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

- anonymous

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

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- anonymous

I have the decay constant and need to find out how long it will take for the element to decay from 5ug to 1ug

- anonymous

Do you know differential equations or calculus?

- anonymous

Yes, but I don't think this question requires it

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## More answers

- anonymous

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?

- anonymous

A sample of*

- anonymous

\[\frac{1}{c}e^{-t \lambda}=ke^{-t \lambda}=N\]

- anonymous

but I want to know how to do it myself

- anonymous

whats c?

- anonymous

Sorry, it deleted half of my answer

- anonymous

lol, ok

- anonymous

I have used a similar equation
\[N = N _{0}e ^{-\lambda t}\]

- anonymous

...in the past

- anonymous

http://upload.wikimedia.org/math/a/8/a/a8a0bd0a12874474352b307d9e919076.png
\[\frac{-1}{\lambda}(lnN+lnk)=\frac{-1}{\lambda}(\ln(kN))=t\]
\[(\ln(kN))=-\lambda t\]
\[kN=e^{-\lambda t}\]
\[N=ce^{-\lambda t}\]

- anonymous

Set c= sample at t=0
That is, \[ c= N_0 \]

- anonymous

That's the derivation of it

- anonymous

ah, so its the same as my equation?

- anonymous

You're given the half life, so I'd convert it into 2^... instead of e^....

- anonymous

Yes

- anonymous

I don't know how to get the values of N and N0 though, because that is the number of atoms, not the mass

- anonymous

hendce why I haven't already used that equation

- anonymous

\[N=N_0 (2^{\log_2(e)})^{-\lambda t}=N_02^{-\log_2(e) \lambda t}\]

- anonymous

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

- anonymous

but N refers to atoms, not mass (which is the 5 and the 1)

- anonymous

So obviously the half-life \[= \log_2(e) \lambda\]

- anonymous

Which element is it?

- anonymous

Na (Sodium)

- anonymous

More specifically Sodium-24

- anonymous

\[Moles= \frac{mass_{grams}}{A_r}\]
\[Atoms= Moles* Avogadro's-number\]

- anonymous

A.N.= 6.0221415 *10^23

- anonymous

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)

- anonymous

So initially the mass (N_0)\[=\frac{5*10^{-6}}{24}6.0221415 *10^{23}\]

- anonymous

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?

- anonymous

both, set an exam question for HW

- anonymous

I'll try working it out with the arogarrdo number (or whatever it is called lol)

- anonymous

Anyway, \[\huge N=125461281250000000*2^{-\frac{t_{hours}}{234}}\]

- anonymous

Sorry, you're right, you don't need AN

- anonymous

You just need\[\large 2^{-\frac{t}{234}}=\frac{1}{5}\]

- anonymous

\[t=-234(\log_2(0.2))\]

- anonymous

I got t = 1955669

- anonymous

:/

- anonymous

=543.33 hours

- anonymous

is my answer right?

- anonymous

Its in seconds

- anonymous

I got 1955992.2

- anonymous

close enough (rounding errors)

- anonymous

So yes. Did you do it using the equation 8 posts ago?

- anonymous

*9

- anonymous

no, I did this...

- anonymous

|dw:1351694706420:dw|

- anonymous

sorry about my terrible handwriting

- anonymous

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\[N=N_0 \frac{1}{5}\]

- anonymous

k=234

- anonymous

ah, i think I see what you have done now

- anonymous

the N0 will cancel

- anonymous

where has half-life came into it?

- anonymous

(it seems a way better way to use it)

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