Question

32 mg of pure ${}_{94}^{238}Pu{O}_2$ has an activity of $6.4\times {10}^7$ dps. Find the half-life of ${}_{94}^{238}\text{Pu}$.

• A. $23.19\times {10}^{11}sec$
• B. $30.92\times {10}^{11}sec$
• C. $15.46\times {10}^{11}sec$
• D. $7.73\times {10}^{11}sec$

Answer 1

The correct option is D $7.73\times {10}^{11}sec$

$MoleofPu{O}_2=\frac{32\times {10}^{-3}}{270}$
$\therefore MoleofPu=\frac{32\times {10}^{-3}}{270}$
$\therefore Atoms\left(N\right)ofPu=\frac{32\times {10}^{-3}}{270}\times 6.023\times {10}^{23}$
Now, $rate=\lambda .N$
$\therefore 6.4\times {10}^7=\frac{\lambda \times 32\times {10}^{-3}\times 6.023\times {10}^{23}}{270}$
$\therefore \lambda =8.97\times {10}^{-13}$
$\therefore t_{1/2}=\frac{0.693}{\lambda }=\frac{0.693}{8.97\times {10}^{-13}}$
$=7.73\times {10}^{11}sec$