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Question
The half-life of 199Au is 2.7 days. (a) Find the activity of a sample containing 1.00 µg of 198Au. (b) What will be the activity after 7 days? Take the atomic weight of 198Au to be 198 g mol−1.
Solution
Given:-
Half-life of 199Au, T1/2= 2.7 days
Disintegration constant, `lambda = 0.693/T_(1"/"2) = 0.639/(2.7 xx 24 xx 60 xx 60) = 2.97 xx 10^-6 "s"^-1`
Number of atoms left undecayed, N = `(1 xx 10^-6 xx 6.023 xx 10^23)/198`
Now, activity, `A_0 = lambdaN`
`= (1 xx 10^-6 xx 6.023 xx 10^23)/198 xx 2.97 xx 10^-6`
= 0.244 Ci
(b) After 7 days Activity,
`A = A_0e^-"λt"`
Here, `A_0 = 0.244` Ci
`therefore A = 0.244 xx e^(-2.97 xx 10^-6 xx 7 xx 3600 xx 24)`
`= 0.244 xx e^(17962.56 xx 10^-4)`
= 0.040 Ci
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