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, $\left(\mathrm{N}_{\mathrm{A}}=6 \times 10^{23} / \mathrm{mol}\right)$
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 $=3.7 \times 10^{10} \mathrm{~Bq}$
$\mathrm{~A}=365 \mathrm{~Bq}$
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