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

[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!

 

[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?

 

[nothing123]

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

 

[malty]

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 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.

 

[nothing123]

Can you explain it a bit more clearly? Thanks.

 

[malty]

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 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}} }

We already had:
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*

:)

 

[nothing123]

Sure no problem.

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

:)

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.

 

[malty]

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 referring 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}}}}

 

[nothing123]

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

 

[malty]

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

Well yes, if

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

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

 

[nothing123]

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

 

[malty]

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...

 

[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

 

[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?

 

[malty]

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

=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*

 

[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.