The half-life of cobalt - 60 is 5.26 years. Calculate the % activity remaining after 4 years.

Explanation

`t^(1//2)=5.26 years`
t=4 years
here to find the % of activity (i.e) to find `(N)/(N_(0))`
`log(N_(0))/(N)=(lambdaxxt)/(2.303)`
`=(0.693)/(5.26)xx(4)/(2.303)`
=0.2288
`(N_(0))/(N)` = antilog (0.2288)
=1.693
% of activity = `0.59 xx100`
=59%