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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

`20` min

B

`28` min

C

`7` min

D

`14` min

Text Solution

Verified by Experts

The correct Answer is:
A

`:' N/N_(0)=[1/2]^(1//T) :. 1/3 =[1/2]^(1_(1)/T)` & `2/3=[1/2]^(1/T)`
`=1/2=[1/2]^((1_(2)-1_(1))1/T)implies 1=((1_(2)-t_(1)))/Timplies T=t_(2)-t_(1)`
`implies t_(2)-t_(1)=20` min
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