Almost all of naturally occurring uranium is (238)/(92) U with a half-life of 4.468 ×10^9 yr. Mo

Almost all of naturally occurring uranium is (238)/(92) U...

Almost all of naturally occurring uranium is $\frac{238}{92} \mathrm{U}$ with a half-life of $4.468 \times 10^{9}$ yr. Most of the rest of natural uranium is $\frac{235}{92} \mathrm{U}$ with a half-life of $7.04 \times 10^{8} \mathrm{yr.}$ Today a sample contains0.720$\%_{0} \frac{235}{92} U$ (a) What was this percentage 1.0 billion years ago? (b) What percentage of$\frac{235}{92} \mathrm{U}$ will remain 100 million years from now?

Almost all of naturally occurring uranium is $\frac{238}{92} \mathrm{U}$ with a
half-life of $4.468 \times 10^{9}$ yr. Most of the rest of natural
uranium is $\frac{235}{92} \mathrm{U}$ with a half-life of $7.04 \times 10^{8} \mathrm{yr.}$ Today a
sample contains0.720$\%_{0} \frac{235}{92} U$ (a) What was this percentage
1.0 billion years ago? (b) What percentage of$\frac{235}{92} \mathrm{U}$ will
remain 100 million years from now?

Key Concepts

Isotopic Abundance Variation

Isotopic abundance variation involves the change in the relative proportions of different isotopes over time due to their differing decay rates. When two isotopes decay at different rates, the ratio of their abundances evolves, which is an important concept in fields such as geology, astronomy, and uranium dating.

Exponential Decay Formula

Exponential decay refers to the mathematical model that describes how the quantity of a radioactive substance decreases exponentially over time. The formula, N(t) = N0 e^(-?t), connects the initial amount of the substance with the remaining amount at time t, using the decay constant ?.

Half-life

The half-life is the time required for half of the quantity of a radioactive substance to undergo decay. It provides a measure of the stability of the isotope; isotopes with longer half-lives decay more slowly than those with shorter half-lives.

Radioactive Decay

Radioactive decay is the process by which an unstable atomic nucleus loses energy by emitting radiation. This process is random at the level of single atoms but predictable in large samples, following a well-defined decay law that dictates how the amount of the substance decreases over time.

Decay Constant

The decay constant is a value that characterizes the rate at which a radioactive substance decays. It is directly related to the half-life through the relationship ? = ln(2) / (half-life), and it plays a central role in the mathematical description of exponential decay.

Transcript

00:01 Okay, so this problem is all about relative natural abundances and how that's going to change over time based on different decay rates.

00:09 So, and the problem we're given that uranium 238 has a half -life of 4 .468 times 10 to the 9 years.

00:29 And i'm just going to label this 8 for the 238.

00:32 So everything within 8 will pertain to that isotope.

00:38 And then t 1 half for the 2351 is 7 .04 times 10 to the 8 years.

00:46 And then it's saying it has a relative abundance.

00:49 So the number of 5 to the total number in the universe.

00:57 That should be a 5 is 0 .072 and what else okay yeah and so now in our goal is defined after for a it um like a billion years ago so oops t equals 10 to the nine years years um what was the relative of abundance.

01:32 And so our goal is to calculate.

01:35 So if like, let's say each of them has been decaying as e to the minus gamma t.

01:43 And so in some sense, we're given information about this number.

01:48 And then we're trying to find sort of what was going on with the original numbers.

01:54 So to put in terms of the original numbers, you can multiply both sides by e to the minus gamma t.

02:00 And by the way, i think i'm going to go through the math kind of fast in this one just because it's like the end of the chapter and to keep these videos not too long and i'm going to assume that you can do most of the math easily.

02:11 So yeah, so here is this e to the gamma t, moving this to the other side.

02:16 And this is the case for either subscript 8 and 5.

02:19 And so we, we, our goal is to get the, oops, i decided to use you in my work.

02:31 Okay, so the goal is to get the new abundance of the fifth one, so n5 over an n8 plus n5.

02:42 And then, oops, that's supposed to be an n.

02:45 Let me make that a little neater.

02:51 And so let's go ahead and, oh yeah, and then the zero point.

02:59 So that's just kind of like at the beginning of time where the beginning of time is 10 of the nine years ago.

03:04 So then plugging in this n -not formula, we can get n5e to the lambda.

03:13 Oh, these are different lambdas.

03:14 So lambda 5t divided by n8, e to the lambda 8t plus n5e to the lambda sub 5t.

03:28 These are subscripts when they have fives and eights.

03:30 And then if we rearrange this equation and sub it in here, so if we sub for n5, then basically what will happen is, is like, right, it'll be like, do some algebra.

03:48 Here, i guess i'll do that out.

03:49 So if you do some algebra, you get that n5 is equal to n8 over.

03:56 I'm just going to use the numbers here.

03:58 I'm going to mix numbers and symbols, but i think it makes it most clear.

04:01 There.

04:02 Oops, that's one minus that.

04:05 And then multiplied by .072, just doing the algebra.

04:12 And then you want to sub that in here...