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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

`25%`

B

`12.5%`

C

`50%`

D

`75%`

Text Solution

Verified by Experts

The correct Answer is:
B
`N = (N_0)/(2^n) n = (180)/(60) = 3 `
`N/N_0 = 1/(2^3) implies N/N_0 xx 100 = 1/8 xx 100 = 12.5 %`
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