Physics quiz question: percent decrease in radioactivity

Physics 205B Quiz 7, Spring Semester 2011
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 29.33(c)

Consider fluoride-18, (which is injected into patients undergoing a positron emission tomography (PET) scan), which decays via β+ decay with a half-life of 110 minutes. What is the percent decrease in activity for a sample of during a PET scan, which takes 30 minutes?
(A) 0.91%.
(B) 1.3%.
(C) 17%.
(D) 24%.

Correct answer: (C)

The activity of a sample is given by:

R = R_0*(1/2)^(t/T),

where T is the half-life. Solving for the activity at t = 30 minutes, assuming that R_0 will be normalized to 1:

(R/R_0) = (1/2)^(t/T) = (1/2)^(30 minutes/110 minutes) = 0.8278,

So the percent decrease in 30 minutes will be 1 - 0.8278 = 0.1722, or 17%.

Response (A) is 1/T; response (B) is the time constant tau = 1/(T*ln(2)); response (D) is 1 - exp^(-30/110), presumably confusing T with tau in the exponential decay formula.

Section 30882
Exam code: quiz07t0LO
(A) : 0 students
(B) : 0 students
(C) : 6 students
(D) : 2 students

"Success level": 75%
Discrimination index (Aubrecht & Aubrecht, 1983): -0.50

Compare to an older version of this question from Fall 2010.