Question

32 mg of pure ${}_{94}^{238}Pu{O}_2$ has an activity of $6.4\times {10}^{7}dps$, the half-life of ${}_{94}^{238}\text{Pu}$ is $x\times {10}^{11}$. Find $x$?

• A. 7.73
• B. 6.73
• C. 5.73
• D. 8.77

Answer 1

Answer: (D)

8.77

$\frac{-dN}{dt}=\lambda N$

$6.4\times {10}^{7}dps=\lambda \times \frac{32}{238\times 1000}\times 6.023\times {10}^{23}$

$\lambda =\frac{6.4\times {10}^{7}dps}{\frac{32}{238\times 1000}\times 6.023\times {10}^{23}}$

$\lambda =7.90\times {10}^{-13}$

$t_{1/2}=\frac{0.693}{\lambda }=\frac{0.693}{7.90\times {10}^{-13}}=8.77\times {10}^{11}=x\times {10}^{11}$

Hence, $x=8.77$.