % of ${}^{240}\text{Pu}=61.05$%; % of ${}^{239}\text{Pu}=38.95$%