- #1

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Thanks!

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- Thread starter nothing123
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- #1

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Thanks!

- #2

malty

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So:

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

Now can you derive it?

- #3

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- #4

malty

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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...

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- #13

malty

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[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

- #14

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- #15

malty

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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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