Half life period of `""_(53)I""^(125)` is 60 days. Percentage of radioactivity preent after 180 days is a).5 b).75 c).36 d).125

Explanation

`t_(1//2) = ` 60 days, t = 180 day
`N = ( t)/( t_((1)/(2))) = ( 180)/( 60) =3 `
`( N )/( N _(0)) = (( 1)/( 2))^(n) rArr N = N_(0) xx ( 1//2) ^(n)`
`=1 00 xx ( 1//2)^(3) = 12.5%`