20101210

Physics quiz question: percent decrease in activity of americium-241

Physics 205B Quiz 7, fall semester 2010
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 29.21.

Am(241,95) is used in smoke detectors, and decays via alpha decay with a half-life of 432 years. The percent decrease in activity for a sample of Am(241,95) in ten years is:
(A) 0.23%.
(B) 0.33%.
(C) 1.6%.
(D) 2.3%.

Correct answer (highlight to unhide): (C)

The activity of a sample is given by:

R = R0·(1/2)(t/T1/2),

where T1/2 is the half-life. Solving for the activity at t = 10 years, assuming that R0 will be normalized to 1:

(R/R0) = (1/2)(t/T1/2) = (1/2)(10 years/432 years) = 0.9840,

So the percent decrease in 10 years will be 1 – 0.984 = 0.016, or 1.6%.

(Response (A) is 1/T1/2; response (B) is the time constant τ = 1/(T1/2·ln(2)); response (D) is 1 – exp(–10/432), presumably confusing T1/2 with τ in the exponential decay formula.)

Student responses
Section 70856
Exam code:quiz07g4mA
(A) : 1 student
(B) : 1 student
(C) : 6 students
(D) : 3 students

Success level: 54%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.67

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