Answer:
[tex]\boxed{\textbf{1700 yr}}[/tex]
Step-by-step explanation:
[tex]-\dfrac{\text{d}L}{\text{d}t} = kL\\\\\dfrac{dL}{L}=-kdt\\\\\ln L = -kt + C\\\text{At t = 0, L = L$_{0}$, so C = $\ln L_{0}$}\\\ln L = -kt + \ln L_{0}\\\ln L_{0} - \ln L = kt\\\\\ln \dfrac{L_{0}}{L} =kt[/tex]
Data:
L₀ = 24 g
L = 20 g
k = 0.000 11 yr⁻¹
Calculation:
[tex]\ln \dfrac{24}{20} =0.000 11t\\\\\ln 1.2 = 0.000 11t\\\\0.1823 = 0.000 11t\\\\t = \dfrac{0.1823}{0.000 11} = \textbf{1700 yr}\\\\\text{It will take } \boxed{\textbf{1700 yr}}\text{ for the polonium to decay to 20 g}[/tex]