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 [in disintegration/second]:

Question

Accepted Answer

Correct answer: D. (d) A= λN = ln2t_1/2= ln23 24 60 60 sec^-1=2.67 10^-6 sec^-1 N = Number of atoms in 2mg Au = 2 10^-3198 6 10^23=6.06 10^15 A= λN=1618 10^13=16.18 10^12 dps

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

Solution

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