The half-life of { }^{198} {Au} is 3 ,days. I | vedclass

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Question

$A =\lambda N$

$\lambda=\frac{\ln 2}{ t _{1 / 2}}=\frac{\ln 2}{3 	imes 24 	imes 60 	imes 60} sec ^{-1}=2.67 	imes 10^{-6} sec ^{-1}$

$N =$ N

Vedclass · Accepted Answer

$16.18$

Vedclass · Answer

$A =\lambda N$

$\lambda=\frac{\ln 2}{ t _{1 / 2}}=\frac{\ln 2}{3 	imes 24 	imes 60 	imes 60} sec ^{-1}=2.67 	imes 10^{-6} sec ^{-1}$

$N =$ Number of atoms in $2 \;mg \;Au$

$=\frac{2 	imes 10^{-3}}{198} 	imes 6 	imes 10^{23}=6.06 	imes 10^{15}$

$A=\lambda N =1.618 	imes 10^{13}=16.18 	imes 10^{12} dps$

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

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