Question

The half-life of a radioisotope is 10 h. Find the total number of disintegrations in the tenth hour measured from a time when the activity was 1 Ci.

Answer 1

$t_{\frac{1}{2}}=10\text{hrs,}{A}_0=1\text{Ci}$

$\text{Activity after 9 hrs}$

$A=A_0{e}^{-\lambda t}$

$=1\times e^{\frac{-6.693}{10}}\times 9$

$=e^{\frac{-6.693}{10}}\times 9$

$=0.5359=0.536\text{Ci}$

$\text{No. of atoms left after}{9}^{th}\text{hour,}$

$A_g=\lambda N_g$

$N_g=\frac{A_0}{\lambda }$

$=\frac{0.536\times 10\times 3.7\times {10}^6\times 3600}{0.693}$

$=28.6176\times {10}^{10}\times 3600$

$=103.023\times {10}^{13}$

Activity after 10 hrs

A = $A_0{e}^{-\lambda t}$

$=1\times e^{\frac{-0.698}{10}}\times 100.5\text{Ci}$

$\text{No. of atoms left after}{10}^{th}\text{hour}$

$A_{10}=\lambda N_{10}$ $N_{10}=\frac{A_{10}}{\lambda }$

$=\frac{0.5\times 3.7\times {10}^{10}\times 3600}{0.693/10}$

$=26.37\times {10}^{10}\times 3600$

$96.103\times {10}^{18}$

$\text{Number of disintegrations}$

$=(103.023-96.103)\times {10}^{18}$

= $6.92\times {10}^{18}$