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## kkid 3 years ago How can I calculate the time taken for an element to decay by a certain amount?

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1. kkid

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

2. henpen

Do you know differential equations or calculus?

3. kkid

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

4. kkid

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?

5. kkid

A sample of*

6. henpen

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

7. kkid

but I want to know how to do it myself

8. kkid

whats c?

9. henpen

Sorry, it deleted half of my answer

10. kkid

lol, ok

11. kkid

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

12. kkid

...in the past

13. henpen

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

14. henpen

Set c= sample at t=0 That is, $c= N_0$

15. henpen

That's the derivation of it

16. kkid

ah, so its the same as my equation?

17. henpen

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

18. henpen

Yes

19. kkid

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

20. kkid

hendce why I haven't already used that equation

21. henpen

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

22. henpen

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

23. kkid

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

24. henpen

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

25. henpen

Which element is it?

26. kkid

Na (Sodium)

27. kkid

More specifically Sodium-24

28. henpen

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

29. henpen

A.N.= 6.0221415 *10^23

30. kkid

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)

31. henpen

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

32. henpen

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?

33. kkid

both, set an exam question for HW

34. kkid

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

35. henpen

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

36. henpen

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

37. henpen

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

38. henpen

$t=-234(\log_2(0.2))$

39. kkid

I got t = 1955669

40. kkid

:/

41. henpen

=543.33 hours

42. kkid

is my answer right?

43. kkid

Its in seconds

44. henpen

I got 1955992.2

45. kkid

close enough (rounding errors)

46. henpen

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

47. henpen

*9

48. kkid

no, I did this...

49. kkid

|dw:1351694706420:dw|

50. kkid

sorry about my terrible handwriting

51. henpen

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$N=N_0 \frac{1}{5}$

52. henpen

k=234

53. kkid

ah, i think I see what you have done now

54. kkid

the N0 will cancel

55. kkid

where has half-life came into it?

56. kkid

(it seems a way better way to use it)

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