Half-life of a radioactive substance is `20` minutes. Difference between points of time when it is `33 %` disintegrated and `67 %` disintegrated is approximate.

Options

A. 10 min
B. 20 min
C. 30 min
D. 40 min

Answer

B

Explanation

`lamda= (0.693)/(T_(1//2)) = (0.693)/(20)= 0.03465`
`N= N_(0)e^(-lamda t) rArr t= (2.303)/(lamda ) "log" (N_(0))/(N)`
Now `t_(1)= (2.303)/(0.03465) "log" (100)/(33)= 32` min
`t_(2) = (2.303)/(lamda) "log" (100)/(67)= 11.6` min
Time difference `=t_(1)-t_(2)=32-11.6= 20.4` min `~~20` min