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Half life period of ""(53)I""^(125) is 6...

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

A

`50%`

B

`20.5%`

C

`12.5%`

D

`25%`

Text Solution

Verified by Experts

The correct Answer is:
C
`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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