[Rutherford]

The element carbon consists of the stable isotopes ¹²C (98.90% of atoms) and ¹³C (1.10% of atoms). In addition, carbon contains a small fraction of the radioisotope ¹⁴C (t_1/2= 5730 years). What is the isotope ¹²C/¹⁴C ratio of carbon, which takes part in the natural CO₂ cycle (1 year=365 days) if the decay rate of carbon 14 is found to be 13.6 disintegrations per minute and gram of carbon.

Because the disintegration is expressed in 1g of carbon, I can assume that I have 1g of a carbon mixture, that is 0.989g of ¹²C and the number of atoms is easily calculated, but how to do it for ¹⁴C?

[Rutherford]

Any idea please? The formula they used to calculate the number of ¹⁴C atom is: N=13.6*t_1/2/ln2. Could someone interpret this formula?

[Borek]

An obvious line of attack seems to be to calculate N₀-N_t from [itex]N_t = N_0 e^{\frac t {t_{1/2}}}[/itex].

[Rutherford]

I don't understand what you wrote. I know this formula of first-order reaction (with natural logarithms):
lnN=lnN₀-λt
λ can be calculated using the t_1/2 ([N]=1/2[N]₀):
λ=ln2/t_1/2
I convert all times into minutes and put λ into the first equation and then I think that I need to calculate N₀, but how?
Where to put the disintegrations?

[Borek]

I know this formula of first-order reaction (with natural logarithms):
lnN=lnN₀-λt

See if you can rearrange it to get equation in terms of N_t and N₀, not in terms of logs.

If you start with N₀ atoms and after time t you are left with N_t atoms, how many atoms decayed?

[Rutherford]

Okay, this formula was given in the problem that I didn't write:
N_t=N₀e^-λt
N₀-Nt decayed but I need N₀. What now?

[Borek]

Now you start thinking on your own. Enough spoon feeding for today.

You are a single rearrangement from the answer.

[Rutherford]

But I don't know how to add the disintegration period. I tried this way: calculated λ, then after 1min 13.6 atoms decay, that would be N_t, so N₀=13.6/e^-λt, t is 1min and λ=2.3*10^-10min^-1. Then N₀=13.6  ?

[Borek]

As you already stated - N₀-N_t decayed. Add to that fact that [itex]N_t = N_0 e^{\frac t {t_{1/2}}}[/itex]. There is one unknown here, all you have to do is to solve.

[Rutherford]

I don't understand how you got that equation.
1)N₀-N_t=13.6    N_t=N₀-13.6
2)N_t=N₀e^-λt
t is 1 so it gets out of equation 2). λ is a very very small number ≈0, so e^-λ=1. Then N₀-13.6=N₀  .

[Borek]

Sorry, my mistake. It should be [itex]N_t=N_0 2^{\frac{t}{t_{1/2}}}[/itex]. Note that they are almost identical, the only difference is the ln(2) constant by which they differ.

You are right these numbers are very close, but try to solve without any approximations.

[Rutherford]

If I don't approximate, then the calculator does it, and it shows 1.00000000 and again I have the same situation. By the way, their answer is 5.91*10¹⁰ atoms of ¹⁴C in 1g of mixture and the formula they used is N=13.6*t_1/2/ln2 which is N=13.6/λ, but I can't think of a way to derive this.

[Borek]

Windows calculator is able to give the correct answer. TI-89 is able to give the correct answer. You need at least 12-13 digits to get the result this way, but it is perfectly doable. Sure, some calculators won't work.

Activity is a first derivative of the number of atoms with respect to time - if you have not started to learn calculus you can't derive the formula they used, you can at best memorize it.

[Rutherford]

Using the math at the level I know, from reactions 1) and 2) without direct calculation I got: N₀=(13.6e^ln2/t_1/2)/(e^ln2/t_1/2-1) here it can be seen that if I use the approximation the denominator is 0, if I don't use it, the denominator is a really small number. Using a calculator here on the net, I got the right answer. I suppose that this requires knowing integrals, so I should skip similar problems for a while. Thanks for helping me through this problem.