Answer:
After 47.9 days, will remain 14.5mg of the isotope
Explanation:
The radioactive decay follows always first-order kinetics where its general law is:
Ln[A] = -Kt + ln[A]₀
Where [A] is actual concentration of the atom, k is rate constant, t is time and [A]₀ is initial concentration.
We can find rate constant from half-life as follows:
[tex]t_{1/2} = \frac{ln2}{K}[/tex]
K = ln 2 / 27.7 days
K = 0.025 days⁻¹
Replacing, initial amount of isotope is 48.0mg = [A]₀ , K is 0.025 days⁻¹ and t = 47.9 days:
Ln[A] = -Kt + ln[A]₀
Ln[A] = -0.025 days⁻¹*47.9 days + ln (48.0mg)
ln [A] = 2.6726
[A] = e^ (2.6726)
[A] = 14.5mg
