the-answers-to-most-of-the-following-exercises-are-in-terms-of-logarithms-and-exponentials-a-calc-18

Question

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?

Bobby Barnes · Accepted Answer

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&#x27;s important to say what the units are, so don&#x27;t forget to do that. So that&#x27;s kay now the mean life. Well, in the previous problem, this is the definition that they give that it&#x27;s one over K. So we just plug in K. So that&#x27;s just gonna reciprocate it again. So it&#x27;s gonna be 2.64 by over natural log of two years and let&#x27;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&#x27;s how long it will take until we have about 95% degradation

Bobby Barnes · Answer

we&#x27;re told that the plutonium isotope has 1/2 life of 24,360 years. Yeah, 10 grams of plutonium released into the atmosphere will determine how long will it take before 80% of the isotope has decayed? No. They tell us in this chapter that whenever we&#x27;re talking about radioactive decay that Paul&#x27;s an exponential growth model. So are amount will be something based off. Why? T is eager to. Why not eat the negative? Katie, Like this here and we can go and solve for Kay, actually, pretty quickly because they tells that half life which I write is t sub 1/2. Does he go to the natural log too? Over K. So if we want to solve Okay, we just need to do K is equal to the natural log to over or half life. Sorry. So we would just plug in at 24,360 jokes natural August 2, Boger 24,360. And the units for this would be per year. Now we can plug all that in to help us soul. For this now, 80% decay means 20% left and We don&#x27;t even need to know how much we have initially to start. Because if we write a 0.2, why not? For how much we have left is equal to Why not our initial and then e to the I&#x27;m just gonna leave its cave right now. Negative, K. T. Um, when we divide by, Why not? Those just counts out each other. So it doesn&#x27;t even matter all that much right now. Let&#x27;s solve for tea. And then after that, we can plug in for Kay. So we would take the natural log on each side and we get natural log 0.2 zero to negative, Katie. Now let&#x27;s divide by negative K on each side so we&#x27;ll get T is equal to the natural log of 0.2 over negative K. All right, now, let&#x27;s plug in what k is. So this is K. So since it&#x27;s one over Kay, we would just reciprocate it. So it should be negative. Natural log of 0.2 times 24,360 over the natural log of two. So this would be in years since Kay was reciprocal years. And let&#x27;s see what this is going to be approximately equal to negative. Natural log of 0.2 is about 1.609 times 24,360 is about 39,000. And then divide that by natural log two and this is going to give somewhere around 56,000 562 years. So this is how long it will take or 80% to decay.

Bobby Barnes · Answer

So we&#x27;re told that the half life of palladium is 139 days. But our sample will be useless after 95% of it has decayed. Well, we want to determine is about how many days after this sample arrives Will our polonium be useless to us? So they tell us in the chapter that radio activity follows exponential decay. So we should have our solution be. Why T is equal to Why not e to the negative Katie since its decay. And they also tell us that half life, which I right as t sub 1/2 is equal to the natural log of two over K. So let&#x27;s go ahead and soul or Okay, so we were just divide by T ha and that will give us K is equal to the natural log of two over our half life. So plugging that in there should be natural. Log on to over 139 and the units with this would be reciprocal dates and hold off and plugging this into our equation just so we don&#x27;t have to write up that traction but to figure out when 95% decays. This is really asking about when is by percent left. So it doesn&#x27;t matter what? Why not? Is all we care about is on the left side. When is it going to be 5% of why not? And so we have e to the negative K t. And we want to solve for teeth. So the why nots for counts out with each other, they would take the natural log on each side, so get natural log of 0.5 is equal to negative k t. Now we&#x27;re going to divide by negative K on each side. To get T is equal to the natural log of 0.5 over negative k l. Let&#x27;s go ahead and plug in K. And since it&#x27;s one over Kay, we would just reciprocate this. So this should be natural log of 0.5 over the natural log of two times 139 days since our unit should get reciprocated as well, which is approximately, Let&#x27;s see, so almost dropped my negative down here. But that mega there should still be there. Otherwise we&#x27;d get a negative time, which, when it makes sense for this problem. So natural law gives there a 0.5 times negative or times 1 39 this or 16 and divide that by negative natural look up to, and that gives us around 601 days. So it will take about that long. Or us, too, or it will. We&#x27;ll get this long of use out of our polonium before we&#x27;ll have to toss it.

Bobby Barnes · Answer

So we&#x27;re told about this mean lifetime being TZ would&#x27;ve won over K for radioactive isotopes and what we want to show is that when t is equal to three over K, which is three mean lifetimes, that 95% of the original nucleus will have disintegrated. So let&#x27;s go ahead and see what we can do for this. Well, let&#x27;s say we weren&#x27;t told that it was equal to tease even 23 K. And let&#x27;s just solve for so at least 90% disintegrate disintegrated will mean that we should have 0.5% left. So we have 0.5 Why not is equal to why not e to the negative k t now we were divided side by Why not those, counselor? Then take the natural log. So we have natural log of 0.5 equal to negative K t. Now we would go ahead and divide each side by negative K would get t is equal to the natural log of that. I&#x27;m gonna rewrite um 0.5 So first, this is 1/20. So that means we can write that as negative natural log of 20 and then it would be over. Negative case of the negatives. Cancel. And we&#x27;re gonna have that. Tea is equal to the natural log of 20. Over. Okay, so now let&#x27;s see. What is this about? So this is approximately, well natural. August 20 is about 2.99 over K, which is less than three over K. So it does seem to be that if we have a time that&#x27;s larger than it, then we should have that 95% actually disintegrated.

Stephen Hobbs · Answer

part a wants us toe. Determine the K value for California in 2 52 which has a half life of 2.645 years dividing the seas. Taking the Ellen on both sides will end up put that one one half is equal to Ellen of e canceled. So we have negative. Okay? And then we could divide that to 0.645 over to left and therefore, R K value is all of this, and we&#x27;ll be using that in other parts of this problem as well. That comes out to zero point to six to Okay, um, the mean life is going to be found by dividing one over K or one over 0.262 262 And that comes out to approximately 11 point. Sorry. Come back to approximately three 0.8 one six years. And then finally, when has 95% of it disintegrated? Um, at that point, we have 5% left 0.5 and therefore we could set this equal t to the negative, okay? Value which we know times t and then solve for T. So we&#x27;ll start by taking the natural log zero 0.5 Dividing yet by r. K value. Okay, And this wonder giving us our t value, which comes out to, uh, this one. Sorry is 11 point for one three years. So 95% of it California to 52 will disintegrate after 11 points or three years.

Bobby Barnes · Answer

So we&#x27;re told that this frozen human mummy was estimated to have died about 5000 years ago. And we want to determine how much of the original carbon 14 remained at the time of his discovery. Well, in example five, they tell us that for C 14 carbon dating, we will have our kay is equal to natural log of two over 5730. And the mount left will follow just the typical exponential decay equation. So for 5000 years ago, that means we&#x27;d want to plug 5000 in to that, so we&#x27;d have why 5000 is equal to. So why not times E to the negative K. So I was going to be natural log of two over 5000 730 and then we multiply this by 5000. So let&#x27;s see what that some natural law goes to times 5000 is about 3465. Then divide by negative 5000 730. So this is going to be why not times e to the negative zero 0.6 about and then exponentially eating. That should give about zero point by for six times. Why not? And so that we multiply it by 100 should be our percent left, which would give us about 54.6 per cent of those C 14 that would be left.

Problem

The answers to most of the following exercises ar…

Problem 40 Medium Difficulty

Answer

Related Courses

University Calculus: Early Transcendentals

Related Topics

Discussion

Top Calculus 2 / BC Educators

Grace H.

Heather Z.

Kristen K.

Michael J.

Calculus 2 / BC Courses

Integration Review - Intro

Definite vs Indefinite

Recommended Videos

Watch More Solved Questions in Chapter 7

Video Transcript

Joel Hass, Christopher Heil, Przemyslaw Bogacki

University Calculus: Early Transcendentals

Grace H.

Heather Z.

Kristen K.

Michael J.

Integration Review - Intro

Definite vs Indefinite

What do you need help with?

Textbooks

Ask a question

Courses

AI Tutor

University Calculus: Early Transcendentals

Grace H.

Heather Z.

Kristen K.

Michael J.

Integration Review - Intro

Definite vs Indefinite

Joel Hass, Christopher Heil, Przemyslaw BogackiUniversity Calculus: Early Transcendentals

Grace H.

Heather Z.

Kristen K.

Michael J.

Joel Hass, Christopher Heil, Przemyslaw Bogacki

University Calculus: Early Transcendentals