Question

The half-life of $A{u}^{198}$ is $2.7$ days. The activity of $1.50\text{mg}$ of $A{u}^{198}$, if its atomic weight is $198{\text{g mol}}^{-1}$is:
$(N_A=6\times {10}^{23}/\text{mol})$.

• A. $252\text{Ci}$
• B. $357\text{Ci}$
• C. $240\text{Ci}$
• D. $535\text{Ci}$

Answer 1

The correct option is B $357\text{Ci}$

Activity is given as: $$A=\lambda N$$$A=\frac{ln2}{T_{1/2}}\times \frac{1.5\times {10}^{-3}}{198}\times 6\times {10}^{23}\text{decay/s}$

$A=\frac{0.693}{2.7\times 24\times 3600}\times \frac{1.5\times {10}^{-3}}{198}\times \frac{6\times {10}^{23}}{3.7\times {10}^{10}}\text{Ci}$

$A=365\text{Ci}$

$\therefore$ Approximately the correct option is (b).