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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Answer & Solution

Correct Answer

(B) \(16.18\)

Detailed explanation

\(A =\lambda N\)

\(\lambda=\frac{\ln 2}{ t _{1 / 2}}=\frac{\ln 2}{3 \times 24 \times 60 \times 60} sec ^{-1}=2.67 \times 10^{-6} sec ^{-1}\)

\(N =\) Number of atoms in \(2 \;mg \;Au\)

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

\(A=\lambda N =1.618 \times 10^{13}=16.18 \times 10^{12} dps\)

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