Fluorine-18 undergoes positron emission with a half-life of 110. minutes. If a patient is given a 248 mg dose for a PET scan, how long will it take for the amount of fluorine-18 to drop to 83 mg? (Assume that none of the fluorine is excreted from the body.)Single choice
A
174 min
B
300. min
C
99 min
D
211 min
E
132 min

Question

Fluorine-18 undergoes positron emission with a half-life of 110. minutes. If a patient is given a 248 mg dose for a PET scan, how long will it take for the amount of fluorine-18 to drop to 83 mg? (Assume that none of the fluorine is excreted from the body.)Single choice

A

174 min

B

300. min

C

99 min

D

211 min

E

132 min

BlackTom AI · Accepted Answer

Radioactive decay follows first-order kinetics with N(t) = N0 · (1/2)^(t / T1/2), where T1/2 is the half-life. Here T1/2 = 110.0 min, N0 = 248 mg, and we seek t when N(t) = 83 mg.

1. Relevant concepts
- Exponential decay governed by the half-life formula N(t) = N0 · (1/2)^(t/T1/2).
- Solving for time requires taking the natural logarithm of both sides.

2. Step-by-step reasoning and algebra
Start with the decay equation:
83 = 248 · (1/2)^(t / 110)

Divide both sides by 248:
83 / 248 = (1/2)^(t / 110)

Compute the ratio:
83 / 248 = 0.334677419...

Take natural logs:
ln(0.334677419) = (t / 110) · ln(1/2)

Solve for t:
t = 110 · ln(0.334677419) / ln(1/2)

3. Intermediate calculations
ln(0.334677419) ≈ -1.094589
ln(1/2) = -0.693147
t = 110 · (-1.094589) / (-0.693147) ≈ 110 · 1.579422 ≈ 173.7364 min

Rounding gives t ≈ 174 min.

4. Why this answer is correct
The calculation uses the exact half-life exponential relationship and logarithms to invert the exponential. The numeric steps show the patient’s fluorine-18 falls from 248 mg to 83 mg in about 173.7 minutes, which rounds to 174 min.

5. Why the other options are incorrect
- 99 min: shorter than one half-life (110 min); after 99 min the amount would still exceed one half-life remaining (~>124 mg), so 99 min is too short.
- 132 min: between 110 and 174 but too small — direct calculation shows 132 min leaves more than 83 mg.
- 211 min and 300 min: both lie beyond the needed time; 211 min would give less than 83 mg (closer to two half-lives), and 300 min is well past that.

Conclusion: The computed time is approximately 174 min, matching the provided correct answer (174 min).

Fluorine-18 undergoes positron emission with a half-life of 110. minutes. If a patient is given a 248 mg dose for a PET scan, how long will it take for the amount of fluorine-18 to drop to 83 mg? (Assume that none of the fluorine is excreted from the body.)Single choice

Similar Questions

[math: Pa91234] undergoes a series of decays to become [math: Rn86222]How many types of what radioactive particles are emitted through this process?

Fluorine-18 undergoes positron emission with a half-life of 1.10 × 102 If a patient is given a 248 mg dose for a PET scan, how long will it take for the amount of fluorine-18 to drop to 83 mg? (Assume that none of the fluorine is excreted from the body.)

A radioactive isotope has a half-life of 10 days. What is the time required for the activity of the isotope to fall to [math: 116]th of its original value?

In 168 seconds the activity of a radioisotope falls to [math: 18] th of its original value. What is its half-life?

A Geiger counter is used to measure the radioactive disintegrations from a sample of a certain radioisotope. The count rate recorded is shown in the graph belowWhat is the half-life of this sample?

The number of radioactive nuclei in a particular sample decreases to one-eighth of its original number in 9 days. What is the half-life of these nuclei?

Question at position 1 Radioactive iodine-131 has a half-life of 8.0 days. What is the rate constant for the first-order decay of this radioactive isotope? 0.087/day 0.12/day 0.032/day 0.55/day 0.67/day

The half life of an isotope is 4 days if 16 days pass how many half life's pass?

More Practical Tools for Students Powered by AI Study Helper

Homework AI Solver

Stylized AI Paper Writer

Plagiarism Checker Assistant

Citation AI Academic Writing Tool

In-Class Translation Assistant

AI Note Generator

AI Quiz Answers

Past Exam Questions from University Test Bank

Smart Practice Assistant

Adaptive Practice

Making Your Study Simpler