± Radioactive Decay Calculations: If a substance is radioactive, this means that

Question

± Radioactive Decay Calculations: If a substance is radioactive, this means that the nucleus is unstable and will therefore decay by any number of processes (alpha decay, beta decay, etc.). The decay of radioactive elements follows first-order kinetics. Therefore, the rate of decay can be described by the same integrated rate equations and half-life equations that are used to describe the rate of first-order chemical reactions: lnAtA0=kt and t1/2=0.693k; where A0 is the initial amount or activity, At is the amount or activity at time t, and k is the rate constant. By manipulation of these equations (substituting 0.693/t1/2 for k in the integrated rate equation), we can arrive at the following formula: fraction remaining=AtA0=(0.5)n; where n is the number of half-lives. The equation relating the number of half-lives to time t is n=tt1/2; where t1/2 is the length of one half-life.

Part A. You are using a Geiger counter to measure the activity of a radioactive substance over the course of several minutes. If the reading of 400. counts has diminished to 100. counts after 53.8 minutes , what is the half-life of this substance? Express your answer with the appropriate units.

Part B. An unknown radioactive substance has a half-life of 3.20 hours . If 16.1 g of the substance is currently present, what mass A0 was present 8.00 hours ago? Express your answer with the appropriate units.

Part C. Americium-241 is used in some smoke detectors. It is an alpha emitter with a half-life of 432 years. How long will it take in years for 39.0 %of an Am-241 sample to decay? Express your answer with the appropriate units.

Part D. A fossil was analyzed and determined to have a carbon-14 level that is 20 % that of living organisms. The half-life of C-14 is 5730 years. How old is the fossil? Express your answer with the appropriate units.

Explanation / Answer

Answers.

Part A:

ln[A]/[Ao] = -kt

ln 400/100 = -(k)(44.1 min)

k = 0.03014 min-1

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t1/2 = 0.693/k

t1/2 = 0.693/0.03014min-1

t1/2 = 22.04 min

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Part B:

t1/2 = 0.693/k

k = 0.693/3.20 hr

k = 0.2165 hr-1

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ln[16.1 g]/[Ao] = -(0.2165 hr-1)(3.20 hr)

ln[16.10 g]/[Ao] = -0.6928

ln[16.10 g]/[Ao] = e-0.6928

[16.10 g]/[Ao] = 0.5001

[Ao] = 32.18 g

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Part C:

[A]/[Ao] = 0.5n

n = t/t1/2

log(0.39) = log(0.5)t/432

t = 1.3584 x 432

t = 586.85 y

another approach:

k = 0.693/432 y

k = 0.0016 y-1

ln[A]/[Ao] = -kt

ln(0.39) = -(0.0016y-1)t

t = 586.97 y

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Part D:

[A]/[Ao] = 0.5n

n = t/t1/2

log(0.2) = log(0.5)t/5730

t = 13304.64 y

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