Question

On your birthday, you measure the activity of the sample 210Bi which has a half-life of 5.01 days. The initial activity that you measure is 1µCi.

1. What is the approximate activity of the sample on your next birthday? Calculate
2. the decay constant
3. the mean life
4. initial number of atoms.

Answer 1

1. A year of 365 days is equivalent to 365 d/5.01 d ≈ 73 half-lives.
   Thus, the activity will be reduced after one year to approximately (1/2)⁷³ (1.000 μCi) ~ 10^-22 μCi.
2. Initial measure R₀ = 1.000 μCi
   = 10^-6 x 3.7 x 10¹⁰
   = 3.7 x 10⁴ Bq
   After 1 year, the measure R = 10^-22 μCi.
   = 10^-22 x 10^-6 x 3.7 x 10¹⁰
   = 3.7 x 10^-18 Bq
   decay constant
   ${\displaystyle \lambda =\frac{1}{\text{t}}\ln \left(\frac{{\text{R}}_0}{\text{R}}\right)=\left(\frac{1}{1\text{year}}\right)\ln \left(\frac{3.7\times {10}^4}{3.7\times {10}^{-18}}\right)}$
   ${\displaystyle =\frac{1}{3.156\times {10}^7}\ln \left({10}^{22}\right)}$
   
   ${\displaystyle \lambda =\frac{50.657}{3.1567\times {10}^7}}$
   
   ${\displaystyle \lambda =1.6\times {10}^{-6}{\text{s}}^{-1}}$
3. Mean life
   ${\displaystyle \tau =\frac{1}{\lambda }=\frac{1}{1.6\times {10}^{-6}}\text{s}[1\text{s}=\frac{1}{86400}\text{days}]}$
   ${\displaystyle \tau =\frac{1}{1.6\times 86400\times {10}^{-6}}=\frac{1}{138240}\times {10}^6}$
   = 7.2337 days
   τ = 7.24 days
4. Initial number of atoms
   R₀ = λN ; N = ${\displaystyle \frac{{\text{R}}_0}{\lambda }}$
   ${\displaystyle =\frac{3.7\times {10}^4}{1.6\times {10}^{-6}}}$; N = 2.31 x 10¹⁰