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How many milligrams of a 16 mg sample of cesium-137 remain after 90 years Express the mass to two significant figures and include the appropriate unlts_ValueUnitsSu...

Question

How many milligrams of a 16 mg sample of cesium-137 remain after 90 years Express the mass to two significant figures and include the appropriate unlts_ValueUnitsSubmitRequest AnswerPart CHow many years are required for 40 Ing of cesium-137 t0 decay to 2.5 Ing of cesium-137? Express the time in years to two significant figures_AzdyoarsSubmitRequest Answer

How many milligrams of a 16 mg sample of cesium-137 remain after 90 years Express the mass to two significant figures and include the appropriate unlts_ Value Units Submit Request Answer Part C How many years are required for 40 Ing of cesium-137 t0 decay to 2.5 Ing of cesium-137? Express the time in years to two significant figures_ Azd yoars Submit Request Answer



Answers

The half-life of cesium-137 is 30 years. Suppose we have a
100-mg sample.
(a) Find the mass that remains after t years.
(b) How much of the sample remains after 100 years?
(c) What is the rate of decay after 100 years?
(d) After how long will only 1 mg remain?

Now is broken. Number of seven. Assuming that the massive caesium is empty, empty at any time equals Sorry, hood there equals and to you, which is mess at year zero Exponential. Okay, so now we need to get the value off K. So m of zero is equal. 100 maybe Graham Exponential. Katie, we have another givings at the half life. Cesium is had 30 years. So Emma 30 equals if off m zero equals, if not, like, by 100 milligrams. So if using question one or function one with function to we get that half lifes off, season is equal to one. Hungry. Okay. Que anti in this articular situation His Sure so dividing both sides by 100 you get that half is equal to expansion. 30 k Now we can have lain for post sides. I'll just move this to the other side. So, sir, TK equals Lynn have which is also equal to negative lane for serviced. So we can say that K is equal to negative Lynn cool over 30 Substituting that in the main function so that we can get the mess at any year. We get that a mufti is equal to 100 exponential negative lento over 30 like Haiti. So this is the main equation we can use to get the mess. Any giving you? So for part B, we want to get to mess a year after 100 years so we can get him 100. I simply substituting t by 100 hungry exponential Negative, Len to the empty halted, like by 100 in this case. So him 100 years would be equal toe nine point? No. You to Mellie grapes half apart. See, we want to get the rate of decay. The rate of decay is a derivative off a mufti with respect to type. So for simplicity off MFP we have already cotton that is equal to 100. Hey, Negative Lynn to I 30 over 30 fighting. So for simplicity, negative exponential off negative lento is equal to yes to, so I m f t is equal to 100 to to the power off Negative me over search. Now, To get the rate of UK, we want to get index key, which is preferred. So it is equal to 100 by two that the power of negative T over 30 multiplied by negative 1/30 Lynn. So is the rate of decay after 100 you his aim Days off 100. So we are going simply to substitute t. Okay, we are going to substitute by 100. You get to I'm hungry. Over 30 You get one over 30 Len too. Then get that M dish off 100 his equal to you Get to point to to nine. Merely crime. Yeah, and that is a logical value because decay means lot off mess. So it's a negative value and it's milligram over a year. So it's year after three of 100 we will lose zero point toe nine off the total mess off sees you No, with part de, we want to get the year at which decision would be one milligram. So we are saying at MLT is equal one milligram and we want to get the value both d the year. So a mufti equals 100 exponential negative then to over 30 and all that is equal toe one military so dividing both sides by 100 you get that e negative then to over 30 like my tea is to one over 100. You can simply take rain for both sides. So negative. Lynn to over 30. Do you? His equal pool, Lynn. One over 100. So simply team would be equal to Lynn. One over 100. Over. Make it too. Lynn. Toe over 30 which is equal Pool 199. Going three. You so to cesium. Or take about 100 99.3 years. Toby decayed from 100 milligrams toe only one milligram. Thank you.

Now for problem on Percival. It's they I m f t If the mess off season in Medic Rams remaining after time t in years, so am with T is equal to a most zero. The phone message season at years here, multiplied by exponential Okay, key. So Mosul is giving to be 100 milligrams exponential k g. Now you need to get the value of K So it's that kink it the message season at any year t so first, to get the value off K, we have another giving that the half life of caesium is 30 years. So a month 30 is actually equal toe have off him of deal. So I am a 30 is equal to 100 exponential Okay, marked by by search losing this two room and the half life cesium is have multiplied by 100. Oh, we can terminate this with 100 with this 100. So we have hey 13 d equal toe 1/2 and we can't take then both sides. So here we will have 30 k equals to lend Hey and let half is also equal toe negative Len tool for simplicity. So the K value would be negative men to over 30. And now we can say that I am with T is the mess it in here is equal to 100 exponential. Negative. Then toe over 30 Gay T shirt over certainty. Okay. And this can be simplified that a mufti is equal toe 2200. Sorry, 200 with the power off Negative t over third cause exponential lento is equal to to So here we have 200. So now we have got that The main function so was part B. We want I don't know after 20 is equal to 100 How much mace remains So we basically saying am off 100 tired Substitution is equal toe 200 with the dollar off negative 100 over search which is equal pool about 9.92 Many grapes now was part See you will try to get the rate of decay So we will start by using the for the functions that we have got Wife he's equal. Do we will have 20 200 Sorry. Five t equals 200 through the park. Negative t over 30. So why dish 50 which is a decay over time. At any year, T is equal to 200 with the power of negative T over 30 multiplied by negative one over 13 Let me to. And now at year 100 Why dash 100? The rate of decay exactly as a year. 100 is equal toe 100 Oh, I'm sorry. Is equal to 200 with the power of negative 100 over 30 multiplied by negative one over 30 thing to which is equal to negative Whole point to to nine milligram were you that is only logical is the negative value your prisons that it is a decayed It is a losing mess And this value is how much mess will be lost at year 100. So this is great. How much milligrams when the cesium lose at year Betty and this And this street is at your 100 now for part d you want to get when with why 50? When will it be equal? Only one milligram. So we will move like backward. Get the value or five t that is equal toe one milligram. So we will say that 100 But we will take one step back will be much easier. 100 e with the power of negative lento over 30 g. So instead of using the value of 100 we took a step back using this exponential off negative then and that is equal No. One. It's divide both sides by 100 giving us e to the port of negative lento over 30 de equal cool, one over 100 and we will take win for both sides. So here we have negative Lynn over 30 de equal tool Lynn one over 100. And now we can get value can get the value of T the year at which decision will be one milligram and that is equal to year 100 99 went three. So it will take decision around two centuries to decay from 100 milligram toe, only one milligram. Thank you

Hi, guys. Welcome back This problem. We know that the initial value is 100 milligrams at the half. Life is 30 years, quite a long half life, actually the 1st 1 to find the equation. So we know that will follow the form change. And why, with respect to t two k y pure to generate both sides. That comes out to why it's equal to the initial value. Y subzero times e raised the Katie. Now we could plug some dollars into this equation. We know. And since we have 1/2 life of 30 are half life will occur when there is 0.5 of the sample left. About 30 years, it will be equal 2.5 times. 100. It will be half of what the initial value us. And that comes out to 50. We have. Why 30 deal of 50. So you plug in 50 for a Y value. Our initial Y value we know is 100. If e race the Katie Que is unknown, R t value is 30 years now. We want to solve for K. We can divide both sides by 100. We're left with 0.5 this evil to e 30 k We can take the natural log of both sides. Upon doing that, we get the natural log of 0.5 is equal to 30 k and this K is equal to the natural log of 0.5 zero point five all over 30. Sorry about that. All over 30. Now we can plug that back in tow our equation. So we know that any time why we have why is equal to y subzero? Just 100 e race to the Ln 0.5 divided by 30 Cumpsty. It'll be our equation for the second part of the problem. The rest to figure out how much of the sample remains. After 100 years, part B t equals 100 t equals 100. What is? Why equal to between this plug this right indoor equation that we just created of why is equal to 100 he raised to the Ellen 0.5 over 30 times? 100? This is all erased all that. And when you do all that math, just quite a bunch of algebra, you're left with nine 0.92 We know that Why? Of 100 nine point nine to no units on that are milligram part C, or as to figure out what happens or how much time it takes with sampled to go down to only one milligram. How much time to sample was only one milligram? So again, we want to use our answer from part A wise he will do 100 e erase the l one point 5/30 times t we're gonna figure out what happens when. Why? Siegel toe one. So we have one is equal to 100. You raised to the Ln 0.5 over 30 t and now we just want to solve for T so we can divide both sides by 100. So we have 1000.0 one x equal to E Ellen 0.5 over 30 times. T we take the natural longer both sides are left with Ellen. 0.1 is able to Ellen 0.5 over 30 times t that would solve for t. You go to 30 times Ln We know one over Ellen 0.5. On doing that, you got an answer of 199 0.3. It will take Sorry. 199.3. It will take 100 in 99 0.3 years. The sample to reduce down to only one milligram or other words. If you were to graft this, this point will be on the graph you have. Why 199 0.3 Xarelto one. And that is the answer to part C. And that concludes all three parts of the problem. Thanks for watching.

Season 1 37 is a beta emitter, so it's going to admit electron and it'll produce something else. That's something else needs to have. Mass number of 1 37 and an atomic number of 56 of the 56 on the one Give us our 55. The atomic number 56 corresponds to bury him. Then if we have 90 years passed and each half life is 30 years, that's a total of 3/2 lives. So the 16 milligrams in the first half life will go to eight milligrams the eight milligrams, and in the second half, life will go to four milligrams and the four milligrams and the third half life will go to to go toe two milligrams. So two milligrams of cesium 1 37 will be left over after 90 years. Do we start with 28 milligrams that we go all the way down to 3.5 milligrams? That's one house life to 14 to half lifes 27 3/2 lifes to 3.5 milligrams, so it'll take 3/2 lives. If the half life is 30 years, then 90 years will pass for 28 milligrams of cesium 1 37 to go to a 3.5 milligrams cesium 1 37


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