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 relative to every 1.30 ¹⁸F.Estimate the half-life of F-18.

Question

Quizplus · Accepted Answer

The decay equation for Fluorine-18 is:

F-18 --> O-18 + e+
where e+ is a positron.

Since the atomic mass number is conserved, the sum of the atomic masses on both sides of the equation must be equal. This gives:

18 = 18 + 0

Since the charges must also balance, the sum of the charges on both sides of the equation must be equal. This gives:

9 = 8 + 1

Therefore, the positron has a positive charge of +1.

The half-life of a radioactive substance is the time it takes for half of the original amount of the substance to decay. The equation for radioactive decay is:

N = N0 * (1/2)^(t/T)
where N is the amount of the substance at time t, N0 is the initial amount of the substance, T is the half-life, and (1/2)^(t/T) is the fraction of the substance that remains after time t.

We are given that after 90.0 minutes, the ratio of oxygen-18 to fluorine-18 is 1.00:1.30. This means that 1.30/2.30 = 0.5652 or 56.52% of the original amount of fluorine-18 has decayed. Using this information, we can solve for the half-life:

0.5652 = (1/2)^(90/T)
Take the logarithm of both sides:
log(0.5652) = log[(1/2)^(90/T)]
Use the property that log(a^b) = b*log(a):
log(0.5652) = (90/T)*log(1/2)
Divide both sides by log(1/2):
log(0.5652)/log(1/2) = 90/T
Solve for T:
T = 90/[log(0.5652)/log(1/2)]
T = 109 minutes

Therefore, the best choice is B, with a half-life of 109 minutes.

Fluorine-18,which Is Often Used in PET Scans to Locate Tumors,decays