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

1. May 11, 2008

### nothing123

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!

2. May 11, 2008

### malty

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

$$-\frac{dN_{(t)}}{dt}=\lambda N_{(t)}$$

Now can you derive it?

3. May 11, 2008

### nothing123

Using what you gave, I am able to derive N = N0e^-kt but the equation I provided is without the decay constant...

4. May 11, 2008

### malty

Well the half life will be when $$N_{(t)}=\frac{N_0}{2}$$. and when $$t=T_{\frac{1}{2}}}$$.

/Not sure how the decay constant disappears tbh.

/No wait I see now.

Last edited: May 11, 2008
5. May 11, 2008

### nothing123

Can you explain it a bit more clearly? Thanks.

6. May 11, 2008

### malty

Sure no problem.

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

So we have:
$$\frac{N_o}{2}=N_o e^{-\lambda T_{\frac{1}{2}}}$$
$$\frac{1}{2}=e^{-\lambda T_{\frac{1}{2}} }$$

$$N_{(t)}=\frac{N_o}{2})$$
but
$$e^{-\lambda T_{\frac{1}{2}}}=\frac{1}{2}$$

so $$-\lambda T_{\frac{1}{2}}= L_n(\frac{1}{2}})$$

$$e^{-\lambda}=\frac{e^{L_n(\frac{1}{2})}}{e^{T_{\frac{1}{2}}}}= \frac{1}{2}e^{-T_{\frac{1}{2}}}$$

Sub that into the equation $$N = N_0e^{-\lambda t}$$
and you got it :D

*Phew that took some time to type*

:)

Last edited: May 11, 2008
7. May 11, 2008

### nothing123

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. May 11, 2008

### malty

Not 100% sure what you are refering to.

Here's how I got that line:

$$-\lambda T_{\frac{1}{2}}= L_n(\frac{1}{2}})$$
and
$${-\lambda}=\frac{L_n(\frac{1}{2})}{{T_{\frac{1}{2}}}}$$

9. May 11, 2008

### nothing123

Right, so does e^(ln(1/2)/T1/2) = e^ln(1/2)/e^T1/2?

10. May 11, 2008

### malty

Well yes, if

$$e^{ab}=e^a*e^b$$

then $$e^{-ab}=e^a*e^{-b}=\frac{e^a}{e^b}$$

11. May 11, 2008

### nothing123

Wait, doesn't e^a*e^b = e^(a+b) not e^ab? e^ab would be (e^a)^b, right?

12. May 11, 2008

### malty

Lol yeah your right is does, I have absolutely no idea what made me think it was well that...

Last edited: May 11, 2008
13. May 11, 2008

### malty

Ahh now I remember the equation I used was

$${-\lambda}=\frac{L_n(\frac{1}{2})}{{T_{\frac{1}{2}}} }$$

I took the exponent of this:

$$e^{-\lambda}=\frac{e^{L_n(\frac{1}{2})}}{e^{T_{\frac{1 }{2}}}}= \frac{1}{2}e^{-T_{\frac{1}{2}}}$$

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. May 11, 2008

### nothing123

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. May 11, 2008

### malty

No this is actually true because

$$=e^{-\lambda}=e^{- \frac{ ln\frac{1}{2}}{T_{\frac{1}{2}}}}=e^{(ln\frac{1}{2})^{-T_{\frac{1}{2}}}}=\frac{1}{2}e^{-T_{\frac{1}{2}}}}$$

Least I think it, *is brain dead*

16. May 11, 2008

### nothing123

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.