Question

The half-life of a radioisotope is 10 hr. The total number of disintegration in the ${10}^{th}$ hr. measured at a time when the activity was 1 Ci is :

• A. $[6.91\times {10}^{12}]$
• B. $[6.91\times {10}^{13}]$
• C. $[6.91\times {10}^{11}]$
• D. $[6.91\times {10}^{14}]$

Answer 1

The correct option is B $[6.91\times {10}^{13}]$

$N=\frac{3.7\times {10}^{10}\times 3.6\times {10}^4}{0.693}$
$-\frac{dN}{dt}=\lambda N\Rightarrow N=N_0{e}^{-\lambda t}$
Now $\Delta N=N_1-N_2=-(N_0{e}^{-\lambda \times 10\times 3600}-N_0{e}^{-\lambda \times 9\times 3600})$
$=\frac{3.7\times {10}^{14}\times 3.6}{0.693}[0.535-0.5]=6.91\times {10}^{13}$