Answer:
[tex]7.3\times 10^2\ days[/tex]
Explanation:
Given that:
Half life = [tex]9.8\times 10^3[/tex] days
[tex]t_{1/2}=\frac{\ln2}{k}[/tex]
Where, k is rate constant
So,
[tex]k=\frac{\ln2}{t_{1/2}}[/tex]
[tex]k=\frac{\ln2}{9.8\times 10^3}\ days^{-1}[/tex]
The rate constant, k = 0.00007 days⁻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 concentration at time t
[tex][A_0][/tex] is the initial concentration
Given:
5 % is lost which means that 0.05 of [tex][A_0][/tex] is decomposed. So,
[tex]\frac {[A_t]}{[A_0]}[/tex] = 1 - 0.05 = 0.95
t = ?
[tex]\frac {[A_t]}{[A_0]}=e^{-k\times t}[/tex]
[tex]0.95=e^{-0.00007\times t}[/tex]
t = 732.76 days = [tex]7.3\times 10^2\ days[/tex]
