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nothing123
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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!
nothing123 said:Using what you gave, I am able to derive N = N0e^-kt but the equation I provided is without the decay constant...
nothing123 said:Can you explain it a bit more clearly? Thanks.
malty said: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
:)
nothing123 said: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.
nothing123 said:Right, so does e^(ln(1/2)/T1/2) = e^ln(1/2)/e^T1/2?
nothing123 said:Wait, doesn't e^a*e^b = e^(a+b) not e^ab? e^ab would be (e^a)^b, right?
nothing123 said:Hmm 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?
In this equation, N represents the final amount of a substance, N0 represents the initial amount of the substance, t represents the time elapsed, and t1/2 represents the half-life of the substance.
This equation is derived from the exponential decay model, which assumes that the rate of decay of a substance is proportional to the amount of the substance present. By solving the differential equation for exponential decay, the equation N = N0(1/2)^t/t1/2 is obtained.
Yes, this equation can be used for any type of substance that follows an exponential decay model, such as radioactive isotopes or pharmaceutical drugs.
To use this equation, simply plug in the initial amount of the substance (N0) and the time elapsed (t) into the equation. The resulting value of N will be the amount of the substance remaining after that amount of time.
This equation is a simplified model that assumes a constant decay rate and does not take into account any external factors that may affect the decay. Therefore, it may not be accurate for all situations, but it can provide a good estimate for many scenarios.