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The unstable isotope $^{40} \mathrm{K}$ is used to date rock samples. Its half-life is $1.28 \times 10^{8}$ years. (a) How many decays occur per second in a sample containing $6.00 \times 10^{-6} \mathrm{g}$ of $^{40} \mathrm{K}$ (b) What is the activity of the sample in curies?

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a. 15.5 $\mathrm{decay} / \mathrm{s}$b. $4.19 \times 10^{-10} \mathrm{Ci}$

Physics 103

Chapter 30

Nuclear and High-Energy Physics

Atomic Physics

Nuclear Physics

Particle Physics

Cornell University

Rutgers, The State University of New Jersey

Simon Fraser University

Hope College

Lectures

02:42

Atomic physics is the field of physics that studies atoms as an isolated system of electrons and an atomic nucleus. It is primarily concerned with the arrangement of electrons around the nucleus and the processes by which these arrangements change. The theory of quantum mechanics, a set of mathematical rules that describe the behaviour of matter and its interactions, provides a good model for the description of atomic structure and properties.

02:26

In physics, nuclear physics is the field of physics that studies the constituents and interactions of atomic nuclei. The most commonly known applications of nuclear physics are nuclear power generation and nuclear weapons technology, but the research has provided application in many fields, including those in nuclear medicine and magnetic resonance imaging, ion implantation in materials engineering, and radiocarbon dating in geology and archaeology.

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The unstable isotope $^{40…

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Suppose that a $8.50 \math…

03:59

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02:14

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for this question were asked to find the decays per second or the activity rate in part A and were asked to put this in units of curry for part B. So the half life of our sample t 1/2 is 1.28 10 10 38 years, which I converted two seconds. That's 4.3 times 10 to 15 seconds. The mass of our sample is six times 10 to the minus six grams or six times 10 to the minus nine kilograms. And we're told that the unstable isotope 40 k has, of course, the mass here, since it's 40 is gonna be 40 times the atomic mass. You what? The atomic mass is just the mass of a single proton. So this is 40 times the mass of a proton. This gives you 6.68 times 10 to the minus 26 kilograms. Okay, so for part A to find the decay rate or the activity A. We simply use the equation that it's the wavelength Lambda Times in here we're in is the, uh in is the number of nuclei in the sample. So Lamda is equal to 0.693 divided by Excuse me, I said, I think I said Lambda Lambda is wavelength. Lambda is not the wavelength. It's the disintegration. Constant land is also commonly used his wavelength. But Lamba, here's our disintegration. Constant 0.693 divided by T 1/2. So obviously it's gonna have units of inverse second, which is why it's not wave link. That's Ah, disintegration value. OK, so that's our value for Lambda on our value for in the number of nuclei that we have is simply equal to uh yeah, the mass of the sample that we have divided by the mass of the isotope. So this is em over m sub k. So now if we go back to our equation for a, we can rewrite this as 0.693 divided by t, 1/2 multiplied by em over m sub K. So if we plug those values into this expression, we find that the activation or the activity here in decays per second is 15 0.5 decays per second. But the case per second is a quarrel, so he can just abbreviate. That is B Q. So we can box set in is their solution. For part a part B wants us to write this in units of Curries. So named after, of course, the famous physicist Marie Curry. So we have our solution from part A, the 15.5 becquerels. We're gonna multiply that by the conversion between becquerels in Curries. So because they measured the same thing, they're just a different unit. So one book aural is equal to 3.7. Claims tend to the nine Curries. Okay, playing those values in this expression and we find this is equal to 4.19 oh times 10 to the minus 10 Curries. So Marie Currie did a lot with X rays during one of the great world wars. And that is why radiation units are named after Marie Currie Comm box set another solution to part B.

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