Question

g if 2 half-lives elapsed in a rock containing Carbon-14 and Nitrogen-14, what is the ratio of parent to daughter isotopes

Answer 1

Answer:

The ratio of parent to daughter isotopes is $\frac{1}{3}$.

Explanation:

We know that Carbon-14 decays in time and transforms into Nitrogen-14, being the latter the "daughter" of the first one. The decay of any isotope is represented by the following ordinary linear differential equation:

$\frac{dm}{dt} = -\frac{m}{\tau}$ (Eq. 1)

Where:

$\frac{dm}{dt}$ - Rate of change of the isotope mass, measured in grams per year.

$\tau$ - Time constant, measured in years.

$m$ - Current mass of the isotope, measured in grams.

The solution of this differential equation is:

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

Where:

$t$ - Time, measured in years.

$m_{o}$ - Initial mass of the isotope, measured in grams.

Time constant can be found as a function of half life. Please notice that half-life of Carbon-14 is 5760 years. The equation of time constant is:

$\tau = \frac{t_{1/2}}{\ln 2}$ (Eq. 3)

Where $t_{1/2}$ is the half-life of the isotope, measured in years.

If we know that $t_{1/2} = 5760\,yr$ and $t = 2\cdot t_{1/2}$, then we have that:

$\tau = \frac{5760\,yr}{\ln 2}$

$\tau \approx 8309.923\,yr$

$m(t) = m_{o}\cdot e^{-\frac{2\cdot (5760\,yr)}{8309.923\,yr} }$

$m = 0.25\cdot m_{o}$

Which means that 75 % of the original mass of Carbon-14 became Nitrogen-14. The parent-to-daughter ratio is:

$r = \frac{0.25\cdot m_{o}}{0.75\cdot m_{o}}$

$r = \frac{1}{3}$

The ratio of parent to daughter isotopes is $\frac{1}{3}$.