1.

The half-life period of ._(53)I^(125) is 60 days. What percent of radioactivity would be present after 180 days

Answer»

0.25
`12.5%`
0.5
0.75

Solution :`N = (N_(0))/(2^(n)) n = (180)/(60) = 3`
`(N)/(N_(0)) = (1)/(2^(3)) rArr (N)/(N_(0)) xx 100 = (1)/(8) xx 100 = 12.5%`


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