The half-life of {}^{198} {Au} is 3 , text{da | vedclass

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(B) The activity $A$ is given by $A = \lambda N$.First, calculate the decay constant $\lambda$:$\lambda = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{3 	ime

Vedclass · Accepted Answer

$16.18$

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(B) The activity $A$ is given by $A = \lambda N$.First, calculate the decay constant $\lambda$:$\lambda = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{3 	imes 24 	imes 3600 \, 	ext{s}} \approx 2.67 	imes 10^{-6} \, 	ext{s}^{-1}$.Next, calculate the number of atoms $N$ in $2 \, 	ext{mg}$ of ${}^{198} {Au}$:$N = \frac{	ext{mass}}{	ext{molar mass}} 	imes N_A = \frac{2 	imes 10^{-3} \, 	ext{g}}{198 \, 	ext{g/mol}} 	imes 6.022 	imes 10^{23} \, 	ext{atoms/mol} \approx 6.08 	imes 10^{18} \, 	ext{atoms}$.(Using $N_A \approx 6 	imes 10^{23}$ as per standard approximation in such problems):$N = \frac{2 	imes 10^{-3}}{198} 	imes 6 	imes 10^{23} \approx 6.06 	imes 10^{18} \, 	ext{atoms}$.Now, calculate the activity $A$:$A = \lambda N = (2.67 	imes 10^{-6} \, 	ext{s}^{-1}) 	imes (6.06 	imes 10^{18}) \approx 16.18 	imes 10^{12} \, 	ext{disintegrations/second}$.

The half-life of ${}^{198} {Au}$ is $3 \, \text{days}$. If the atomic weight of ${}^{198} {Au}$ is $198 \, \text{g/mol}$, then the activity of $2 \, \text{mg}$ of ${}^{198} {Au}$ is ..... $\times 10^{12} \, \text{disintegration/second}$.

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