Question

2) If a rock has 70.7% Uranium-235 (parent) to 29.3% Lead-207 (daughter) isotopes, what is the age of that rock

Answer 1

The age of that rock is approximately 353.3 million years.

Step-by-step explanation:

According to the statement, 29.3 percent of the original mass of Uranium-235 became Lead-207. We know that decay of isotopes is represented by the following ordinary linear differential equation:

`(dm)/(dt) = -(m)/(\tau)` (Eq. 1)

Where:

`(dm)/(dt)` - Rate of change of mass in time, measured in grams per year.

`m` - Current mass of the isotope, measured in grams.

`\tau` - Time constant, measured in years.

The solution of this differential equation is:

`m(t) = m_(o)\cdot e^{-(t)/(\tau) }` (Eq. 2)

Where:

`m_(o)` - Initial mass of the isotope, measured in grams.

`t` - Time, measured in years.

And we solve the expression for time herein:

`(m(t))/(m_(o)) = e^{-(t)/(\tau) }`

`t = -\tau\cdot \ln (m(t))/(m_(o))`

Besides, time constant can be calculated as a function of half-life. Please notice that half-life of Uranium-235 is 704 million years. The equation is presented below:

`\tau = (t_(1/2))/(\ln 2)` (Eq. 3)

Where
`t_(1/2)` is the half-life of the isotope, measured in years.

If we know that
`(m(t))/(m_(o)) = 0.707` and
`t_(1/2) = 704* 10^(6)\,yr`, then:

`\tau = (704* 10^(6)\,yr)/(\ln2)`

`\tau \approx 1.016* 10^(9)\,yr`

`t = -(1.019* 10^(9)\,yr)\cdot \ln 0.707`

`t \approx 353.312* 10^(6)\,yr`

The age of that rock is approximately 353.3 million years.