Answer:
The half-life of the radioactive isotope is 346 years.
Explanation:
The decay rate of the isotope is modelled after the following first-order linear ordinary differential equation:
[tex]\frac{dm}{dt} = -\frac{m}{\tau}[/tex]
Where:
[tex]m[/tex] - Current isotope mass, measured in kilograms.
[tex]t[/tex] - Time, measured in years.
[tex]\tau[/tex] - Time constant, measured in years.
The solution of this differential equation is:
[tex]m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }[/tex]
Where [tex]m_{o}[/tex] is the initial mass of the isotope. It is known that radioactive isotope decays at a yearly rate of 0.2 % annually, then, the following relationship is obtained:
[tex]\%e = \frac{m(t)-m(t+1)}{m(t)}\times 100\,\% = 0.2\,\%[/tex]
[tex]1 - \frac{m(t+1)}{m(t)} = 0.002[/tex]
[tex]1 - \frac{m_{o}\cdot e^{-\frac{t+1}{\tau} }}{m_{o}\cdot e^{-\frac{t}{\tau} }}=0.002[/tex]
[tex]1 - e^{-\frac{1}{\tau} } = 0.002[/tex]
[tex]e^{-\frac{1}{\tau} } = 0.998[/tex]
[tex]-\frac{1}{\tau} = \ln 0.998[/tex]
The time constant associated to the decay is:
[tex]\tau = -\frac{1}{\ln 0.998}[/tex]
[tex]\tau \approx 499.500\,years[/tex]
Finally, the half-life of the isotope as a function of time constant is given by the expression described below:
[tex]t_{1/2} = \tau \cdot \ln 2[/tex]
If [tex]\tau \approx 499.500\,years[/tex], the half-life of the isotope is:
[tex]t_{1/2} = (499.500\,years)\cdot \ln 2[/tex]
[tex]t_{1/2}\approx 346.227\,years[/tex]
The half-life of the radioactive isotope is 346 years.
