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

User Bechir
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Answer:

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.

User Ioums
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