Answer:
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:
[tex]N(t) = N_0 e^{-\frac{t}{\tau}}[/tex]
where [tex]N_0[/tex] is the initial quantity of substance and [tex]\tau[/tex] 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 [tex]N_0[/tex].
The quantity of U-235 after a time t will be:
[tex]^{U-235}N(t) = N_0 e^{-\frac{t}{\tau_{235}}}[/tex]
and for U-238
[tex]^{U-238}N(t) = N_0 e^{-\frac{t}{\tau_{238}}}[/tex]
Lets call the ratio between this two r. r will be:
[tex]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} } } }[/tex]
[tex]r(t) = \frac{ e^{ -\frac{t}{ \tau_{235} } } }{ e^{ -\frac{t}{ \tau_{238} } } }[/tex]
[tex]r(t) = e^{ -\frac{t}{ \tau_{235} } + \frac{t}{ \tau_{238} } } [/tex]
[tex] ln ( r(t) ) = \frac{t}{ \tau_{238} } - \frac{t}{ \tau_{235} } [/tex]
[tex] ln ( r(t) ) = t ( \frac{1}{ \tau_{238} } } -\frac{1}{ \tau_{235} } ) [/tex]
[tex]\frac{ln ( r(t) )}{ ( \frac{1}{ \tau_{238} } -\frac{1}{ \tau_{235} } } = t[/tex]
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:
[tex]r(t_{present}) =\frac{0.720 \ \%}{99.274 \ \%} = 7.2526 \ 10^{-3}[/tex]
the mean lifetime of U-235 is
[tex]\tau_{235} = 1.016 \ 10^9 years[/tex]
and the mean lifetime of U-238 is
[tex]\tau_{238} = 6.445 \ 10^9 years[/tex]
so
[tex]\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}[/tex]
Taking all this in consideration, we get:
[tex] t_{present}=\frac{ln ( r(t_{present}) )}{ ( \frac{1}{ \tau_{238} } -\frac{1}{ \tau_{235} } )} [/tex]
[tex] t_{present}=\frac{ln ( 7.2526 \ 10^{-3} )}{-8.2909 \ 10^{-10} \frac{1}{year}} [/tex]
[tex] t_{present}=\frac{ 4.926 }{-8.2909 \ 10^{-10} \frac{1}{year}} [/tex]
[tex] t_{present}= 5.942 \ 10^9 years [/tex]
