# Age of the earth - uranium radioactive decay

1. Mar 1, 2006

### A_I_

Natural Uranium found in the Earth's crust contains the isotopes A=235 and A=238 in the atom ratio of 7.3*10^-3 to 1. Assuming that the time of formation of the Earth these two isotopes were formed equally, and that the mean lives are 1.03*10^9 years and 6.49*10^9 years respectively, show that the age of Earth is 5.15*10^9 years.

ok firt i set up the equation for the Uranium 238 decay:
N = N(o) e^(-lambda*t)

N/N(0) = 7.3*10^-3 / 1 = e^-(t/1.03*10^9)

solving for t (using the natural log function)

i got: t = 5.06*10^9 years.
which is pretty close to the value in the problem.

I want to know if the way i solved is right or if i have to consider the decay of Uranium 235 also thus we will have to lambda's in the exponential function. If we use both we get another value which is close to 6.02*10^9 years.
I would like to take the opinion of few people here.
And Thanks for the help :)

Joe

2. Mar 1, 2006

### Tide

How can you address the ratio of U-235 to U-238 without considering the the decay of both of them?

3. Mar 1, 2006

### A_I_

so we have to consider the two meanlifes.
and we get the 6.02*10^9 years?
(which contradicts the age in the problem, well our teacher does a lot of mistakes.. so probably it's one of them? but i do know that the age of the Earth is estimated to be 5 billions years).

Do i make any sense?
Thanks again

4. Mar 4, 2006

### Staff: Mentor

The decay constant $\lambda$ = 0.693/t1/2.

One must use the ratio of the atoms in this problem.

N235(T)/N238(T) = 0.0073 =

(N235(0) $e^{-\lambda_{235}T}$)/(N238(0) $e^{-\lambda_{238}T}$), but

N235(0) = N238(0), so

0.0073 = $e^{-\lambda_{235}T}$/$e^{-\lambda_{238}T}$