The half life period of `._(53)I^(125)` is 60 days. What % of radioactivity would be present after 240 days.

Answer

`6.25 %`

Explanation

No. of half life`=(240)/(60)=4=("Given time")/(t_(1//2))`
`N_(t)=(N_(0))/(2^(n))=(N_(0))/(2^(4))`
`100xx(N_(t))/(N_(0))=(1)/(16)xx100rArr 6.25%`