The half-life of Au^{198} is 2.7 , days. The | vedclass

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(B) The activity $A$ is given by $A = \lambda N$, where $\lambda = \frac{\ln 2}{T_{1/2}}$.First, convert the half-life to seconds: $T_{1/2} = 2.7 	im

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(B) The activity $A$ is given by $A = \lambda N$, where $\lambda = \frac{\ln 2}{T_{1/2}}$.First, convert the half-life to seconds: $T_{1/2} = 2.7 	imes 24 	imes 3600 \, s = 233280 \, s$.The number of atoms $N$ is given by $N = \frac{m}{M} 	imes N_A = \frac{1.5 	imes 10^{-3} \, g}{198 \, g/mol} 	imes 6 	imes 10^{23} \, mol^{-1} \approx 4.545 	imes 10^{18} \, atoms$.The decay constant $\lambda = \frac{0.693}{233280 \, s} \approx 2.97 	imes 10^{-6} \, s^{-1}$.The activity $A = \lambda N = (2.97 	imes 10^{-6} \, s^{-1}) 	imes (4.545 	imes 10^{18}) \approx 1.35 	imes 10^{13} \, Bq$.Since $1 \, Ci = 3.7 	imes 10^{10} \, Bq$, the activity in Curies is $A = \frac{1.35 	imes 10^{13}}{3.7 	imes 10^{10}} \approx 365 \, Ci$.

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

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