Question

What is the half-life of polonium-210 if it takes 828 days for a sample to be reduced to 1.6% of its original mass?

Answer 1

Answer:

The half-life of polonium-210 is approximately 138.792 days.

Explanation:

We must remember that the decay of a radioisotope is modelled by this ordinary differential equation:

$\frac{dm}{dt} = -\frac{m(t)}{\tau}$

Where:

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

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

Whose solution is:

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

Where $m_{o}$ is the initial mass of the isotope, measured in grams.

Our first step is to determine the value of the time constant:

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

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

If we know that $\frac{m(t)}{m_{o}} = 0.016$ and $t = 828\,days$, then the time constant of the radioisotope is:

$\tau = -\frac{828\,days}{\ln 0.016}$

$\tau \approx 200.234\,days$

And lastly we find the half-life of polonium-210 ($t_{1/2}$), measured in days, by using this expression:

$t_{1/2} = \tau \cdot \ln 2$

$t_{1/2} = (200.234\,days)\cdot \ln 2$

$t_{1/2}\approx 138.792\,days$

The half-life of polonium-210 is approximately 138.792 days.