Question

10. A doctor can obtain images of organs and bones by
injecting a patient with a solution of Tc-99m. The
half-life of the metastable Tc-99m is six hours.
Determine the fraction of an original sample of
metastable Tc-99m that remains unchanged after 24
hours

Answer 1

Answer:

6.25% remains after 24 hours

Explanation:

Isotope decay of an atom follows the equation:

Ln[A] = -kt + Ln[A]₀

Where [A] is amount of isotope after time t,

k is rate constant = ln 2 / Half-life = 0.11552hours⁻¹

[A]₀ is initial amount of A = 100%

Replacing:

Ln[A] = -0.1152hours⁻¹*24hours + Ln[100%]

Ln[A] = 1.8326

[A] = 6.25% remains after 24 hours