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malawi_glenn
#8
Jun6-08, 07:58 AM
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Quote Quote by SReinhardt View Post
Wouldn't there have to be a negative sign in there somewhere >_>

I believe you have to use integrals to solve that, which I haven't done yet.
yeah it should have a minus sign, good! :-)

Solving this:

[tex] \int N ^{-1}dN = - \int \lambda dt [/tex]

[tex] \ln(N(t)) - \ln(N(0)) = -\lambda t [/tex]

[tex] \ln(N(t)/N(0)) = -\lambda t [/tex]

[tex] N(t)/N(0) = e^{-\lambda t } [/tex]

[tex] N(t) = N(0) e^{-\lambda t } [/tex]

Lambda is the number of decays per unit time, is related to half life by:
[tex] \lambda = \frac{\ln 2}{T_{1/2}} [/tex]