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

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Question

$A =\lambda N$

$N = nN _{ A }$$\quad\left( t _{1 / 2}=\frac{\ln 2}{\lambda}ight)$

$N =\left(\frac{1.5 	imes 10^{-3}}{198}ight) N _{ A }$

Vedclass · Accepted Answer

$357$

Vedclass · Answer

$A =\lambda N$

$N = nN _{ A }$$\quad\left( t _{1 / 2}=\frac{\ln 2}{\lambda}ight)$

$N =\left(\frac{1.5 	imes 10^{-3}}{198}ight) N _{ A }$

$A=\left(\frac{\ln 2}{t_{1 / 2}}ight) N$

$1$ Curie $=3.7 	imes 10^{10}\, Bq$

$A =365\, Bq$

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 .......$Ci$ $\left( N _{ A }=6 \times 10^{23}\, / mol \right)$

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