1.

The half-life of cobalt - 60 is 5.26 years. Calculate the % activity remaining after 4 years.

Answer»

SOLUTION :`t^(1//2)=5.26 years`
t=4 years
here to findthe % of activity (i.e) to find`(N)/(N_(0))`
`log(N_(0))/(N)=(lambdaxxt)/(2.303)`
`=(0.693)/(5.26)xx(4)/(2.303)`
=0.2288
`(N_(0))/(N)` = ANTILOG (0.2288)
=1.693
% of activity = `0.59 XX100`
=59%


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