Certain types of atoms (e.g., carbon-14, xenon-133, lead-210) are inherently uns

Question

Certain types of atoms (e.g., carbon-14, xenon-133, lead-210) are inherently unstable. They exhibit random transitions to a different atom while emitting radiation in the process. Based on experimental evidence, the number, N, of atoms in a radioactive substance can be described by the equation dN/dt = - lambda N where t is measured in years and lambda > 0 is known as the decay constant. The decay constant is found experimentally by measuring the half-life, tau, of the radioactive substance (i.e., the time it takes for half of the substance to decay). Use this information in Problems 28 to 32. Find a solution to the decay equation assuming that N (0) =N_0. For xenon-133, the half-life is five days. Find lambda. Assume t is measured in days. For carbon-14, the half-life is 5,568 years. Find the decay constant lambda. Assume t is measured in years. How old is a piece of human bone which contains just 60% of the amount of carbon-14 expected in a sample of bone from a living person? Assume the half-life of carbon-14 is 5,568 years. The Dead Sea Scrolls were written on parchment at about 100 BC. Given that the half-life of carbon-14 is 5,568 years, what percentage of carbon-14 originally contained in the parchment remained when the scrolls were discovered in 1947?

Explanation / Answer

In a previous chapter we made an observation about a special property of the function y = f(x) = e x namely, that dy dx = e x = y so that this function satisfies the relationship dy dx = y. We call this a differential equation because it connects one (or more) derivatives of a function with the function itself. In this chapter we will study the implications of the above observation. Since most of the applications that we examine will be time-dependent processes, we will here use t (for time) as the independent variable. Then we can make the following observations: 1. Let y be the function of time: y = f(t) = e t Then dy dt = e t = y With this slight change of notation, we see that the function y = e t satisfies the differential equation dy dt = y. v.2005.1 - September 4, 2009 1 Math 102 Notes Chapter 9 2. Now consider y = e kt . Then, using the chain rule, and setting u = kt, and y = e u we find that dy dt = dy du du dt = e u · k = kekt = ky. So we see that the function y = e kt satisfies the differential equation dy dt = ky. 3. If instead we had the function y = e kt we could similarly show that the differential equation it satisfies is dy dt = ky. 4. Now suppose we had a constant in front, e.g. we were interested in the function y = 5e kt . Then, by simple differentiation and rearrangement we have dy dt = 5 d dte kt = 5(kekt) = k(5e kt) = ky. So we see that this function with the constant in front also satisfies the differential equation dy dt = ky. 5. The conclusion we reached in the previous step did not depend at all on the constant out front. Indeed, if we had started with a function of the form y = Cekt where C is

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.