Answer:
[tex]t \approx 142.514\,days[/tex]
Step-by-step explanation:
The half-life of the phosphorus-32 is 14.3 days. The time constant of the isotope is:
[tex]\tau = \frac{14.3\,days}{\ln 2}[/tex]
[tex]\tau = 20.631\,days[/tex]
The decay formula is:
[tex]\frac{m}{m_{o}} = e^{-\frac{t}{\tau} }[/tex]
The time related to the decayed proportion of the radioactive isotope is:
[tex]-\frac{t}{\tau} = \ln \left(\frac{m}{m_{o}} \right)[/tex]
[tex]t = -\tau \cdot \ln \left(\frac{m}{m_{o}} \right)[/tex]
[tex]t = - (20.631\,days)\cdot \ln 0.001[/tex]
[tex]t \approx 142.514\,days[/tex]