Part A You are using a Geiger counter to measure the activity of a radioactive s
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Question
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 44.1minutes , what is the half-life of this substance?
Express your answer with the appropriate units.
An unknown radioactive substance has a half-life of 3.20hours . If 19.8g of the substance is currently present, what mass A0 was present 8.00 hours ago?
Express your answer with the appropriate units.
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: In At / A0 = -kt and t1 / 2 = 0.693 / k 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 / t ½ for k in the integrated rate equation), we can arrive at the following formula: fraction remaining = At / A0 = (0.5)n where n is the number of half-lives. The equation relating the number of half-lives to time t is n = t / t1 / 2 where tl/2 is the length of one half-life. 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 44.1 minutes , what is the half-life of this substance? Express your answer with the appropriate units. An unknown radioactive substance has a half-life of 3.20hours . If 19.8g of the substance is currently present, what mass Aq was present 8.00 hours ago? Express your answer with the appropriate unitsExplanation / Answer
1) as decay is first order we have ln(A0 / At) = Kt
therefore ln(400/100) = K (44.1)
therefore K = 0.0314
therefore t0.5 = 0.693 / K = 0.693 / 0.0314 = 22.07 minutes
2) given half life = 3.20 hours
therefore K = 0.693 / t0.5 = 0.693 / 3.20 = 0.2166
given present mass At = 19.8 g and present time t = 8 hours
therefore ln(A0 / 19.8) = 0.2166 * 8
therefore by solving we get A0 = 111.99 grams