Argon radioactive decay dating

[bmxicle]

Homework Statement

The technique known as potassium-argon dating is used to date old lava flows. The potassium isotope ⁴⁰K has a 1.28 billion year half-life and is naturally present as very low levels. ⁴⁰K decays by beta emission into ⁴⁰Ar. Argon is a gas, and there is no argon in flowing lava because the gas escapes. Once the lava solidifies, any argon produced in the decay is trapped inside and cannot escape. A geologist brings you a piece of solidified lava in which you find the ⁴⁰Ar/⁴⁰K ratio to be 0.12. What is the age of the rock

Homework Equations

N(t) = N(0)(1/2)^t/t_1/2
r = ln2/t_1/2[/SUP]

The Attempt at a Solution

Here's what i have so far.
N_Ar is the number of argon atoms in the sample
N_K is the number of pottasium atoms in the sample
_K0 is the original number of pottassium atoms in the sample.

N_Ar/N_K =0.12 ===> N_Ar = 0.12N_K
N_k0 = N_Ar + N_k
These two equations combined give N_K0 = 1.12N_k
and since N_k = N_Ar/0.12
===> N_K0 = N_Ar1.12/0.12

Plugging this into the equation gives:

N_Ar = N_Ar1.12/0.12(0.5)^t/t_1/2
0.12/1.12 = (0.5)^t/t_1/2
T = T_1/2(ln(0.12/1.12)/ln(0.5))

This gives me 4.1 billion years which isn't right.

 

[ehild]

Homework Equations

N(t) = N(0)(1/2)^t/t_1/2

What is N(t)?

ehild

 

[bmxicle]

ah sorry i guess that's not normal notation. N(t) is just the number of decayed particles as a function of time, so in this case it's N_Ar

 

[ehild]

Do you think that the number of Ar atoms is No at the beginning of the decay, and it decreases with time? Is not it just the opposite? ehild

 

[bmxicle]

I don't think that's what i have up there--and if i do that would explain the wrong answer. The number N_K0 ie. the initial number of potassium atoms in the rock when the lava cooled is going to be equal to the number of Argon atoms (atoms that underwent decay) plus the number of undecayed potassium atoms N_k.

 

[ehild]

It is the number of the potassium atoms that decreases exponentially, so your N(t) means the number of ⁴⁰K atoms still present at time t.

ehild

 

[bmxicle]

Yup that's what I was doing wrong; don't know why i got so flipped around on that. Thanks for your help.