Question

The activity of a radioactive nuclide $\left(X^{100}\right)$ is $6.023$ curie at a certain time $‘t^{\prime }$. If its disintegration constant is $3.7\times {10}^4{s}^{-1}$ the mass of $X$ after t sec is:

• A. $6.022\times {10}^{6}g$
• B. ${10}^{-13}g$
• C. ${10}^{-15}g$
• D. ${10}^{-17}g$

Answer 1

The correct option is C ${10}^{-15}g$

Acitivity = $\lambda \times N$
$1C$ = $3.7\times {10}^{10}Dps$
threfore
$6.203\times 3.7\times {10}^{10}=3.7\times {10}^4\times N$
$N=6.023\times {10}^6$
Number of moles = $\frac{6.023\times {10}^6}{6.023\times {10}^{23}}$
Mass of $X$ after time t= $\frac{6.023\times {10}^6}{6.023\times {10}^{23}}\times 100$
$={10}^{-15}g$