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

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

Step-by-step 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^{-(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^{-(t)/(\tau_(235))}`

and for U-238

`^(U-238)N(t) = N_0 e^{-(t)/(\tau_(238))}`

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

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

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

`r(t) =  e^{ -(t)/( \tau_(235) )   + (t)/( \tau_(238) )  }`

`ln ( r(t) ) =   (t)/( \tau_(238) )  - (t)/( \tau_(235) )`

`ln ( r(t) ) =   t   ( (1)/( \tau_(238) ) }  -(1)/( \tau_(235) )  )`

`(ln ( r(t) ))/( ( (1)/( \tau_(238) )   -(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)) =(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

`(1)/( \tau_(238) ) }  -(1)/( \tau_(235) ) =  (1)/( 6.445 \ 10^9 years ) }  -(1)/( 1.016 \ 10^9 years )  = -8.2909 \ 10^(-10) (1)/(year)`

Taking all this in consideration, we get:

`t_(present)=(ln ( r(t_(present)) ))/( ( (1)/( \tau_(238) )  -(1)/( \tau_(235) )  ))`

`t_(present)=(ln ( 7.2526 \ 10^(-3) ))/(-8.2909 \ 10^(-10) (1)/(year))`

`t_(present)=( 4.926  )/(-8.2909 \ 10^(-10) (1)/(year))`

`t_(present)= 5.942 \ 10^9 years`