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 " day ", " "T=180 " day "`
`:. " " n=(T)/(t_(1//2))=(180)/(60)=3`
`:. " "` % of radioactivity left after 3 halves `=(N_(0))/(2^(3))=(100)/(2^(3))=12.5%`