Question

14. A sample that originally contained 2.5 g of rubidium-87 now contains 1.25 g. The half-life of rubidium-87 is 6 x 10^10 years. How old is the sample? Is this possible? Why or why not?​

Answer 1

Answer:

The age of the sample is $t \approx 6\times 10^{10}\,yr$. It is possible, since the definition of half-life is the time taken by the isotope to halve its mass.

Explanation:

All radioactive isotopes decays exponentially, the decay is represented by the following formula:

$m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }$ (1)

Where:

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

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

$t$ - Time, measured in years.

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

Now we clear the time of the isotope within the formula:

$t = -\tau\cdot \ln \frac{m(t)}{m_{o}}$

In addtion, the time constant can be calculated in terms of the half-life ($t_{1/2}$), measured in years:

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

If we know that $m_{o} = 2.5\,g$, $m(t) = 1.25\,g$ and $t_{1/2} = 6\times 10^{10}\,yr$, then the age of the isotope is:

$\tau = \frac{6\times 10^{10}\,yr}{\ln 2}$

$\tau \approx 8.656\times 10^{10}\,yr$

$t = -(8.656\times 10^{10}\,yr)\cdot \ln \frac{1.25\,g}{2.5\,g}$

$t \approx 6\times 10^{10}\,yr$

The age of the sample is $t \approx 6\times 10^{10}\,yr$. It is possible, since the definition of half-life is the time taken by the isotope to halve its mass.