**The half-life of Au¹⁹⁸ is 2.7 days. The activity of 1.50 mg of Au¹⁹⁸ if its atomic weight is 198 g mol⁻¹ is, (Nₐ = 6*10²³ / mol)**

Option 1 - 240 Ci

Option 2 - 252 Ci

Option 3 - 535 Ci

Option 4 - 357 Ci

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Answered by

Alok Kumar Singh | Contributor-Level 10

6 months ago

Correct Option - 4

Detailed Solution:

A = Activity = λN = (ln (2)/t? /? ) N
N = (1.5*10? ³ / 198) * 6*10²³
A = (0.693 / (2.7*24*3600) * (1.5*10? ³ / 198) * 6*10²³ disintegration/s
A ≈ 1.32 * 10¹³ Bq = (1.32*10¹³ / 3.7*10¹? ) Ci ≈ 357 Ci

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