Moon Rocks In one of the rocks brought back from the Moon,...

Moon Rocks In one of the rocks brought back from the Moon, it is found that $80.5 \%$ of the initial potassium-40 in the rock has decayed to argon-40. (a) If the half-life for this decay is $1.20 \times 10^{9}$ years, how old is the rock? (b) How much longer will it take before only $10.0 \%$ of the original potassium-40 is still present in the rock?

Key Concepts

Radiometric Dating

Radiometric dating is a method used to determine the age of materials by measuring the relative abundances of radioactive isotopes and their decay products. This technique relies on the known half-lives of isotopes, making it a foundational tool in fields such as geology and planetary science for dating rocks and fossils.

Exponential Decay Equation

The exponential decay equation mathematically describes how the quantity of a radioactive substance decreases over time. This equation relates the remaining quantity to the initial quantity using the decay constant and time, providing a framework to solve for unknown variables in decay processes.

Decay Constant

The decay constant is a parameter that quantifies the probability per unit time that a nucleus will decay. It is inversely related to the half-life and is a key component in the exponential decay equation, allowing for the calculation of decay rates and time elapsed.

Radioactive Decay

Radioactive decay is the process by which an unstable atomic nucleus loses energy by emitting radiation. In physics and chemistry, this spontaneous transformation involves a predictable statistical decay over time, and it underpins techniques for age determination in various materials.

Half-life

The half-life is the period required for half of the original quantity of a radioactive isotope to decay. It is a fundamental concept in nuclear physics that provides a measure of the stability of isotopes, and is used to calculate the decay over time in radioactive decay processes.

Transcript

00:01 Question 74 here, it has to do with moon rocks.

00:05 In one of the rocks brought back for the moon, it is found that 80 .5 % of the initial potassium 40 in the rock has decayed to argon 40.

00:13 If the half -life of this decay is 1 .2 times 10 to 9 years, how old is a rock? that's question a.

00:20 And question b, how much longer will it take before only 10 % of the original potassium 40 to still be present in the rock? so it's easy for this question to be tricked right off the bat, which i was the first time i tried to solve the problem, is that it states we want to know how long it takes for 80 .5 % to decay, not 0 .805 remaining.

00:44 So we want to look for this scenario where 19 .5 % of the remaining potassium 40 exists.

00:53 So this is what we're trying to solve for here.

00:58 So we want to know at what time frame this ratio exists.

01:03 So first, well, we know that.

01:05 The ratio or the path that these member of particles follows is based on their typical decay equation.

01:13 For those to work, we need lambda here, which we know we can find if we just take ln2 over the half -life of our sample, not half, there we go, and we're given our half -life, which i did not write down here, but in the question states that it is 1 .2 times 10 to the 9 years, which gives the decay constant of 5 .78 times 10 to the negative 10 per year.

02:07 So we can use this following equation now.

02:10 We take the ratio of n over n -not, the natural longer than both sides, and we'll actually write it out, why not? n over n -not, the scenario, we want that value to be 0 .195, e -negative lambda -t, lambda now we know.

02:34 So by taking the natural algorithm, we can find that the time required for this ratio to occur in is simply the natural algorithm of 0 .195 over lambda negative because the ln of that value will be negative because it is less than 1.

02:57 So we have 0 .585 .78 times 10 to the negative 10 per year on our new denominator here such that the time required would be 2 .83.

03:14 Times 10 to the 9 years.

03:19 There you go.

03:24 So part b, which i'll do on a separate page here, part b again, it asks, how much longer will it take before only 10 % of the original potassium 40 to still be present? well, it's the same idea as the previous question...

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