Question

Fluorine-18, which is often used in PET scans to locate tumors, decays to form a positron and oxygen-18. After 90.0 minutes, 1.00 18O is present for every 1.30 18F. Estimate the half-life of F-18.

Answer 1

The rate constant for the decomposition of fluorine-18 is approximately 0.0063 min^-1. After 5.59 hours, approximately 23.6% of the radioactivity will remain. It takes approximately 439.6 minutes for 99.99% of the fluorine-18 to decay.

Step-by-step explanation:

The rate constant for the decomposition of fluorine-18 can be determined using the half-life of the isotope. The half-life of fluorine-18 is given as 109.7 minutes. The rate constant (k) can be calculated using the formula:

k = 0.693 / t1/2

Substituting the half-life of fluorine-18 into the formula:

k = 0.693 / 109.7 = 0.0063 min-1

Therefore, the rate constant for the decomposition of fluorine-18 is approximately 0.0063 min-1.

To calculate the percent of radioactivity remaining after a certain time, we can use the formula:

Percent remaining = (1 - (e-kt)) x 100

Substituting the values into the formula, where t=5.59 hours = 335.4 minutes:

Percent remaining = (1 - (e-(0.0063 x 335.4))) x 100 = 23.6%

Therefore, after 5.59 hours, approximately 23.6% of the radioactivity will remain.

The time it takes for 99.99% of fluorine-18 to decay can be calculated using the formula:

t = (ln(1 - (percent remaining / 100))) / -k

Substituting the values into the formula, where percent remaining = 0.01:

t = (ln(1 - (0.01 / 100))) / -0.0063 = 439.6 minutes

Therefore, it takes approximately 439.6 minutes for 99.99% of the fluorine-18 to decay.