How can I calculate the time taken for an element to decay by a certain amount?

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How can I calculate the time taken for an element to decay by a certain amount?

Physics
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I have the decay constant and need to find out how long it will take for the element to decay from 5ug to 1ug
Do you know differential equations or calculus?
Yes, but I don't think this question requires it

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The question I'm tryingt o answer is: TA sample of 24Na has a half-life of 234 hours, How much time elapses before a 5μg sample contains 1μg of undecayed atoms?
A sample of*
\[\frac{1}{c}e^{-t \lambda}=ke^{-t \lambda}=N\]
but I want to know how to do it myself
whats c?
Sorry, it deleted half of my answer
lol, ok
I have used a similar equation \[N = N _{0}e ^{-\lambda t}\]
...in the past
http://upload.wikimedia.org/math/a/8/a/a8a0bd0a12874474352b307d9e919076.png \[\frac{-1}{\lambda}(lnN+lnk)=\frac{-1}{\lambda}(\ln(kN))=t\] \[(\ln(kN))=-\lambda t\] \[kN=e^{-\lambda t}\] \[N=ce^{-\lambda t}\]
Set c= sample at t=0 That is, \[ c= N_0 \]
That's the derivation of it
ah, so its the same as my equation?
You're given the half life, so I'd convert it into 2^... instead of e^....
Yes
I don't know how to get the values of N and N0 though, because that is the number of atoms, not the mass
hendce why I haven't already used that equation
\[N=N_0 (2^{\log_2(e)})^{-\lambda t}=N_02^{-\log_2(e) \lambda t}\]
N_0= 5μg Just call it '5'- It doesn't matter which units you use (I suppose moles or no. of atoms would be most natural, though) as long as you're consistent
but N refers to atoms, not mass (which is the 5 and the 1)
So obviously the half-life \[= \log_2(e) \lambda\]
Which element is it?
Na (Sodium)
More specifically Sodium-24
\[Moles= \frac{mass_{grams}}{A_r}\] \[Atoms= Moles* Avogadro's-number\]
A.N.= 6.0221415 *10^23
I thought about that, but in my questions, avogadros number isn't mentioned until the next part (therefore, due to my experience with exam paper questions, it isn't needed until then)
So initially the mass (N_0)\[=\frac{5*10^{-6}}{24}6.0221415 *10^{23}\]
I'm not sure if there's any other (fundamental- i.e. not derived from it) way of working it out. Is this Homework or an exam paper?
both, set an exam question for HW
I'll try working it out with the arogarrdo number (or whatever it is called lol)
Anyway, \[\huge N=125461281250000000*2^{-\frac{t_{hours}}{234}}\]
Sorry, you're right, you don't need AN
You just need\[\large 2^{-\frac{t}{234}}=\frac{1}{5}\]
\[t=-234(\log_2(0.2))\]
I got t = 1955669
:/
=543.33 hours
is my answer right?
Its in seconds
I got 1955992.2
close enough (rounding errors)
So yes. Did you do it using the equation 8 posts ago?
*9
no, I did this...
|dw:1351694706420:dw|
sorry about my terrible handwriting
I suppose that's possible, but it's not necessary to use Avoardo's number. All you need is \[N=N_0 2^{-t/k}\] and\[N=N_0 \frac{1}{5}\]
k=234
ah, i think I see what you have done now
the N0 will cancel
where has half-life came into it?
(it seems a way better way to use it)

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