Question

The activity of $30\mu g$ of ${}^{247}Cm$ is $2.8nCi\left({10}^{-9}curie\right)$. Calculate the disintegration rate constant and the half-life of ${}^{247}Cm$:

• A. $1.551\times {10}^{7}years$
• B. $7.551\times {10}^{7}years$
• C. $2.551\times {10}^{7}years$
• D. None of these

Answer 1

The correct option is A $1.551\times {10}^{7}years$

$2.8nCi$ $=2.8\times {10}^{-9}Ci$

$=2.8\times {10}^{-9}\times 3.7\times {10}^{10}dps$

$=103.6dps$

$103.6=-\frac{dN}{dt}=\lambda \times \frac{30\times {10}^{-6}\times 6.02\times {10}^{23}}{247}$

$\lambda =1.417\times {10}^{-15}{s}^{-1}$

$\therefore t_{50}=\frac{0.693}{\lambda }=4.891\times {10}^{14}s$

$=1.551\times {10}^{7}years$