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Radioactive isotope X has a half-life of 500,000 years. This isotope may be found in some types of volcanic rocks. A particular sample of volcanic rock taken from a layer that covered up some of the earliest known human-like footprints contains 0.125 mg of isotope X. The volcanic rock sample originally contained 8.00 mg of isotope X. How long ago were these footprints made. I need the equation. Thanks. please I am stuck on this question. PLEASE

User Blagae
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Answer:

The age of the earliest known human-like footprints is approximately 3000000 years.

Step-by-step explanation:

Decay of isotopes is represented by the following ordinary differential equation:


(dm)/(dt) = -(m)/(\tau) (Eq. 1)

Where:


(dm)/(dt) - First derivative of mass with respect to time, measured in miligrams per year.


m - Current mass of the isotope, measured in miligrams.


\tau - Time constant, measured in years.

Now we proceed to obtain the solution of the differential equation:


\int\limits {(dm)/(m) } = -(1)/(\tau)\int dt


\ln m = -(t)/(\tau)+C


m(t) = e^{-(t)/(\tau)+C }


m(t) = m_(o)\cdot e^{-(t)/(\tau) } (Eq. 2)

Where:


m_(o) - Initial mass of the isotope, measured in miligrams.


t - Time, measured in years.


\tau - Time constant, measured in years.

We proceed to clear time within the formula presented above:


\ln (m(t))/(m_(o)) = -(t)/(\tau)


t = -\tau \cdot \ln (m(t))/(m_(o))

In addition, time constant can be found as a function of half-life:


\tau = (t_(1/2))/(\ln 2) (Eq. 3)

If we know that
t_(1/2) = 500000\,yr,
m_(o) = 8\,mg and
m(t) = 0.125\,mg, the age of the earliest known human-like footprints is:


\tau = (500000\,yr)/(\ln 2)


\tau \approx 721347.520\,yr


t = -(721347.520\,yr)\cdot \ln \left((0.125\,mg)/(8\,mg) \right)


t \approx 3000000\,yr

The age of the earliest known human-like footprints is approximately 3000000 years.

User GolfARama
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