Ask Me Help Desk - How much potassium-40 did the rock contain 1.3 x 109 years ago?

how much potassium-40 did the rock contain 1.3 x 109 years ago?

1.If a rock sample was found to contain 5.60 x 10-7 mol of argon-40, how much potassium-40 did the rock contain 1.3 x 109 years ago?

2.A wooden door lintel from an excavated site in Mexico would be expected to have what ratio of carbon-14 to carbon-12 atoms if the lintel is 9.0 x 103 yr old?

How can I find like that type of question?
I didn't learn yet. I really want to know how to solve that two problems.Please help
But I really want to know how to do like that problems
IF have any equation or like that please help me
Thanks

You need to use exponential decay (f(t) = A + B*exp(C*t)). You also need the halflife of $^{40}K$ and $^^{14}C$ - do ypu know about all of this?

I really don't know how to do it..

The rate of radipoactive decay is called "half life," and is the length of time it takes for one half of a particular isotope to turn into its daughter isotope. Here you have K-40, which turns into Ar-40 with a half life of 1.25 x 10^9 years. The half life formula is this:

$<br /> C(t) = C_0 (\frac 1 2 ) ^ {t/H}<br />$

where $C(t)$ is the concentration of K-40 as a function of time, $C_0$ is the initial concentration, $t$ is time (in years) and $H$ is the half life (also in years).

So for the first problem: You know that t = 1.3 x 10^9 (which is awfully close to the half life value - I wonder if your text book uses this as the half life value rather than 1.25 x 10^9 ?), and H = 1.25 x 10^9. Therefore:

$<br /> \frac C {C_1} = ( \frac 1 2 ) ^{(\frac {1.3 x 10^9} {1.25 x 10^9})} = 0.48.<br />$

Hence 48% of the K-40 is remaining, which means 52% of the K-40 has turned into Ar-40. Since you now have 5.6 x 10-7 mole of Ar-40, that means you started with:

5.6 x 10^(-7)/ 0.52 = 1.08 x 10-6 moles K-40.

Now - can you do the second problem?

Sorry.. your answer is not correct.. anyway thanks

Quote:

Originally Posted by hslove142331

Sorry..your answer is not correct..anyway thanks

So you figured out what the correct answer is? Please post it, so I can see where I went wrong. Thanks.

Quote:

Originally Posted by ebaines

So you figured out what the correct answer is? Please post it, so I can see where I went wrong. Thanks.

Oh, ebaines, I need to tell you. Hsolve is using a program I think on the computer and the program requires you to insert the answer. Then it will tell you whether it's right or wrong.

Btw, the code for the multiply symbol is '\times' ebaines.

It's perhaps as you suggested ebaines, I mean, the program uses 1.3x10^9, making the final answer as 1.12x10^-6.