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The half-life of

Question:

The half-life of ${ }^{198} \mathrm{Au}$ is 3 days. If atomic weight of ${ }^{198} \mathrm{Au}$ is $198 \mathrm{~g} / \mathrm{mol}$ then the activity of $2 \mathrm{mg}$ of ${ }^{198} \mathrm{Au}$ is [in disintegration/second] :

  1. $2.67 \times 10^{12}$

  2. $6.06 \times 10^{18}$

  3. $32.36 \times 10^{12}$

  4. $16.18 \times 10^{12}$


Correct Option: , 4

Solution:

$\mathrm{A}=\lambda \mathrm{N}$

$\lambda=\frac{\ln 2}{t_{1 / 2}}=\frac{\ln 2}{3 \times 24 \times 60 \times 60} \mathrm{sec}^{-1}=2.67 \times 10^{-6} \mathrm{sec}^{-1}$

$\mathrm{N}=$ Number of atoms in $2 \mathrm{mg} \mathrm{Au}$

$=\frac{2 \times 10^{-3}}{198} \times 6 \times 10^{23}=6.06 \times 10^{15}$

$A=\lambda N=1.618 \times 10^{13}=16.18 \times 10^{12} \mathrm{dps}$

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