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Two species of radioactive atoms are mix...

Two species of radioactive atoms are mixed in equal number. The disintegration of the first species is `lambda` and of the second is `lambda`/3. After a long time the mixture will behave as a species with mean life of approximately

A

`(0.70)/(lamda)`

B

`(2.10)/(lamda)`

C

`(1.00)/(lamda)`

D

`(0.52)/(lamda)`

Text Solution

Verified by Experts

The correct Answer is:
B
Avg life `= (int t dN_(1) + tdN_(2))/(2N_(0)) rArr dN_(2)= (lamda)/(3) N_(0) e^((-lamda t)/(3)) dt`
Avg life `= (underset(0)overset(oo)int t(lamda N_(0) e^(-lamda t) dt + (lamda)/(3) N_(0) e^((-lamda t)/(3)) dt))/(2N_(0))`
On Integration, we get
Average life `= (2)/(lamda) ~~ (2.10)/(lamda)`
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