Question

Uranium-235 decays to form lead-207. The half-life of uranium-235 is 704,000,000 years. What is the approximate age of an igneous rock with a 1:1 ratio of uranium-235 atoms to lead-207 atoms?

Answer 1

Answer:

The approximate age of the igneous rock is $7.042\times 10^{6}$ years.

Explanation:

Chemically speaking, the Uranium-235 decays to form Lead-207 throughout time. All isotopes decay exponentially by means of the following model:

$m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }$ (1)

In addition, we can determine the time constant of the former isotope in terms of its half life, that is:

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

Where:

$m_{o}$ - Initial mass of Uranium-235, in atoms.

$m(t)$ - Current mass of Uranium-235, in atoms.

$\tau$ - TIme constant, in years.

$t_{1/2}$ - Half-life of Uranium-235, in years.

Please remind that a 1 : 1 ratio of Uranium-235 to Lead-207 atoms means that current mass of Uranium-235 is a half of its initial mass. If we know that $m(t) = 0.5\cdot m_{o}$ and $t_{1/2} = 7.04\times 10^{8}\,yr$, then the approximate age of the igneous rock is:

$\tau = \frac{7.04\times 10^{8}\,yr}{\ln 2}$

$\tau \approx 1.016\times 10^{9}\,yr$

$t = -\tau \cdot \ln \left(\frac{m(t)}{m_{o}} \right)$

$t = -(1.016\times 10^{9}\,yr)\cdot \ln \frac{1}{2}$

$t \approx 7.042\times 10^{6}\,yr$

The approximate age of the igneous rock is $7.042\times 10^{6}$ years.