1.

Half-life of a radioactive substance is `20` minutes. Difference between points of time when it is `33 %` disintegrated and `67 %` disintegrated is approximate.A. 10 minB. 20 minC. 30 minD. 40 min

Answer» Correct Answer - B
(b) `lamda = (0.693)/(T_(1//2)) = (0.693)/(20) = 0.03465`
Now time od decay `t = (2.303)/(lamda) log ((N_0)/(N))`
`rArr t_1 = (2.303)/(0.03465) log((100)/(67)) = 11.6 min`
and `t_2 = (2.303)/(0.03465) log ((100)/(33)) = 32 min`
Thus time difference between points of time
=`t_1 - t_2 = 32 - 11.6 = 20.4 min ~~ 20 min`.


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