If the half-life of a radioactive nucleus is 3 days, nearly what fraction of the initial number of nuclei will decay on the 3rd day? (Given, `3sqrt(0.25)~~ 0.63`).

Options

A. 0.63
B. 0.37
C. 0.5
D. 0.13

Answer

D

Explanation

We know, in case of radioactive decay,
`N=N_0e^(-lambdat)`
Again, `lambda="In2"/T_(1//2)`
Given, `T_(1//2)` = 3 day
`therefore lambda="In2"/3`
The fraction of the initial number of nuclei will decay on the 3rd day
`=(N_0e^(-"In2"/3xx2)-N_0e^(-"In2"/3xx3))/N_0 =e^(-"2In2"/3) -e^("3In2"/3)`
`=2^(-2/3)-2^(-1)` = 0.13