Question

Rutherford once assumed that when the earth was first formed, it contained equal amounts of U-235 and U-238. From this, he was able to determine the age of the earth. Find the Rutherford age of the earth.

Answer 1

Answer:

• the Rutherford age of the Earth is 5.942 x 10^9 years

Explanation:

As the radioactive decay is an exponential decay, lets first remember how to solve an exponential decay problem.

In an exponential decay the quantity of substance N at time t is given by:

$N(t) = N_0 e^{-\frac{t}{\tau}}$

where $N_0$ is the initial quantity of substance and $\tau$ is the mean lifetime of the substance.

For our problem we start with the same quantity of U-235 and U 238. Lets call this quantity as $N_0$.

The quantity of U-235 after a time t will be:

$^{U-235}N(t) = N_0 e^{-\frac{t}{\tau_{235}}}$

and for U-238

$^{U-238}N(t) = N_0 e^{-\frac{t}{\tau_{238}}}$

Lets call the ratio between this two r. r will be:

$r(t) = \frac{^{U-235}N(t)}{^{U-238}N(t)} = \frac{ N_0 e^{ -\frac{t}{ \tau_{235} } } }{ N_0 e^{ -\frac{t}{ \tau_{238} } } }$

$r(t) = \frac{ e^{ -\frac{t}{ \tau_{235} } } }{ e^{ -\frac{t}{ \tau_{238} } } }$

$r(t) = e^{ -\frac{t}{ \tau_{235} } + \frac{t}{ \tau_{238} } }$

$ln ( r(t) ) = \frac{t}{ \tau_{238} } - \frac{t}{ \tau_{235} }$

$ln ( r(t) ) = t ( \frac{1}{ \tau_{238} } } -\frac{1}{ \tau_{235} } )$

$\frac{ln ( r(t) )}{ ( \frac{1}{ \tau_{238} } -\frac{1}{ \tau_{235} } } = t$

Now, in the present time the abundance of U-235 is 0.720% and the abundance of U-238 is 99.274%. This gives us a ratio of:

$r(t_{present}) =\frac{0.720 \ \%}{99.274 \ \%} = 7.2526 \ 10^{-3}$

the mean lifetime of U-235 is

$\tau_{235} = 1.016 \ 10^9 years$

and the mean lifetime of U-238 is

$\tau_{238} = 6.445 \ 10^9 years$

so

$\frac{1}{ \tau_{238} } } -\frac{1}{ \tau_{235} } = \frac{1}{ 6.445 \ 10^9 years } } -\frac{1}{ 1.016 \ 10^9 years } = -8.2909 \ 10^{-10} \frac{1}{year}$

Taking all this in consideration, we get:

$t_{present}=\frac{ln ( r(t_{present}) )}{ ( \frac{1}{ \tau_{238} } -\frac{1}{ \tau_{235} } )}$

$t_{present}=\frac{ln ( 7.2526 \ 10^{-3} )}{-8.2909 \ 10^{-10} \frac{1}{year}}$

$t_{present}=\frac{ 4.926 }{-8.2909 \ 10^{-10} \frac{1}{year}}$

$t_{present}= 5.942 \ 10^9 years$