Answer:
96.9 years would pass before the strontium-90 concentration would drop to 1.0 ppm.
Explanation:
Given that:
Half life = 28.8 years
[tex]t_{1/2}=\frac {ln\ 2}{k}[/tex]
Where, k is rate constant
So,
[tex]k=\frac {ln\ 2}{t_{1/2}}[/tex]
[tex]k=\frac {ln\ 2}{28.8}\ years^{-1}[/tex]
The rate constant, k = 0.024067 years⁻¹
Using integrated rate law for first order kinetics as:
[tex][A_t]=[A_0]e^{-kt}[/tex]
Where,
[tex][A_t][/tex] is the final concentration= 1.0 ppm
[tex][A_0][/tex] is the initial concentration = 10.3 ppm
Time = ?
So,
[tex]\frac{1.0}{10.3}=e^{-0.024067\times t}[/tex]
[tex]\ln \left(\frac{1}{10.3}\right)=-0.024067t[/tex]
[tex]t=\frac{\ln \left(10.3\right)}{0.024067}\ years[/tex]
t = 96.9 years
96.9 years would pass before the strontium-90 concentration would drop to 1.0 ppm.
