Answer: The half-life of 27Mg is 9.50 min. (a) Initially
Chapter 19, Problem 95P(choose chapter or problem)
The half-life of \({ }^{27} \mathrm{Mg}\) is 9.50 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 min later? (b) Calculate the \({ }^{27} \mathrm{Mg}\) activities (in Ci) at t = 0 and 1 = 30.0 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?
Unfortunately, we don't have that question answered yet. But you can get it answered in just 5 hours by Logging in or Becoming a subscriber.
Becoming a subscriber
Or look for another answer