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Answer:
The approximate age of the igneous rock is [tex]7.042\times 10^{6}[/tex] years.
Explanation:
Chemically speaking, the Uranium-235 decays to form Lead-207 throughout time. All isotopes decay exponentially by means of the following model:
[tex]m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }[/tex] (1)
In addition, we can determine the time constant of the former isotope in terms of its half life, that is:
[tex]\tau = \frac{t_{1/2}}{\ln 2}[/tex] (2)
Where:
[tex]m_{o}[/tex] - Initial mass of Uranium-235, in atoms.
[tex]m(t)[/tex] - Current mass of Uranium-235, in atoms.
[tex]\tau[/tex] - TIme constant, in years.
[tex]t_{1/2}[/tex] - 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 [tex]m(t) = 0.5\cdot m_{o}[/tex] and [tex]t_{1/2} = 7.04\times 10^{8}\,yr[/tex], then the approximate age of the igneous rock is:
[tex]\tau = \frac{7.04\times 10^{8}\,yr}{\ln 2}[/tex]
[tex]\tau \approx 1.016\times 10^{9}\,yr[/tex]
[tex]t = -\tau \cdot \ln \left(\frac{m(t)}{m_{o}} \right)[/tex]
[tex]t = -(1.016\times 10^{9}\,yr)\cdot \ln \frac{1}{2}[/tex]
[tex]t \approx 7.042\times 10^{6}\,yr[/tex]
The approximate age of the igneous rock is [tex]7.042\times 10^{6}[/tex] years.