Q.

The activity of a sample is 64 x 10-5 Ci. Its half—life is 3 days. The activity will become 5 x 10-6 Ci after

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a

12 days

b

7 days

c

18 days

d

21 days

answer is D.

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Detailed Solution

A=A012t/T1/2⇒5 x 10-6=64 x 10-512t/3 ⇒1128=12t/3⇒t=21 days
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