1.

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

Answer»

`50%`
`20.5%`
`12.5%`
`25%`

Solution :`t_(1//2)=60 "days", "and" t=180 "days"`
Therefore, `t//t_(1//2) = (180)/(60)=3.0 " Or"N=(N_(0))/(2^(3.0))=(1)/(8)N_(0)=(1)/(8)XX100` Precent of `N_(0)=12.5%`


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