An air sample is contaminated with ¹⁵O,which has a half-life of 2.03 min.One possible way to minimize its hazard is to pass it through a long pipe to allow it to decay inside the pipe until it can be safely released into the atmosphere.If the oxygen moves at a speed of 1.1 m/s in the pipe,how long must the pipe be for the sample to have decayed to 3.0% of its original activity just as it leaves the pipe?

Question

Quizplus · Accepted Answer

The decay of ¹⁵O follows first-order kinetics, which means that the rate of decay is proportional to the amount of the substance present. The half-life of the substance is given as 2.03 min, which means that after 2.03 min, the activity (or the amount of the substance) will be reduced to half its original value. We want to find the length of the pipe that will reduce the activity of the sample to 3.0% of its original value. Let's call the original activity A0 and the final activity after passing through the pipe Af. We know that Af/A0 = 0.03. We can use the decay equation to relate Af to A0 and the time t it takes to pass through the pipe: Af = A0*e^(-kt) where k is the decay constant (ln2/T1/2), and t is the time. We can rewrite the equation as: ln(Af/A0) = -kt Substituting the values we know: ln(0.03) = -(ln2/2.03 min)*t t = 2.03 min * ln(1/0.03)/ln2 Now we can use the speed of the oxygen in the pipe to find the length of the pipe: distance = speed*time distance = 1.1 m/s * (2.03 min * ln(1/0.03)/ln2 * 60 s/min) distance = 680 m Therefore, the pipe must be 680 m long for the sample to have decayed to 3.0% of its original activity just as it leaves the pipe. Answer choice A) is correct.

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An Air Sample Is Contaminated with ¹⁵O,which Has a Half-Life