Problem: Fluorine-18 is a radioactive isotope that decays by positron emission to form oxygen-18 with a half-life of 109.7 min. (A positron is a particle with the mass of an electron and a single unit of positive charge; the equation is 9 18F⟶ 8 18O + +10e)Physicians use 18F to study the brain by injecting a quantity of fluoro-substituted glucose into the blood of a patient. The glucose accumulates in the regions where the brain is active and needs nourishment.(b) If a sample of glucose containing radioactive fluorine-18 is injected into the blood, what percent of the radioactivity will remain after 5.59 h?

FREE Expert Solution

We can determine [A]t, but lack the value of k.

can be obtained from the equation of half-life for 1st order reactions:


Step 1: We rearrange the half-life equation to solve for k :

k=ln(2)t12

k=ln(2)109.7 min=6.32×10-3 min-1

Step 2: We convert the time elapsed to minutes:

t=5.59 h(60 min1 h)=335.4 min


Step 3: We determine [A]t in the 1st Order Rate Law:

ln[A]t=ln[A]o-ktln[A]t=ln(1)-(6.32×10-3 min-1)(335.4 min)

ln [A]t = -2.119728 

Ln has a base of e. When you divide a number by ln you will take e to the power of the number.

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Problem Details

Fluorine-18 is a radioactive isotope that decays by positron emission to form oxygen-18 with a half-life of 109.7 min. (A positron is a particle with the mass of an electron and a single unit of positive charge; the equation is 9 18F⟶ 8 18O + +10e)
Physicians use 18F to study the brain by injecting a quantity of fluoro-substituted glucose into the blood of a patient. The glucose accumulates in the regions where the brain is active and needs nourishment.

(b) If a sample of glucose containing radioactive fluorine-18 is injected into the blood, what percent of the radioactivity will remain after 5.59 h?

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