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Derivation for this equation: N = N0(1/2)^t/t1/2?

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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!
 

Answers and Replies

  • #2
malty
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Well, we know that the rate of radioactive decay is proportional to the Number of particles at a time t.
So:

[tex] -\frac{dN_{(t)}}{dt}=\lambda N_{(t)} [/tex]

Now can you derive it?
 
  • #3
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Using what you gave, I am able to derive N = N0e^-kt but the equation I provided is without the decay constant...
 
  • #4
malty
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Using what you gave, I am able to derive N = N0e^-kt but the equation I provided is without the decay constant...
Well the half life will be when [tex]N_{(t)}=\frac{N_0}{2}[/tex]. and when [tex]t=T_{\frac{1}{2}}}[/tex].

/Not sure how the decay constant disappears tbh.

/No wait I see now.
 
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  • #5
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Can you explain it a bit more clearly? Thanks.
 
  • #6
malty
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Can you explain it a bit more clearly? Thanks.
Sure no problem.

Half life is the when the number of particles is reduced by half.
Hence this occurs when [tex] N_{(t)}=\frac{N_o}{2})[/tex] No is the original number of particles.
The time which this occurs will be the half life and we call it [tex] T_{\frac{1}{2}}[/tex]

So we have:
[tex]\frac{N_o}{2}=N_o e^{-\lambda T_{\frac{1}{2}}}[/tex]
[tex]\frac{1}{2}=e^{-\lambda T_{\frac{1}{2}} }[/tex]

We already had:
[tex] N_{(t)}=\frac{N_o}{2})[/tex]
but
[tex] e^{-\lambda T_{\frac{1}{2}}}=\frac{1}{2}[/tex]

so [tex] -\lambda T_{\frac{1}{2}}= L_n(\frac{1}{2}}) [/tex]

[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

*Phew that took some time to type*

:)
 
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  • #7
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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

:)
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.
 
  • #8
malty
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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.
Not 100% sure what you are refering to.

Here's how I got that line:

[tex] -\lambda T_{\frac{1}{2}}= L_n(\frac{1}{2}}) [/tex]
and
[tex] {-\lambda}=\frac{L_n(\frac{1}{2})}{{T_{\frac{1}{2}}}}[/tex]
 
  • #9
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Right, so does e^(ln(1/2)/T1/2) = e^ln(1/2)/e^T1/2?
 
  • #10
malty
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Right, so does e^(ln(1/2)/T1/2) = e^ln(1/2)/e^T1/2?
Well yes, if

[tex] e^{ab}=e^a*e^b [/tex]

then [tex] e^{-ab}=e^a*e^{-b}=\frac{e^a}{e^b}[/tex]
 
  • #11
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Wait, doesn't e^a*e^b = e^(a+b) not e^ab? e^ab would be (e^a)^b, right?
 
  • #12
malty
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Wait, doesn't e^a*e^b = e^(a+b) not e^ab? e^ab would be (e^a)^b, right?
Lol yeah your right is does, I have absolutely no idea what made me think it was well that...

:redface:
 
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  • #13
malty
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Ahh now I remember the equation I used was

[tex] {-\lambda}=\frac{L_n(\frac{1}{2})}{{T_{\frac{1}{2}}} }[/tex]

I took the exponent of this:




[tex] e^{-\lambda}=\frac{e^{L_n(\frac{1}{2})}}{e^{T_{\frac{1 }{2}}}}= \frac{1}{2}e^{-T_{\frac{1}{2}}}[/tex]

And that is still correct, I believe, my whole escapade with the awful exponent rules was well becasue...I'm tired (and dreaming) :p

:smile:
 
  • #14
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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?
 
  • #15
malty
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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?
No this is actually true because

[tex]=e^{-\lambda}=e^{- \frac{ ln\frac{1}{2}}{T_{\frac{1}{2}}}}=e^{(ln\frac{1}{2})^{-T_{\frac{1}{2}}}}=\frac{1}{2}e^{-T_{\frac{1}{2}}}}[/tex]

Least I think it, *is brain dead*
 
  • #16
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Ok, I think I figured it out but I don't think the equation you just wrote is correct. e^(ln1/2)/T1/2 = 1/2^(T1/2^-1). Now, subbing this into our original equation, we get N = N0(1/2)^t/t1/2. Thanks so much for your help anyways, it got my brain going.
 

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