Posted by Lee on Sunday, November 8, 2009 at 7:47pm.
a) Let the half life in years be H
2^-(1/H) = 0.935
Solve that for H.
-1/H log 2 = log 0.935
-1/H = -0.09696
H = 10.31 years
b) Solve 2^-(t/10.31) = 0.10
t will be the time required to decasy to 10% activity
You could also solve
(0.935)^t = 0.1
The answer should be the same either way you do it: about 30 years.
Thank you!
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