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

1. 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

2. anonymous

Do you know differential equations or calculus?

3. anonymous

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

4. 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?

5. anonymous

A sample of*

6. anonymous

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

7. anonymous

but I want to know how to do it myself

8. anonymous

whats c?

9. anonymous

Sorry, it deleted half of my answer

10. anonymous

lol, ok

11. anonymous

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

12. anonymous

...in the past

13. anonymous

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

14. anonymous

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

15. anonymous

That's the derivation of it

16. anonymous

ah, so its the same as my equation?

17. anonymous

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

18. anonymous

Yes

19. 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

20. anonymous

hendce why I haven't already used that equation

21. anonymous

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

22. 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

23. anonymous

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

24. anonymous

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

25. anonymous

Which element is it?

26. anonymous

Na (Sodium)

27. anonymous

More specifically Sodium-24

28. anonymous

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

29. anonymous

A.N.= 6.0221415 *10^23

30. 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)

31. anonymous

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

32. 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?

33. anonymous

both, set an exam question for HW

34. anonymous

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

35. anonymous

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

36. anonymous

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

37. anonymous

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

38. anonymous

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

39. anonymous

I got t = 1955669

40. anonymous

:/

41. anonymous

=543.33 hours

42. anonymous

43. anonymous

Its in seconds

44. anonymous

I got 1955992.2

45. anonymous

close enough (rounding errors)

46. anonymous

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

47. anonymous

*9

48. anonymous

no, I did this...

49. anonymous

|dw:1351694706420:dw|

50. anonymous

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

52. anonymous

k=234

53. anonymous

ah, i think I see what you have done now

54. anonymous

the N0 will cancel

55. anonymous

where has half-life came into it?

56. anonymous

(it seems a way better way to use it)