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The half life of a radioactive substance...

The half life of a radioactive substance is `20` minutes . The approximate time interval `(t_(2) - t_(1))` between the time `t_(2)` when `(2)/(3)` of it had decayed and time `t_(1)` when `(1)/(3)` of it had decay is

A

7 min

B

14 min

C

20 min

D

28 min

Text Solution

Verified by Experts

The correct Answer is:
C
`N= N_(0) e^(- lamda t)`
`(2)/(3 ) N_(0)= N_(0) e^(- lamda t_(2))` ....(i)
`(1)/(3) N_(0) = N_(0) e^(- lamda t_(1))` ...(ii)
by dividing (i) by (ii)
`rArr 2= e^(-lmada (t_(2)-t_(1))) rArr |t_(2)- t_(1)|= (l n 2)/(lamda)= 20` min
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