The half-life

Question:

The half-life of $\mathrm{Au}^{198}$ is $2.7$ days. The activity of $1.50 \mathrm{mg}$ of $\mathrm{Au}^{198}$ if its atomic weight is $198 \mathrm{~g} \mathrm{~mol}^{-1}$ is, $\left(\mathrm{N}_{\mathrm{A}}=6 \times 10^{23} / \mathrm{mol}\right)$

1. (1) $240 \mathrm{Ci}$

2. (2) $357 \mathrm{Ci}$

3. (3) $535 \mathrm{Ci}$

4. (4) $252 \mathrm{Ci}$

Correct Option: , 2

Solution:

(2)

$\mathrm{A}=\lambda \mathrm{N}$

$\mathrm{N}=\mathrm{nN}_{\mathrm{A}}$

$\mathrm{N}=\left(\frac{1.5 \times 10^{-3}}{198}\right) \mathrm{N}_{\mathrm{A}}$

$\mathrm{A}=\left(\frac{\ln 2}{\mathrm{t}_{1 / 2}}\right) \mathrm{N}$

1 Curie $=3.7 \times 10^{10} \mathrm{~Bq}$

$\mathrm{~A}=365 \mathrm{~Bq}$