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Chapter 19: Problem 95

The half-life of \({ }^{27} \mathrm{Mg}\) is \(9.50 \mathrm{~min}\). (a) Initially there were \(4.20 \times 10^{12}{ }^{27} \mathrm{Mg}\) nuclei present. How many \({ }^{27} \mathrm{Mg}\) nuclei are left \(30.0 \mathrm{~min}\) later? (b) Calculate the \({ }^{27} \mathrm{Mg}\) activities \((\) in \(\mathrm{Ci})\) at \(t=0\) and \(t=30.0 \mathrm{~min}\). (c) What is the probability that any one \({ }^{27} \mathrm{Mg}\) nucleus decays during a 1 -s interval? What assumption is made in this calculation?

Short Answer

Expert verified
The remaining \(^{27}Mg\) nuclei after 30.0 minutes is calculated in step 1. The activities of \(^{27}Mg\) at \(t=0\) and at \(t=30.0\) minutes are obtained in step 2. The decay probability of any \(^{27}Mg\) nucleus within a 1-second interval is found in step 3. The assumption made here is that the decay process is constant and continuous, that is, the probability of decay does not change over time.

Step by step solution

01

Calculate the remaining magnesium nuclei after 30 minutes

The formula for the decay of a radioactive element is given by \(N = N_0 * 0.5^{(t/T)}\), where \(N\) is the final quantity, \(N_0\) is the initial quantity, \(t\) is the elapsed time, and \(T\) is the half-life. Plugging the given values, we find \(N = 4.20 * 10^{12} * 0.5^{(30.0/9.5)}\).
02

Calculate the activities at the specified times

The formula for activity is given by \(A = \frac{ln(2) * N}{T}\) where \(A\) is the activity, \(N\) is the number of remaining nuclei, and \(T\) is the half-life. For \(t = 0\), \(N = N_0\). For \(t = 30.0\) minutes, \(N\) is the number calculated in step 1. Remember to convert minutes to seconds for the calculation, as activity is typically stated in disintegrations per second.
03

Determine the decay probability within a 1-second interval

The probability \(P\) that any one magnesium nucleus will decay during any 1s interval is given by \(P = \frac{A}{N}\), where \(A\) is the activity and \(N\) is the number of remaining nuclei. Use \(N\) and \(A\) at \(t = 30.0\) minutes. Note, the assumption made in this calculation is that the probability of decay is constant over time.

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