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[-I5 Points]DETAILSAUFEXC4 10.5.028.MY NOTESASK YOUR TEACHERPRACTICE ANOTHERIsotope technetlumUsedprepare ImadesInterna body organs Thls Isotope halfIlfe (time requ...

Question

[-I5 Points]DETAILSAUFEXC4 10.5.028.MY NOTESASK YOUR TEACHERPRACTICE ANOTHERIsotope technetlumUsedprepare ImadesInterna body organs Thls Isotope halfIlfe (time requlred for half the materlal erode) approxlmately patlent Injected wlth 30 mg of this isotone what will be the technetium level the patient after h? Use the function A(t) 38(%) where the technetium level millirdins tne patient after houns_ Round the nearest tenth of & lcnoondinNeed Help?Koar

[-I5 Points] DETAILS AUFEXC4 10.5.028. MY NOTES ASK YOUR TEACHER PRACTICE ANOTHER Isotope technetlum Used prepare Imades Interna body organs Thls Isotope halfIlfe (time requlred for half the materlal erode) approxlmately patlent Injected wlth 30 mg of this isotone what will be the technetium level the patient after h? Use the function A(t) 38(%) where the technetium level millirdins tne patient after houns_ Round the nearest tenth of & lcnoondin Need Help? Koar



Answers

A common isotope used in medical imaging is technetium-99m, which emits gamma rays.

A sample initially containing 0.500 mg of technetium-99m is monitored as a function of time. Based on its rate of gamma ray emission, a graph, showing the mass of active technetium-99m as a function of time, is prepared. Study the graph and answer the questions that follow.

a. What is the mass of technetium-99m present at 200 minutes? At 400 minutes?

b. What is the half-life of technetium-99m in minutes? In hours?

c. If a patient is given a 2.0-mg dose of technetium-99m, how much of it is left in the patient's body after 10 hours? (For this problem, assume that the technetium-99m is not biologically removed from the body.)

Answer this question. We simply need to take any two set of points from the table. So I'm just gonna take the 1st 2 points and the activity, then after four hours, is 3150 in terms of emissions of photons per second. Well, originally, it was 5000. That then will be equal to negative. Katie, the time that passed with four hours between the 1st 2 points. It's okay, then is gonna be equal 2.1161 over ours and the half life. Using the half life equation in the K value, I just determined, will be six hours. Then the fraction that will be left over after two hours will be equal to negative K, which I determined, and then the two hour period Natural law. Ogden of the fraction is negative 0.231 or the fraction then that remains is 0.794 So the fraction that was lost in the two hour period will be one minus. That or 0.206 Convert that to a percentage 20.6, or about 21% of lost in two hours

From the data, we can use any Uh two lines um To calculate the half life, I'm going to use the first line in the second line. So after four hours, it goes from 5000 photons per second to 30 150 photons per second. So a natural log of the ratio of the photon emissions starting within the numerator. The photon emission per second at time. T divided by The photon emission at time zero will be equal to negative K. Multiplied by the time that has elapsed four hours, Gives us AK value of .1161 over ours. Half life is simply K divided into natural log of two, given us six hours. So after two hours we'll take natural log of the fraction that has left will be equal to negative K. Multiplied by T the time that passes two hours And we get x equal 2.793, which means 79 3% is still left, or 27% has decayed.

If we have Technetium 99 meta stable with a half life of 6.0 hours and we start with 80 mg of the substance. After one half life we'll have half of 80 mg or 40 mg after two half lives will have half of 80 half of half of 80 which would be 20 mg. First half life is 42nd half life is half of that 20 and then for c 18 hours is three half lives, so we'll have half of half of half of 80 or 10 mg. And for the last one 1.5 days we do a simple conversion is going to correspond to six half lives, so we'll take half of 86 times or one half race to the sixth, multiplied by 80 gives us 1.25 mg. Technetium 99 left over.

In those question we've beaten told that Technician ninety nine has a half life off six house and the initial amount that is their moment before any time t it is zoo point zero five zero, Milica. And after time t, the number ofthe rejected knew Clyde as reduced to one point zero times ten to the negative to really grow up. So we know that using that integrated Rachel no, no natural law off anti divided by and not is equal to negative, Katie, where Kay is the rate constant and t is the time it takes for you not to gets to Auntie So you can see from this equation that we have nt and not but what we don't always ke anti. And from the data we've been provided, we can gets cave that I relate in the half life to kill you use in the half life in question. So we know that half life it's equal to zero point six nine three divided by K in the half life of off technetium ninety nine a. M has been given to us as sex sells so divided by speaks. And that is going to get Kay to zero point one one five five The Okay, So now let me have completed Kay. Now we can go ahead and calculates forty, which is what the question is asking us for. So what we do is we put in all that all the data we have So anti azi called too One times every times ten. The party needs you three Milica, if I don't buy a lot which is an zero point zero five What? So really grab which is it? Caught a negative. So find one one five five Ah see, it is a time t so find Auntie, we'LL be going to land off one point four times ten is the point of it is being really divided by one zero zero zero five Really all divided by negative So point one one by five they are And that is going to get you thirty three point tonight Which is how pros Millie therefore Ah Sylar timing to take four zero point zero five milligram of technetium ninety nine a m to get to one point zero times ten ways to the Bynegative Fury Milligram is there for hours


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