anonymous
  • anonymous
How can I calculate the time taken for an element to decay by a certain amount?
Physics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • 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
  • anonymous
Do you know differential equations or calculus?
anonymous
  • anonymous
Yes, but I don't think this question requires it

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.