Question

4. Phosphorus-32 has a half-life of 14.3days. How long would it take a 50-mg sample to decay to 3 mg

Answer 1

Answer:

58.0 days

Step-by-step explanation:

The equation for the radioactive decay is

$m(t)=m_0 (\frac{1}{2})^{-t/t_{1/2}}$

where

m(t) is the mass of the sample left at time t

$m_0$ is the initial mass of the sample

$t_{1/2}$ is the half-life

For the phosporus-32 isotope in the problem, we have:

$t_{1/2}=14.3 d$ (half-life in days)

$m_0 = 50 mg$ is the initial mass

$m(t)=3 mg$ is the mass at time t

Solving for t, we find the time needed for the sample to reduce to 3 mg:

$\frac{m(t)}{m_0} = (\frac{1}{2})^{-t/t_{1/2}}\\t=-t_{1/2} ln_{1/2} (\frac{m}{m_0})=-(14.3) ln_{1/2} (\frac{50}{3})=58.0 d$