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

Decay constant, `lamda=(0.693)/(T_(1//2))=(0.693)/(20)=0.03465`
Now, time of decay, `t= (2.303)/(lamda)log"(N_(0))/(N)`
`implies" "t_(1)=(2.303)/(0.03465)log"(100)/(67)=11.6` min
and `t_(2)=(2.303)/(0.03465)log"(100)/(33)=32` min
So, difference between points of time
`=t_(1)-t_(2)=32-11.6=20.4min ~~20min`