Answer:
The age of that rock is approximately 353.3 million years.
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:
[tex]\frac{dm}{dt} = -\frac{m}{\tau}[/tex] (Eq. 1)
Where:
[tex]\frac{dm}{dt}[/tex] - Rate of change of mass in time, measured in grams per year.
[tex]m[/tex] - Current mass of the isotope, measured in grams.
[tex]\tau[/tex] - Time constant, measured in years.
The solution of this differential equation is:
[tex]m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }[/tex] (Eq. 2)
Where:
[tex]m_{o}[/tex] - Initial mass of the isotope, measured in grams.
[tex]t[/tex] - Time, measured in years.
And we solve the expression for time herein:
[tex]\frac{m(t)}{m_{o}} = e^{-\frac{t}{\tau} }[/tex]
[tex]t = -\tau\cdot \ln \frac{m(t)}{m_{o}}[/tex]
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:
[tex]\tau = \frac{t_{1/2}}{\ln 2}[/tex] (Eq. 3)
Where [tex]t_{1/2}[/tex] is the half-life of the isotope, measured in years.
If we know that [tex]\frac{m(t)}{m_{o}} = 0.707[/tex] and [tex]t_{1/2} = 704\times 10^{6}\,yr[/tex], then:
[tex]\tau = \frac{704\times 10^{6}\,yr}{\ln2}[/tex]
[tex]\tau \approx 1.016\times 10^{9}\,yr[/tex]
[tex]t = -(1.019\times 10^{9}\,yr)\cdot \ln 0.707[/tex]
[tex]t \approx 353.312\times 10^{6}\,yr[/tex]
The age of that rock is approximately 353.3 million years.
