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

Options

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

Answer

D

Explanation

Let `N_(0)` be the number of atoms of a radioactive nucleus initially and N be the number at the end of `3^(rd)` day.
Using `((N)/(N_(0))) = ((1)/(2))^(t//T_(1//2))`
No. of nucleus at the `2^(nd)` day, `(N_(2))/(N_(0)) = ((1)/(2))^(2//3) rArr N_(2) = ((1)/(4))^(1//3) N_(0)`
No. of nucleus at the end of `3^(rd)` days, `N_(3) = ((1)/(2))N_(0)`
`therefore` Fraction of the initial number of nuclei will decay on `3^(rd)` days, `N' = (N_(3) - N_(2))/(N_(0)) ~~ 0.13`.