Share of Uranium isotopes

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?

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