Question

A sample of the lava flow within the outcrop has 707 Uranium-235 atoms, and 293 Lead-207 atoms. How old is the lava flow?

Answer 1

Answer:

The lava flow is approximately 351.925 million years.

Explanation:

We find that atoms of Uranium-235 decays and turns into Lead-207 and that half-life of the former one is $703.8\times 10^{6}$ years. The initial amount of Uranium-235 is the sum of current atoms of Uranium-235 and Lead-207. The decay of isotopes is modelled by the following ordinary differential equation:

$\frac{dn}{dt} = -\frac{n}{\tau}$ (Eq. 1)

Where:

$\frac{dn}{dt}$ - Rate of change of the amount of Uranium-235 atoms, measured in atoms per year.

$n$ - Current amount of Uranium-235 atoms, measured in atoms.

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

The solution of this differential equation is described below:

$n = n_{o}\cdot e^{-\frac{t}{\tau} }$ (Eq. 2)

Where:

$n_{o}$ - Initial amount of Uranium-235 atoms, measured in atoms.

$t$ - Time, measured in years.

In addition, we can calculate the time constant in terms of the half-life:

$\tau = \frac{t_{1/2}}{\ln 2}$ (Eq. 3)

If we know that $n_{o} = 1000\,atoms$, $n = 707\,atoms$ and $t_{1/2} = 703.8\times 10^{6}\,yr$, then the age of the lava flow is:

From (Eq. 2):

$t = -\tau \cdot \ln \frac{n}{n_{o}}$

By (Eq. 3);

$\tau = \frac{703.8\times 10^{6}\,yr}{\ln 2}$

$\tau \approx 1.015\times 10^{9}\,yr$

$t = -(1.015\times 10^{9})\cdot \ln \frac{707\,atoms}{1000\,atoms}$

$t \approx 351.925\times 10^{6}\,yr$

The lava flow is approximately 351.925 million years.