Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
kkid
Group Title
How can I calculate the time taken for an element to decay by a certain amount?
 2 years ago
 2 years ago
kkid Group Title
How can I calculate the time taken for an element to decay by a certain amount?
 2 years ago
 2 years ago

This Question is Closed

kkid Group TitleBest ResponseYou've already chosen the best response.1
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 years ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
Do you know differential equations or calculus?
 2 years ago

kkid Group TitleBest ResponseYou've already chosen the best response.1
Yes, but I don't think this question requires it
 2 years ago

kkid Group TitleBest ResponseYou've already chosen the best response.1
The question I'm tryingt o answer is: TA sample of 24Na has a halflife of 234 hours, How much time elapses before a 5μg sample contains 1μg of undecayed atoms?
 2 years ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
\[\frac{1}{c}e^{t \lambda}=ke^{t \lambda}=N\]
 2 years ago

kkid Group TitleBest ResponseYou've already chosen the best response.1
but I want to know how to do it myself
 2 years ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
Sorry, it deleted half of my answer
 2 years ago

kkid Group TitleBest ResponseYou've already chosen the best response.1
I have used a similar equation \[N = N _{0}e ^{\lambda t}\]
 2 years ago

kkid Group TitleBest ResponseYou've already chosen the best response.1
...in the past
 2 years ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
http://upload.wikimedia.org/math/a/8/a/a8a0bd0a12874474352b307d9e919076.png \[\frac{1}{\lambda}(lnN+lnk)=\frac{1}{\lambda}(\ln(kN))=t\] \[(\ln(kN))=\lambda t\] \[kN=e^{\lambda t}\] \[N=ce^{\lambda t}\]
 2 years ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
Set c= sample at t=0 That is, \[ c= N_0 \]
 2 years ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
That's the derivation of it
 2 years ago

kkid Group TitleBest ResponseYou've already chosen the best response.1
ah, so its the same as my equation?
 2 years ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
You're given the half life, so I'd convert it into 2^... instead of e^....
 2 years ago

kkid Group TitleBest ResponseYou've already chosen the best response.1
I don't know how to get the values of N and N0 though, because that is the number of atoms, not the mass
 2 years ago

kkid Group TitleBest ResponseYou've already chosen the best response.1
hendce why I haven't already used that equation
 2 years ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
\[N=N_0 (2^{\log_2(e)})^{\lambda t}=N_02^{\log_2(e) \lambda t}\]
 2 years ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
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
 2 years ago

kkid Group TitleBest ResponseYou've already chosen the best response.1
but N refers to atoms, not mass (which is the 5 and the 1)
 2 years ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
So obviously the halflife \[= \log_2(e) \lambda\]
 2 years ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
Which element is it?
 2 years ago

kkid Group TitleBest ResponseYou've already chosen the best response.1
More specifically Sodium24
 2 years ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
\[Moles= \frac{mass_{grams}}{A_r}\] \[Atoms= Moles* Avogadro'snumber\]
 2 years ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
A.N.= 6.0221415 *10^23
 2 years ago

kkid Group TitleBest ResponseYou've already chosen the best response.1
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)
 2 years ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
So initially the mass (N_0)\[=\frac{5*10^{6}}{24}6.0221415 *10^{23}\]
 2 years ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
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?
 2 years ago

kkid Group TitleBest ResponseYou've already chosen the best response.1
both, set an exam question for HW
 2 years ago

kkid Group TitleBest ResponseYou've already chosen the best response.1
I'll try working it out with the arogarrdo number (or whatever it is called lol)
 2 years ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
Anyway, \[\huge N=125461281250000000*2^{\frac{t_{hours}}{234}}\]
 2 years ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
Sorry, you're right, you don't need AN
 2 years ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
You just need\[\large 2^{\frac{t}{234}}=\frac{1}{5}\]
 2 years ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
\[t=234(\log_2(0.2))\]
 2 years ago

kkid Group TitleBest ResponseYou've already chosen the best response.1
I got t = 1955669
 2 years ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
=543.33 hours
 2 years ago

kkid Group TitleBest ResponseYou've already chosen the best response.1
is my answer right?
 2 years ago

kkid Group TitleBest ResponseYou've already chosen the best response.1
Its in seconds
 2 years ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
I got 1955992.2
 2 years ago

kkid Group TitleBest ResponseYou've already chosen the best response.1
close enough (rounding errors)
 2 years ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
So yes. Did you do it using the equation 8 posts ago?
 2 years ago

kkid Group TitleBest ResponseYou've already chosen the best response.1
no, I did this...
 2 years ago

kkid Group TitleBest ResponseYou've already chosen the best response.1
dw:1351694706420:dw
 2 years ago

kkid Group TitleBest ResponseYou've already chosen the best response.1
sorry about my terrible handwriting
 2 years ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
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\[N=N_0 \frac{1}{5}\]
 2 years ago

kkid Group TitleBest ResponseYou've already chosen the best response.1
ah, i think I see what you have done now
 2 years ago

kkid Group TitleBest ResponseYou've already chosen the best response.1
the N0 will cancel
 2 years ago

kkid Group TitleBest ResponseYou've already chosen the best response.1
where has halflife came into it?
 2 years ago

kkid Group TitleBest ResponseYou've already chosen the best response.1
(it seems a way better way to use it)
 2 years ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.