A pure 50 mg sample of lead-212 has half-life 10.6 h. How much lead-212 remains after 53 h?

Solution & explanation

1. Given initial mass 50 mg, elapsed time 53 h and half-life 10.6 h.

2. Count the half-lives; both times use hours.

   $$n=\frac{53}{10.6}=5$$

3. Remaining fraction is (1/2) to this power.

   $$\frac{m}{m_0}={\left(\frac{1}{2}\right)}^5=\frac{1}{32}$$

4. Multiply initial mass by remaining fraction.

   $$m=\frac{50}{32}$$

5. Successive halves are 25, 12.5, 6.25, 3.125 and 1.5625 mg.

6. This is original radionuclide remaining, not the total mass of all daughter products.

   $$m=1.5625$$

1.5625 mg of lead-212 remains