Question

As a result of being exposed to the radiation released during the Japan nuclear accident in every recent recent Sunami, the dose if iodine-131 in a person's body is 7.4 mCi $(1mCi=1\times {10}^{-3}Ci)$. Calculate the number of atome of iodine-131 $(T_{50}=8.1days)$.

• A. $N=2.765\times {10}^{14}atoms$
• B. $N=4.755\times {10}^{14}atoms$
• C. $N=6.765\times {10}^{14}atoms$
• D. None of these

Answer 1

The correct option is A $N=2.765\times {10}^{14}atoms$

$7.4mCi=7.4\times {10}^{-3}Ci=7.4\times {10}^{-3}\times 3.7\times {10}^{10}dps$
$=2.738\times {10}^{8}dps$
$\therefore -\frac{dN}{dt}=\lambda N$
$t_{50}=8.1days$
$=8.1\times 24\times 3600s$
$\therefore \lambda =\frac{0.693}{t_{50}}$
$2.738\times {10}^8=\frac{0.693N}{8.1\times 24\times 3600}$
$N=2.765\times {10}^{14}atoms$