Question

Fluorine-18 is a radioactive isotope that decays by positron emission to form oxygen-18 with a half-life of 109.8 min. (A positron is a particle with the mass of an electron and a single unit of positive charge.) What is the rate constant (in min−1) for the decomposition of fluorine-18?

Answer 1

Answer:

k = 6.31 x 10⁻³ min⁻¹

Explanation:

The equation required to solve this question is:

k = 0693 / t half-life

This equation is derived from the the equation from the radioctive first order reactions:

ln At/A₀ = -kt

where At is the number of isoopes after a time t , and A₀ is the number of of isotopes initially. The half-life is when the number of  isotopes has decayed by a half, so

ln(1/2) = -kt half-life

-0.693 = - k t half-life

t half-life = 109.8 min

⇒ k = 0.693 / t half-life = 0.693 / 109.8 min = 6.31 x 10⁻³ min⁻¹