Given that the half-life of ^ 198 Au is 2.7 days , determine the activity of a sample which contains 1.00 of ^ 198 Au . Consider the atomic weight of ^ 198 Au to be 198 g mol^ -1 .

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Quantrex Academy · Accepted Answer

Correct answer: C. 0.244 Ci

The number of atoms in the sample is given by:

N = \dfrac{m}{M} \times N_A = \dfrac{1.00 \times 10^{-6}}{198} \times 6.022 \times 10^{23} \approx 3.041 \times 10^{15} \text{ atoms}

The decay constant is:

\lambda = \dfrac{\ln 2}{T_{1/2}} = \dfrac{0.693}{2.7 \times 24 \times 3600} \approx 2.97 \times 10^{-6} \text{ s}^{-1}

The activity of the sample is:

A = \lambda N = (2.97 \times 10^{-6}) \times (3.041 \times 10^{15}) \approx 9.03 \times 10^9 \text{ Bq}

Converting the activity to Curies (

1 \text{ Ci} = 3.7 \times 10^{10} \text{ Bq}

):

A = \dfrac{9.03 \times 10^9}{3.7 \times 10^{10}} \approx 0.244 \text{ Ci}

Given that the half-life of ^ 198 Au is 2.7 days , determine the...

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