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Anyone know the derivation for this equation: N = N0(1/2)^t/t1/2? I can understand it plugging numbers in but I don't really know how it was derived in the first place.
Thanks!
Thanks!
Well the half life will be when [tex]N_{(t)}=\frac{N_0}{2}[/tex]. and when [tex]t=T_{\frac{1}{2}}}[/tex].Using what you gave, I am able to derive N = N0e^-kt but the equation I provided is without the decay constant...
Sure no problem.Can you explain it a bit more clearly? Thanks.
Thanks for the reply but I'm not entirely sure how you got the above lines. My exponent rules might be a big hazy but I don't think e^a/b is the same as e^a/e^b if you know what I'm saying.Sure no problem.
[tex] e^{-\lambda}=\frac{e^{L_n(\frac{1}{2})}}{e^{T_{\frac{1}{2}}}}= \frac{1}{2}e^{-T_{\frac{1}{2}}}[/tex]
Sub that into the equation [tex]N = N_0e^{-\lambda t} [/tex]
and you got it :D
:)
Not 100% sure what you are refering to.Thanks for the reply but I'm not entirely sure how you got the above lines. My exponent rules might be a big hazy but I don't think e^a/b is the same as e^a/e^b if you know what I'm saying.
Well yes, ifRight, so does e^(ln(1/2)/T1/2) = e^ln(1/2)/e^T1/2?
Lol yeah your right is does, I have absolutely no idea what made me think it was well that...Wait, doesn't e^a*e^b = e^(a+b) not e^ab? e^ab would be (e^a)^b, right?
No this is actually true becauseHmm I don't know if I'm just tired too or what but we seem to be back at square one. Didn't we conclude that e^(ln(1/2)/T1/2) does not equal e^ln(1/2)/e^T1/2?