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Do you know differential equations or calculus?

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

A sample of*

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

but I want to know how to do it myself

whats c?

Sorry, it deleted half of my answer

lol, ok

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

...in the past

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

That's the derivation of it

ah, so its the same as my equation?

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

Yes

hendce why I haven't already used that equation

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

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

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

Which element is it?

Na (Sodium)

More specifically Sodium-24

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

A.N.= 6.0221415 *10^23

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

both, set an exam question for HW

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

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

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

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

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

I got t = 1955669

:/

=543.33 hours

is my answer right?

Its in seconds

I got 1955992.2

close enough (rounding errors)

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

*9

no, I did this...

|dw:1351694706420:dw|

sorry about my terrible handwriting

k=234

ah, i think I see what you have done now

the N0 will cancel

where has half-life came into it?

(it seems a way better way to use it)