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Problem 49 Hard Difficulty

A patient swallows a radiopharmaceutical tagged with phosphorus - 32 $\left(\begin{array}{c}{32} \\ {15}\end{array} P\right), \quad a \beta^{-}$ emitter with a half - life of 14.3 days. The average kinetic energy of the emitted electrons is $7.00 \times 10^{2} \mathrm{keV}$ If the initial activity of the sample is 1.31 MBq, determine (a) the number of electrons emitted in a 10.0-day period, (b) the total energy deposited in the body during the 10.0 days, and (c) the absorbed dose if the electrons are completely absorbed in $1 \times 10^{2} \mathrm{g}$ of tissue.

Answer

a. 8.97 $\times 10^{11} \text { decays }$
b. 0.100 \mathrm{J}
c. 100 \mathrm{rad}

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Top Physics 103 Educators
Christina K.

Rutgers, The State University of New Jersey

Andy C.

University of Michigan - Ann Arbor

Aspen F.

University of Sheffield

Meghan M.

McMaster University

Video Transcript

for number 49. Phosphate 32 is used as a tracer in some kind of medical procedure, and that has 1/2 life of 14.3 days. And it's a beta emitter, so it's going to go through beta decay, and each electron submitted has it connect energy of 700 killer electron volts, and initial activity was 1.31 mega back row. So this was my original are and important everyone to find how many electrons have been omitted in 10 days. Well, each time one of these nuclei decays, it gives off one electron so we can figure out how many nuclear decayed we know how many life trends are. So basically, we're trying to figure out how many, uh, what the different value of end waas. So if we know how many end what end is after 10 days and subtract, would original end was, that would be how many of these got decayed and in turn. But that will be the number of electrons. So that's my strategy. Um, I like to always use this equation. There's other equations that are derived, but I usually like to give him with this one the number of new clay is the original number times 1/2 raised to the number of half lifes. So these are the ends that I want to subtract. I don't know the original end, but I know the original activity, and I know how they related. Remember that, Um if you take and times your decay constant, that is your are so in turn, my are provided by my decay rate. Well, be my M. So I'm gonna substitute that. You re arrange that and put that in here. So if I take my original are divided by my decay rate, that is my m. So I know that end. I'm going to figure out this end by using this someone half race to the number of half Life's. Well, we went for 10 days, so that's 10. Divided by the 14.3 is it's going to be less than 1/2 life. Okay, so now I just need to come up with what my decay rate is. Remember, my decay constant. I mean, my dick a constant this lambda, remember, that is the half life, uh, the natural log of two through 6.93 Always divided by It's Kate, right? So you can tell my decay, or it can be 6.93 divided by my half life. But for this, my half life has to be in seconds. So you wanna start plating stuff in here? So I'm looking for the end. My original activity was this in mega background. You put that back row, So 1.31 times 10 to the sixth for that mega. And I'm dividing a play the decay constant, which will be the point 693 divided by the half. Life in seconds, though. So that's the 14.3. I'm just gonna put my conversions right in here. So if that's days I'm going to get take it two hours and learning the multiply baby 3600 to take it two seconds. That's just my conversion factor. So that this is in seconds. So this whole thing was this still times 1/2 raised to the ton over 14.3. Remember, this whole thing is this or that I want No, this whole thing is the end that I want. So I'm going to calculate that and figure that out too, because I want to find this end, but I won't find this original and also so and equals. When I duel with this, I get to point 336 times 10 to the 12th. That's still gonna be multiplied by 1/2. Raise to the 10 of the 14 23. And when I multiply that by that, then I get one point 438 times 10 to 12. So this is my original end. This is my new an Remember up here I said about its attractive. That'll be my number of electrons. So I'm taking my number of electrons will be this 2.33 six times 10 to the 12th minus 1.438 Time Stone was 12 and I get 8.975 times 10 to the 11th. That's the number of nuclei that have decayed. So that is also the number of electrons that will be given off for Part B. We want to know what's the total energy in those left trends After the Hubble 10 days, we just figured how many electron are theirs? 8.95 times 10 to the 11th. Electrons just confused that was my electric. And over here from the beginning, I was told that each electron has 700 killer electron volts of kinetic energy. Um, so I'm just gonna convert that. I know that each electron in a per election there's 700. I'm gonna put that in electron volts. So 700 killa. So three more zeros. That's when the electron volts there are so electron volts is a unit of energy, but usually anything Connecticut or should put that in jewels instead. So I'm just gonna convert that now. One electron volts is 1.60 I'm stunned. You 19 Jules. So electrons cancel. Electron volts canceled, and I'm gonna be in jewels. I get point 10 Jules. And then in part, See, we wonder, what's the dose if all that energy goes into just 100 grams of the tissue? We wanted the dose and that be measured. Red. Remember what Wrede is though red is when you have 10 donated to Jules per kilogram. So we want to know how many of those many rads there are in our situation. One our situation. We know that we have. We just found that we had a point 10 Jules. For every 100 grams. So if I put that in kilograms, that would be quaint. 10 kilograms. And we're gonna figure out how many rads that it's over for you. How many times does this go into that? So we'll divide, tend the ***, too. Journals per kilogram. And this is 100. You can tell us going to 100 cancels is one on top. The negative exponents on the bottom is a positive. Tops attended the two for 100 red.

University of Virginia
Top Physics 103 Educators
Christina K.

Rutgers, The State University of New Jersey

Andy C.

University of Michigan - Ann Arbor

Aspen F.

University of Sheffield

Meghan M.

McMaster University