Question

The half-life of Thorium 230 is about 75000 years, while that of Uranium 234 is about 245000 years. A certain sample of ancient coral has a Thorium/uranium ratio of 10 percent. How old is the coral?

How do I go about solving this problem? The topic is ODE by the way.

Answer 1

Some physics/chemistry knowledge is needed for this I guess. I know that Uranium decays into Thorium but I am stuck there and not sure how to form the required ODEs.

Answer 2

Nah, this is a straight decay problem.

Amount of isotope left at time \(t\) is \(2^{-t/75000}\) for Th and \(2^{-t/245000}\) for U.

Set the quotient of those two equal to the desired ratio and solve for \(t\).

Answer 3

My lecturer gave me the clue that i needed to include the fact that Uranium decays into thorium so the amount of thorium depends on the amount of Uranium that has decayed

Answer 4

Okay, then he is assuming some more physics knowledge and my solution won't work.

Answer 5

Without accounting for that pesky detail, you get an answer of about 359,000 years.

Answer 6

To type out what he said,

Let U(t) be the mount of Uranium at time t,
Let T(t) be the amount of Thorium at time t.

Assume that each decay of uranium atom produced one thorium atom.
hence thorium atoms are being born at exactly the same rate at which uranium atom die.

i.e.
birth rate of thorium = (-1) death rate of Uranium

Answer 7

Oh and he gave the answer as 40,000 years

Answer 8

We can solve this problem using ODE's because an elements half-life expresses the rate of change of that element. The expression for the half-life concentration of an element is:
\[N(t)=N _{0}\left( 0.5 \right)^{\frac{ t }{ t _{0.5} }}\]
In the problem statement, we are given
\[t _{0.5 thorium} = 75000\]
\[t _{0.5 uranium} = 245000\]
So we can write this as a second-order ODE that expresses the relationship between the rates over time as:
\[Ratio = 0.1 = \frac{ N(t) _{thorium} }{ N(t) _{uranium} }\]
Now if we plug in our time constants to the half-life equation and plug those into the ratio expression, then we can write:
\[(0.1)N _{U initial}\left( 0.5 \right)^{\frac{ t }{ 245000 }} -N _{T initial}\left( 0.5 \right)^{\frac{ t }{ 75000 }}=0\]
You will need some time-dependent information about the initial ratio or a relationship the governs how they interact to solve for the initial levels.

Answer 9

That's a very nicely formatted version of what I did, but it doesn't account for the fact that some of the U-234 decays into Th-230 which then decays during the time of interest...

Answer 10

hmm in that case,

\[\frac{ dU }{ dT } = -k _{u}U\]
and
\[\frac{ dT }{ dt } = k _{u}U - k _{T}T\]

Answer 11

I think i got it now

Answer 12

let me try

Answer 13

Yes, sorry - I didn't see the physics-y-ness of it. My apologies.

Answer 14

I was hoping after all that typing that you had it for us :-)

Answer 15

No, I'm just a slow typer. I think irkiz has the correct relationship already though.

Answer 16

Yeah. Got the answer.

Answer 17

Thanks for the help anyway guys

Answer 18

Hey, where's the writeup?!

Answer 19

Will take me awhile to type it all up. I hope my phone can take a good picture of my working....

Answer 20

I'm guessing this makes the number of years to get to the 1:10 point be somewhat shorter...

Answer 21

Well, I'd love to see it, if convenient. If not, them's the breaks...

Answer 22

well starting with the initial ODE that i got, i integrated both of them to get an equation of T and U. Then i divided them to get T/U and used the half life equations to get the answer. That is the gist of what i did

Answer 23

I'll try to upload the picture when i have a cable for you guys

Answer 24

typing is gonna take too long it seems

Answer 25

I'll get a notification when you post something here, no rush. Thanks!