Asked by William
The activity of a sample of Ba-137m has decreased by 75% from its initial activity after 5.10 minutes. What is the half-life of Ba-137m? What fraction of activity remains after 6.25 minutes?
(A=Aoe (-kt); k = .693/t1/2)
(A=Aoe (-kt); k = .693/t1/2)
Answers
Answered by
DrBob222
ln(No/N) = kt
No = 100
N = 100-75 = 25
k = ?
t = 5.10 minutes.
Solve for k, then
k = 0.693/(t<sub>1/2></sub>
No = 100
N = 100-75 = 25
k = ?
t = 5.10 minutes.
Solve for k, then
k = 0.693/(t<sub>1/2></sub>
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