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

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kkidBest ResponseYou've already chosen the best response.1
I have the decay constant and need to find out how long it will take for the element to decay from 5ug to 1ug
 one year ago

henpenBest ResponseYou've already chosen the best response.1
Do you know differential equations or calculus?
 one year ago

kkidBest ResponseYou've already chosen the best response.1
Yes, but I don't think this question requires it
 one year ago

kkidBest ResponseYou've already chosen the best response.1
The question I'm tryingt o answer is: TA sample of 24Na has a halflife of 234 hours, How much time elapses before a 5μg sample contains 1μg of undecayed atoms?
 one year ago

henpenBest ResponseYou've already chosen the best response.1
\[\frac{1}{c}e^{t \lambda}=ke^{t \lambda}=N\]
 one year ago

kkidBest ResponseYou've already chosen the best response.1
but I want to know how to do it myself
 one year ago

henpenBest ResponseYou've already chosen the best response.1
Sorry, it deleted half of my answer
 one year ago

kkidBest ResponseYou've already chosen the best response.1
I have used a similar equation \[N = N _{0}e ^{\lambda t}\]
 one year ago

henpenBest ResponseYou've already chosen the best response.1
http://upload.wikimedia.org/math/a/8/a/a8a0bd0a12874474352b307d9e919076.png \[\frac{1}{\lambda}(lnN+lnk)=\frac{1}{\lambda}(\ln(kN))=t\] \[(\ln(kN))=\lambda t\] \[kN=e^{\lambda t}\] \[N=ce^{\lambda t}\]
 one year ago

henpenBest ResponseYou've already chosen the best response.1
Set c= sample at t=0 That is, \[ c= N_0 \]
 one year ago

henpenBest ResponseYou've already chosen the best response.1
That's the derivation of it
 one year ago

kkidBest ResponseYou've already chosen the best response.1
ah, so its the same as my equation?
 one year ago

henpenBest ResponseYou've already chosen the best response.1
You're given the half life, so I'd convert it into 2^... instead of e^....
 one year ago

kkidBest ResponseYou've already chosen the best response.1
I don't know how to get the values of N and N0 though, because that is the number of atoms, not the mass
 one year ago

kkidBest ResponseYou've already chosen the best response.1
hendce why I haven't already used that equation
 one year ago

henpenBest ResponseYou've already chosen the best response.1
\[N=N_0 (2^{\log_2(e)})^{\lambda t}=N_02^{\log_2(e) \lambda t}\]
 one year ago

henpenBest ResponseYou've already chosen the best response.1
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
 one year ago

kkidBest ResponseYou've already chosen the best response.1
but N refers to atoms, not mass (which is the 5 and the 1)
 one year ago

henpenBest ResponseYou've already chosen the best response.1
So obviously the halflife \[= \log_2(e) \lambda\]
 one year ago

kkidBest ResponseYou've already chosen the best response.1
More specifically Sodium24
 one year ago

henpenBest ResponseYou've already chosen the best response.1
\[Moles= \frac{mass_{grams}}{A_r}\] \[Atoms= Moles* Avogadro'snumber\]
 one year ago

kkidBest ResponseYou've already chosen the best response.1
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)
 one year ago

henpenBest ResponseYou've already chosen the best response.1
So initially the mass (N_0)\[=\frac{5*10^{6}}{24}6.0221415 *10^{23}\]
 one year ago

henpenBest ResponseYou've already chosen the best response.1
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?
 one year ago

kkidBest ResponseYou've already chosen the best response.1
both, set an exam question for HW
 one year ago

kkidBest ResponseYou've already chosen the best response.1
I'll try working it out with the arogarrdo number (or whatever it is called lol)
 one year ago

henpenBest ResponseYou've already chosen the best response.1
Anyway, \[\huge N=125461281250000000*2^{\frac{t_{hours}}{234}}\]
 one year ago

henpenBest ResponseYou've already chosen the best response.1
Sorry, you're right, you don't need AN
 one year ago

henpenBest ResponseYou've already chosen the best response.1
You just need\[\large 2^{\frac{t}{234}}=\frac{1}{5}\]
 one year ago

henpenBest ResponseYou've already chosen the best response.1
\[t=234(\log_2(0.2))\]
 one year ago

kkidBest ResponseYou've already chosen the best response.1
close enough (rounding errors)
 one year ago

henpenBest ResponseYou've already chosen the best response.1
So yes. Did you do it using the equation 8 posts ago?
 one year ago

kkidBest ResponseYou've already chosen the best response.1
sorry about my terrible handwriting
 one year ago

henpenBest ResponseYou've already chosen the best response.1
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\[N=N_0 \frac{1}{5}\]
 one year ago

kkidBest ResponseYou've already chosen the best response.1
ah, i think I see what you have done now
 one year ago

kkidBest ResponseYou've already chosen the best response.1
where has halflife came into it?
 one year ago

kkidBest ResponseYou've already chosen the best response.1
(it seems a way better way to use it)
 one year ago
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