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

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I have the decay constant and need to find out how long it will take for the element to decay from 5ug to 1ug

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Do you know differential equations or calculus?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yes, but I don't think this question requires it

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The 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?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\frac{1}{c}e^{t \lambda}=ke^{t \lambda}=N\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but I want to know how to do it myself

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Sorry, it deleted half of my answer

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I have used a similar equation \[N = N _{0}e ^{\lambda t}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0http://upload.wikimedia.org/math/a/8/a/a8a0bd0a12874474352b307d9e919076.png \[\frac{1}{\lambda}(lnN+lnk)=\frac{1}{\lambda}(\ln(kN))=t\] \[(\ln(kN))=\lambda t\] \[kN=e^{\lambda t}\] \[N=ce^{\lambda t}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Set c= sample at t=0 That is, \[ c= N_0 \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0That's the derivation of it

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ah, so its the same as my equation?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0You're given the half life, so I'd convert it into 2^... instead of e^....

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I don't know how to get the values of N and N0 though, because that is the number of atoms, not the mass

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hendce why I haven't already used that equation

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[N=N_0 (2^{\log_2(e)})^{\lambda t}=N_02^{\log_2(e) \lambda t}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0N_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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but N refers to atoms, not mass (which is the 5 and the 1)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So obviously the halflife \[= \log_2(e) \lambda\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0More specifically Sodium24

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[Moles= \frac{mass_{grams}}{A_r}\] \[Atoms= Moles* Avogadro'snumber\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0A.N.= 6.0221415 *10^23

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I 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)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So initially the mass (N_0)\[=\frac{5*10^{6}}{24}6.0221415 *10^{23}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I'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?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0both, set an exam question for HW

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I'll try working it out with the arogarrdo number (or whatever it is called lol)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Anyway, \[\huge N=125461281250000000*2^{\frac{t_{hours}}{234}}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Sorry, you're right, you don't need AN

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0You just need\[\large 2^{\frac{t}{234}}=\frac{1}{5}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[t=234(\log_2(0.2))\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0close enough (rounding errors)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So yes. Did you do it using the equation 8 posts ago?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1351694706420:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0sorry about my terrible handwriting

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I 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\[N=N_0 \frac{1}{5}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ah, i think I see what you have done now

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0where has halflife came into it?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0(it seems a way better way to use it)
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