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Question

Half life of a radioactive substance is 20 minutes. The time between 20 % and 80% decay will be:

A
40 minutes
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B
20 minutes
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C
25 minutes
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D
30 minutes
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Solution

The correct option is A 40 minutes
Let the initial activity of the radioactive substance be Ro and activity be R after time t.
Given : T1/2=20 min
Using λt=ln(RoR) where λ=0.693T1/2

0.69320t=ln(RoR)

OR 0.69320(t2t1)=ln(R1R2) .........(1)
Let time be t1 for 20 % decay i.e R1=Ro0.2Ro=0.8Ro
Also the time be t2 for 80 % decay i.e R2=Ro0.8Ro=0.2Ro

0.69320(t2t1)=ln(0.80.2)

(t2t1)=200.693×ln4=40 min (ln4=1.386)

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