Answer: The amount of I-125 after the given time is 5.002 mg.
Explanation:
All the radioactive reactions follows first order kinetics.
The equation used to calculate half life for first order kinetics:
[tex]t_{1/2}=\frac{0.693}{k}[/tex]
We are given:
[tex]t_{1/2}=60days[/tex]
Putting values in above equation, we get:
[tex]k=\frac{0.693}{60}=0.01155days^{-1}[/tex]
Rate law expression for first order kinetics is given by the equation:
[tex]k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}[/tex]
where,
k = rate constant = [tex]0.01155days^{-1}[/tex]
t = time taken for decay process = 240 days
[tex][A_o][/tex] = initial amount of the reactant = 80 mg
[A] = amount left after decay process = ?
Putting values in above equation, we get:
[tex]0.01155days^{-1}=\frac{2.303}{240days}\log\frac{80}{[A]}[/tex]
[tex][A]=5.002mg[/tex]
Hence, the amount of I-125 after the given time is 5.002 mg.
