Question

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

Answer 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

Answer 2

Do you know differential equations or calculus?

Answer 3

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

Answer 4

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?

Answer 5

A sample of*

Answer 6

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

Answer 7

but I want to know how to do it myself

Answer 8

whats c?

Answer 9

Sorry, it deleted half of my answer

Answer 10

lol, ok

Answer 11

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

Answer 12

...in the past

Answer 13

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

Answer 14

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

Answer 15

That's the derivation of it

Answer 16

ah, so its the same as my equation?

Answer 17

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

Answer 18

Yes

Answer 19

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

Answer 20

hendce why I haven't already used that equation

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

Which element is it?

Answer 26

Na (Sodium)

Answer 27

More specifically Sodium-24

Answer 28

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

Answer 29

A.N.= 6.0221415 *10^23

Answer 30

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)

Answer 31

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

Answer 32

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?

Answer 33

both, set an exam question for HW

Answer 34

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

Answer 35

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

Answer 36

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

Answer 37

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

Answer 38

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

Answer 39

I got t = 1955669

Answer 40

:/

Answer 41

=543.33 hours

Answer 42

is my answer right?

Answer 43

Its in seconds

Answer 44

I got 1955992.2

Answer 45

close enough (rounding errors)

Answer 46

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

Answer 47

*9

Answer 48

no, I did this...