# Homework Help: Share of Uranium isotopes

1. Apr 17, 2014

### skrat

1. The problem statement, all variables and given/known data
Current share of Uranium isotope on Earth is 99.28% ($^{238}U$) and 0.72% ($^{235}U$), half-life times are $7.04\cdot 10^8 years$ ($^{235}U$) and $4.468\cdot 10^9 years$ ($^{238}U$). Calculate the ratio between the isotopes $4.5\cdot 10^9 years$ ago.

2. Relevant equations

3. The attempt at a solution

If hope it is ok to say that $N(t)=N_0e^{-\frac{t}{\tau }}$.

Let's now say that $N_{238}(t)=Ae^{-\frac{t}{\lambda }}$ where $\lambda =\frac{t_{1/2}^{238}}{ln2}$ and
$N_{235}(t)=Be^{-\frac{t}{\mu }}$ where $\mu =\frac{t_{1/2}^{235}}{ln2}$.

Now we know that $\frac{N_{238}(t)}{N_{238}(t)+N_{235}(t)}=0.9928$ and also $\frac{N_{235}(t)}{N_{238}(t)+N_{235}(t)}=0.0072$

Knowing this, I can write:

$\frac{N_{238}(t)}{N_{238}(t)+N_{235}(t)}=\frac{Ae^{-\frac{t}{\lambda }}}{Ae^{-\frac{t}{\lambda }}+Be^{-\frac{t}{\mu }}}=0.9928$ and

$\frac{Be^{-\frac{t}{\mu }}}{Ae^{-\frac{t}{\lambda }}+Be^{-\frac{t}{\mu }}}=0.0072$

Dividing last two gives me:

$\frac{Ae^{-\frac{t}{\lambda }}}{Be^{-\frac{t}{\mu }}}=\frac{0.9928}{0.0072}$ and

$A=Be^{-t/\mu +t/\lambda }\frac{0.9928}{0.0072}$

Inserting this into $\frac{Be^{-\frac{t}{\mu }}}{Ae^{-\frac{t}{\lambda }}+Be^{-\frac{t}{\mu }}}=0.0072$ leaves me with the result that $B=1$.

Knowing this also gives me a result for $A$, therefore $A=459.8$.

So.... If I am not mistaken, than following ratios should be the result I am searching:

$\frac{B}{B+A}=0.9978$ for Uranium 235 and $\frac{A}{B+A}=0.0022$ for Uranium 238.

Which to me is a bit confusing, so I would kindly ask somebody to tell me where I got it all wrong?

2. Apr 17, 2014

### BvU

Is not correct. It leaves you with the result that 0.0072 = 0.0072, which is not extremely useful.

B = 1 is a strange answer anyway, right ?
And: you should become suspicious when your answer stipulates that the fastest decaying isotope was less abundant in the past than the other one!

And now for some more constructive stuff:
First you want to check that 238U doesn't decay into 235U, which it doesn't.
So both can be considered independent.

You have expressions for the current ratio N238 / N235.
With $N(t)=N_0e^{-\frac{t}{\tau }}$ you also have expressions for $N(t)/N_0$ for both.
All you have to do is fill in the numbers !

3. Apr 17, 2014

### skrat

Fill in the numbers into what?

All I have is $\frac{N_{238}(t)}{N_{238}(t)+N_{235}(t)}=0.9928$ and $\frac{N_{235}(t)}{N_{238}(t)+N_{235}(t)}=0.0072$, of course dividing those two brings me to current ratio, just like you said but... ?

I don't get it what the next step is supposed to be..

4. Apr 17, 2014

### dauto

You reached the equation
$A=Be^{-t/\mu +t/\lambda }\frac{0.9928}{0.0072}$
which is correct. You should now plug in the known values for t, λ, and μ. and solve for A/B.

5. Apr 17, 2014

### skrat

Ok...

$A=3.30B$.