Monday

May 30, 2016
Posted by **Andrew** on Thursday, June 15, 2006 at 6:38pm.

Argon 40 / Potassium 40 ratio is 1.55?

The half life of Potassium 40 is 1.28 x 10^9 years

I am confussed on this question theres no examples in my book but i have to know it can someone explain to me how to get to the answer?

You don't say if this is mass ratio or not. I will assume it is an atom ratio.

Start with any number you wish. Let's start with 100 atoms.

Let X = # atoms Ar at some time.

then 100 - x = # atoms K at that time.

Then Ar atoms/K atoms = (X)/(100-X) =1.55.

Solve for X to give # atoms of Ar and 100-X to give # atoms of K.

Then ln(No/N) = kt

k = 0.693/(t1/2) = 0.693/1.28 x 10^9 yrs.

No will be 100

N will be X

Solve for t.

Check my thinking. Check my work.

ok i get up to this part

k = 0.693/(t1/2) = 0.693/1.28 x 10^9 yrs.

which is the decay constant

but i when u say let Nzero = 100 and N = x im not sure where to plug them in at would it be:

-((ln(x/100)/k)) = T ?

Andrew/utramos/???

Let me start over a little. My first post and this post will give the same answer but this one should be a little easier to understand.

Let's start with 100 atoms of K40 many moons ago. That will decay to Ar40. We will call the final # atoms K40 as X and that is what we have today. That means Ar40 will be 100-X today.

So the ratio today is Ar/K = 1.55 as stated in the problem.

Substituting in the ratio we have

100-X/X = 1.55

Solve for X = 39.2 atoms K.

Ar = 100-X = 60.7.

You have determined k = 0.693/half life and I obtained 5.42 x 10^-10 years^-1.

So we start with 100 atoms K many moons ago and today we have 39.2. Therefore, No = number of K atoms initially = 100 and N = number of K atoms today = 39.2. Plug that into the decay equation.

ln(No/N) = kt

ln(100/39.2) = 5.42 x 10^-10 t

solve for t. I get something like 10^9 years.

In your second question you asked if the following was correct.

-((ln(x/100)/k)) = T ?

I hope this helps. Repost if you still have a question about what I have done.

I think i am just not sure on how u got this

Solve for X = 39.2 atoms K.

Ar = 100-X = 60.7.

Actually, it doesn't matter what the Ar is because you are using No/N for K and using 100 for No, which we started with, and 39.2 for N (for K) for when it finished decaying (that's what we calculated N to be). However, the 60.7 for Ar comes from this:

If we started with 100 atoms K40, it decays over the ages and ends up X (which we later determined X to be 39.2). So if Ar is 100-X, then 100-39.2 = 60.7. OK? or not?