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:
[tex]\frac{dm}{dt} = -\frac{m(t)}{\tau}[/tex]
Where:
[tex]m(t)[/tex] - Current mass of the isotope, measured in grams.
[tex]\tau[/tex] - Time constant, measured in days.
Whose solution is:
[tex]m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }[/tex]
Where [tex]m_{o}[/tex] is the initial mass of the isotope, measured in grams.
Our first step is to determine the value of the time constant:
[tex]-\frac{t}{\tau} = \ln \frac{m(t)}{m_{o}}[/tex]
[tex]\tau = -\frac{t}{\ln \frac{m(t)}{m_{o}} }[/tex]
If we know that [tex]\frac{m(t)}{m_{o}} = 0.016[/tex] and [tex]t = 828\,days[/tex], then the time constant of the radioisotope is:
[tex]\tau = -\frac{828\,days}{\ln 0.016}[/tex]
[tex]\tau \approx 200.234\,days[/tex]
And lastly we find the half-life of polonium-210 ([tex]t_{1/2}[/tex]), measured in days, by using this expression:
[tex]t_{1/2} = \tau \cdot \ln 2[/tex]
[tex]t_{1/2} = (200.234\,days)\cdot \ln 2[/tex]
[tex]t_{1/2}\approx 138.792\,days[/tex]
The half-life of polonium-210 is approximately 138.792 days.
