Answer:
2.79 mg iron-59 is left in the sample after 267 days.
Explanation:
Given that:
Half life = 44.5 days
[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}{44.5}\ days^{-1}[/tex]
The rate constant, k = 0.0156 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 = 180 mg
Time = 267 days
So,
[tex][A_t]=180\ mg\times e^{-0.0156\times 267}[/tex]
[tex][A_t]=180\times e^{-0.0156\times 267}\ mg[/tex]
[tex][A_t]=2.79\ mg[/tex]
2.79 mg iron-59 is left in the sample after 267 days.
