04-12-2018, 05:25 AM

The Problem:

One of the isotopes of potassium, 40K, occurs at 120 ppm in natural samples. This 0.012% naturally decays with a 1.28X10^9 year half life into argon-40 (440Ar).

If a rock stays crystallized the argon formed will be trapped in the rock matrix. After bringing the same to the lab the argon and potassium can be measured experimentally by isotope ratio mass spectrometry.

Assuming no argon-40 is present in rock when it is formed, if the following plot [see drive.google.com/file/d/0B-BclHOW2R0QemZKOFM0WXJJOFc0WS0zay1saTVzYXZ4Qmdr/view?usp=sharing and drive.google.com/file/d/0B-BclHOW2R0QaEEwc1NpSDJQZURkOWJQSFZXcHNDOU93SHFz/view?usp=sharing] represents the amount of potassium-40 and argon-40 in a sample of sediment, what is the age of the rock?

The Equations Provided:

ln ([A] / [A]_o) = - kt

t_1/2=0.693/k

The Attempt at a Solution:

k = 0.693/t_1/2

k = 0.693/(1.29X10^9)

k = 5.41X10^-10

Ar/K = 0.00052/0.001 :rarrow: I found this using peak heights from above spectrum; I'm mainly confused about what I'm supposed to be plugging in to the equation and how to find that information from the spectrum provided.

= 52%

ln([A]/[A]_o)=-kt

ln([52][0.012]_o)=-(5.41X10^-10)t

t = -1.55X10^10 which doesn't make any sense

I know this is basic stuff and I should know how to do it, but I can't seem to figure out what I'm doing wrong.

Using the isotope ratio mas spectrum, I found the heights of both peaks and divided the height of the 40Ar peak by the height of the 40K peak. This ends up being 52%. I'm relatively certain that I wasn't supposed to plug this number in to the equation; here is where I'm stumped. If most natural samples are 0.012% 40K, should I multiple 48% (100-52=48) by 0.012% to get the percentage of the sample that is still 40K? Even then, the number ends up being too large. But I thought that I'm supposed to be using concentrations in the ln equation. In any case, the percentage of 40K should have gone down because some should have decayed, right? I'm lost and there's going to be a problem like this on our exam tomorrow. I appreciate any help. Thank you!

One of the isotopes of potassium, 40K, occurs at 120 ppm in natural samples. This 0.012% naturally decays with a 1.28X10^9 year half life into argon-40 (440Ar).

If a rock stays crystallized the argon formed will be trapped in the rock matrix. After bringing the same to the lab the argon and potassium can be measured experimentally by isotope ratio mass spectrometry.

Assuming no argon-40 is present in rock when it is formed, if the following plot [see drive.google.com/file/d/0B-BclHOW2R0QemZKOFM0WXJJOFc0WS0zay1saTVzYXZ4Qmdr/view?usp=sharing and drive.google.com/file/d/0B-BclHOW2R0QaEEwc1NpSDJQZURkOWJQSFZXcHNDOU93SHFz/view?usp=sharing] represents the amount of potassium-40 and argon-40 in a sample of sediment, what is the age of the rock?

The Equations Provided:

ln ([A] / [A]_o) = - kt

t_1/2=0.693/k

The Attempt at a Solution:

k = 0.693/t_1/2

k = 0.693/(1.29X10^9)

k = 5.41X10^-10

Ar/K = 0.00052/0.001 :rarrow: I found this using peak heights from above spectrum; I'm mainly confused about what I'm supposed to be plugging in to the equation and how to find that information from the spectrum provided.

= 52%

ln([A]/[A]_o)=-kt

ln([52][0.012]_o)=-(5.41X10^-10)t

t = -1.55X10^10 which doesn't make any sense

I know this is basic stuff and I should know how to do it, but I can't seem to figure out what I'm doing wrong.

Using the isotope ratio mas spectrum, I found the heights of both peaks and divided the height of the 40Ar peak by the height of the 40K peak. This ends up being 52%. I'm relatively certain that I wasn't supposed to plug this number in to the equation; here is where I'm stumped. If most natural samples are 0.012% 40K, should I multiple 48% (100-52=48) by 0.012% to get the percentage of the sample that is still 40K? Even then, the number ends up being too large. But I thought that I'm supposed to be using concentrations in the ln equation. In any case, the percentage of 40K should have gone down because some should have decayed, right? I'm lost and there's going to be a problem like this on our exam tomorrow. I appreciate any help. Thank you!