The half life of a radioactive substance is `20` minutes . The approximate time interval `(t_(1) - t_(2))` 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
`14 min `
B
`20 min `
C
`28 min`
D
`7 min `
Text Solution
Verified by Experts
The correct Answer is:
B
Number of undecayed atom after time `t_(1)` `(N_(0))/(3) = N_(0) e^(-lambda_(t_2))` ….(i) Number of undecayed atom after time `t_(1)` `(2N_(0))/(3) = N_(0) e^(-lambda_(t_2))` ….(ii) from (i) ` e^(-lambda_(t)_2)) = (1)/(3)` rArr - lambda t_(2) = log _(e) ((1)/(3))` ....(ii) ` from (ii) ` e^(-lambda_(t_2)) = (2)/(3)` rArr -lambda_(t_2)) ((2)/(3))` .....(iv) solving (iii) and (iv) , we get `t_(1) - t_(2) = 20 min`
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