F-18 has the half-life of 109.8 min what percentage of F-19 sample remains after

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Explanation / Answer

Radioactive disintegration is first order.

thus the rate law is

k = (2.303/t) log (a/a-x)

given t1/2 = 109.8 min , hence k = 0.693/t1/2  = 0.693/109.8 min-1

i) k = (2.303/t) log (a/a-x)

0.693/ 109.8 = (2.303/24min) log (a/a-x)

hence (a/a-x) = 1.163

since a = 1

1/1-x = 1.163

and x= 0.14 and percentae = 0.14x100= 14%

After 24 min 14% of the sample is disintegrated.

hence 100-14 = 86% ample remains

ii) in a similar manner after 24 hours

0.693/ 109.8 = (2.303/24x 60min) log (a/a-x)

1/1-x = 8837

and x= 0.9999 and percentage of disintegrated = 99.99%

So percentage of sample remaining is 0.01%

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