Answer:
58.0 days
Step-by-step explanation:
The equation for the radioactive decay is
[tex]m(t)=m_0 (\frac{1}{2})^{-t/t_{1/2}}[/tex]
where
m(t) is the mass of the sample left at time t
[tex]m_0[/tex] is the initial mass of the sample
[tex]t_{1/2}[/tex] is the half-life
For the phosporus-32 isotope in the problem, we have:
[tex]t_{1/2}=14.3 d[/tex] (half-life in days)
[tex]m_0 = 50 mg[/tex] is the initial mass
[tex]m(t)=3 mg[/tex] is the mass at time t
Solving for t, we find the time needed for the sample to reduce to 3 mg:
[tex]\frac{m(t)}{m_0} = (\frac{1}{2})^{-t/t_{1/2}}\\t=-t_{1/2} ln_{1/2} (\frac{m}{m_0})=-(14.3) ln_{1/2} (\frac{50}{3})=58.0 d[/tex]
