kkid
How can I calculate the time taken for an element to decay by a certain amount?
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kkid
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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
henpen
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Do you know differential equations or calculus?
kkid
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Yes, but I don't think this question requires it
kkid
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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?
kkid
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A sample of*
henpen
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\[\frac{1}{c}e^{-t \lambda}=ke^{-t \lambda}=N\]
kkid
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but I want to know how to do it myself
kkid
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whats c?
henpen
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Sorry, it deleted half of my answer
kkid
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lol, ok
kkid
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I have used a similar equation
\[N = N _{0}e ^{-\lambda t}\]
kkid
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...in the past
henpen
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Set c= sample at t=0
That is, \[ c= N_0 \]
henpen
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That's the derivation of it
kkid
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ah, so its the same as my equation?
henpen
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You're given the half life, so I'd convert it into 2^... instead of e^....
henpen
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Yes
kkid
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I don't know how to get the values of N and N0 though, because that is the number of atoms, not the mass
kkid
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hendce why I haven't already used that equation
henpen
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\[N=N_0 (2^{\log_2(e)})^{-\lambda t}=N_02^{-\log_2(e) \lambda t}\]
henpen
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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
kkid
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but N refers to atoms, not mass (which is the 5 and the 1)
henpen
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So obviously the half-life \[= \log_2(e) \lambda\]
henpen
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Which element is it?
kkid
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Na (Sodium)
kkid
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More specifically Sodium-24
henpen
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\[Moles= \frac{mass_{grams}}{A_r}\]
\[Atoms= Moles* Avogadro's-number\]
henpen
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A.N.= 6.0221415 *10^23
kkid
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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)
henpen
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So initially the mass (N_0)\[=\frac{5*10^{-6}}{24}6.0221415 *10^{23}\]
henpen
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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?
kkid
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both, set an exam question for HW
kkid
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I'll try working it out with the arogarrdo number (or whatever it is called lol)
henpen
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Anyway, \[\huge N=125461281250000000*2^{-\frac{t_{hours}}{234}}\]
henpen
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Sorry, you're right, you don't need AN
henpen
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You just need\[\large 2^{-\frac{t}{234}}=\frac{1}{5}\]
henpen
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\[t=-234(\log_2(0.2))\]
kkid
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I got t = 1955669
kkid
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:/
henpen
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=543.33 hours
kkid
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is my answer right?
kkid
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Its in seconds
henpen
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I got 1955992.2
kkid
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close enough (rounding errors)
henpen
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So yes. Did you do it using the equation 8 posts ago?
henpen
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*9
kkid
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no, I did this...
kkid
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|dw:1351694706420:dw|
kkid
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sorry about my terrible handwriting
henpen
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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\[N=N_0 \frac{1}{5}\]
henpen
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k=234
kkid
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ah, i think I see what you have done now
kkid
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the N0 will cancel
kkid
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where has half-life came into it?
kkid
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(it seems a way better way to use it)