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Q.

A radioactive sample is undergoing α decay. At any time t1, its activity is A and another time t2 , the activity is A5 . What is the average life time for the sample?

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a

t2−t1ln 5

b

ln 5t2−t1

c

t1−t2ln 5

d

ln(t2+t1)2

answer is A.

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Detailed Solution

Activity, A=A0e−λt At time t=t1,A=A0e−λt1 At time t=t2,A5=A0e−λt2 Dividing, 5=eλt2−t1⇒ln⁡5=λt2−t1⇒λ=ln⁡5t2−t1 Now, average life =T=1λ=t2−t1ln⁡5
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