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The answers to most of the following exercises are in terms of logarithms and exponentials. A calculator can be helpful, enabling you to express the answers in decimal form.What costs $\$ 60$ million per gram and can be used to treat brain cancer, analyze coal for its sulfur content, and detect explosives in luggage? The answer is californium- $252,$ a radioactive isotope so rare that only $8 \mathrm{g}$ of it have been made in the Western world since its discovery by Glenn Seaborg in $1950 .$ The half-life of the isotope is 2.645 years- -long enough for a useful service life and short enough to have a high radioactivity per unit mass. One micro gram of the isotope releases 170 million neutrons per minute.a. What is the value of $k$ in the decay equation for this isotope?b. What is the isotope's mean life? (See Exercise $39 .$ )b. What is the isotope's mean life? (See Exercise $39 .$ )c. How long will it take $95 \%$ of a sample's radioactive nuclei to disintegrate?

Calculus 1 / AB

Calculus 2 / BC

Chapter 7

Integrals and Transcendental Functions

Section 2

Exponential Change and Separable Differential Equations

Functions

Trig Integrals

Oregon State University

University of Michigan - Ann Arbor

Idaho State University

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so were given some information about California, and they first ask us to determine for our radioactive decay equation. What is K equal to given that the half life is 2.645 years now, they actually tell us in this chapter that half life, which I write is T sub 1/2 is equal to the natural log of two over K. So we can just use this here to solve for Kay, as opposed to having to plug everything into this formula here. So we would just multiply by K Divide by 2.6 for five years. So okay is going to be equal to the natural log off too. Divided by 2.645 analogy, reciprocal years for our units. And it's important to say what the units are, so don't forget to do that. So that's kay now the mean life. Well, in the previous problem, this is the definition that they give that it's one over K. So we just plug in K. So that's just gonna reciprocate it again. So it's gonna be 2.64 by over natural log of two years and let's see what this is approximately equal to so natural log of two is about 20.69 to buy that by or other way around. Divide that into 2.645 now, and that gives us about three point eight two years. So this is approximately what are mean life will be. Now they say, How long? For 95% degradation so we could plug in. Why is he go to 0.5? Why not and then solve for tea? But also in number 39 Number 39. We showed that why of our mean life Oh, third mean life three over. Kay is approximately equal to 0.5 Um, why not? And because we showed that we can just go ahead and plug in that for tea. So we have t over three k, which is going to be war mean. Life was supposed to be 3.82 so this is really just going to be approximately equal to three times 3.8 two years, which is around 11 point for five years. So that's how long it will take until we have about 95% degradation

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