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You are asked to prepare a sample of ruthenium-106 for a radiation treatment: Its half-life is 373.59 days, it is beta emitter; its atomic weight is 106 glmol, and ...

Question

You are asked to prepare a sample of ruthenium-106 for a radiation treatment: Its half-life is 373.59 days, it is beta emitter; its atomic weight is 106 glmol, and its density at room temperature is 12.45 glem? . How many grams are needed to prepare a sample having an activity rate of 560 uCi?quantity needed:If the sample is a spherical droplet; what will be its radius?radius:cm

You are asked to prepare a sample of ruthenium-106 for a radiation treatment: Its half-life is 373.59 days, it is beta emitter; its atomic weight is 106 glmol, and its density at room temperature is 12.45 glem? . How many grams are needed to prepare a sample having an activity rate of 560 uCi? quantity needed: If the sample is a spherical droplet; what will be its radius? radius: cm



Answers

Bismuth-210 is a beta emitter with a half-life of 5.0 days. If a sample
contains 1.2 $\mathrm{g}$ of $\mathrm{Bi}-210$ (atomic mass $=209.984105 \mathrm{amu}$ ), how many beta emissions occur in 13.5 days? If a person's body intercepts 5.5$\%$ of
those emissions, to what dose of radiation (in Ci) is the person exposed?

All right, So this problem has a lot of spits. It's asking us to solve Hominy Curries. Is the person exposed to if they only take in 5.5% of emissions. So the half life is five days. My original amount is 12.2 grams. The atomic mass of business to 10 is 200.984105 So the first thing we have to do is find the cave value of K constant. So we take t 1/2 equals 0.693 divided by K and sulphuric A cause you know, the half life is five days, so we get 0.1386 per day. Then I have to find the amount left over. So I used the Ln of the leftover mass divided by the original mass. It was negative, Katie. And when I saw throw that I get 0.186 grams left over. So now that I know that I have 0.18 grams left over, I could get the change in matters of business, which is this Oops. So I get the change in mass and then I convert that mass into particles by dividing by the molar Mass and while trying by the avocados number. And that gives me 2.9 e to the 11 particles. One curry equals 3.7 each of the 10 to case for second, so I can convert my number of particles into decays per second. So I take the party because I found divided by the number of days that were given and multiply that by one day converted into seconds as well. So I get 2.49 each of the 15 decays per second now that I have to case per second, and that's going to cancel out the units for decades per second. With the current constant, I multiply my 2.49 each of the 15 times the decimal percentage that they gave us, which was 5.5 and undermined by the curry costume, which is 3.7 each of the six. And I end up with 3700

In this problem I can write the value of A. Is equal to linda end which is equal to 0.693 by T half multiplication. M by a multiplication. A. B now going forward and simplifying this problem. So here I can like half life is equal to 24100 year. Changing the unit. I just multiplied by this multiplication. 24 Harvard per day multiplication, 36 00 2nd per Harvard which is equal to 7.6 multiplication and 10 to depart 11 seconds now going forward and calculating the activity of the sample. So it is given by A. Which is equal to 0.693 by T half multiplication and M by a multiplication and A. B. Which is equal to 0.693 by 7.6 multiplication and 10 to depart. 11 2nd multiplication and 1000 grand by 239 grandpa. All mole multiplication, 6.2 multiplication. 10 to the power 23 nuclear. On solving it, I get the value of A is equal to 2.29 multiplication and 10 to the power to well decades pass again. So changing the unit I just multiply it by this number. On further simplification, I get the value of age equal to 2.29 multiplication tend to the part two will make you as our final answer.

For this problem. We're being asked to figure out how many Alfa emissions occur in 25 minutes from polonium 2 18 which has 1/2 life of three minutes. If we have a sample of 55 milligrams which has on that polonium 2 18 has an atomic mass of to 18.8965 a m. U. Um, And then we're also being asked to figure out what that dose of radiation is in Curie. So first thing we have to do is figure out our, um are rate constants. Okay, Um And to do that, we're going to go ahead and switch these minutes into seconds. So, um, three times 60. It's gonna turn our minutes in two seconds. This is 180 seconds. And for our 25 minutes, we're gonna get there That's 1500 stuck in. The reason we're doing that is because we want to have it in seconds later when we have to determine Arcuri. Okay, so first off to find Kay, we're going to rearrange our equation here, but we'll have our half. Life is 180 seconds equals zero point 693 divided by K. And we find that K comes out to be 3.8508 times 10 to the negative. Third per second. Um, now we can go in the head and figure out from there. How much? Um, mass is remaining at time equals 25 minutes. Um, And to do that, we're gonna use this equation here, so and start off with the natural log of and t divided by 55 milligrams equals negative. Three point 8508 times 10 to the negative. Third time time, which is 1500 seconds, um, to rearrange and solve for anti. We're gonna take the inverse here. So? So this will, under being anti equals 55 the inverse negative 3.8508 times 10 to the negative. Third times 1500. And we get that are the mass after is going to equal zero point 1705 three milligrams. So that's how much is left after decay. So to figure out how much has decayed, we have to subtract that from our original value. So we're gonna 55 minus 0.17053 now we get that 54.8 to 947 milligrams decayed. Now to figure out the emissions, we're going to take that value. The how many have decayed, how many milligrams decayed and converted into Adam's. So we will take this 55 or 54 0.8 to 947 and multiply that by one mole of, um, polonium to 18 is equal to, uh, to 18.8 965 grams of polonium 2 18 and then multiply that by 6.22 times 10 to the 23rd Adams her mole, Um, and we get that there were a total of for one point 514 five times, 10 to the 20th Adams of polonium to 18. That's how many Adams were admitted during that time span. So to find the rate, we will now take that value 1.5145 times tend to the 20th and divide that by our time span, which is 50 hunt 1500 seconds, and we come out with 1.97 times, 10 to the 17th Adams of polonium 2 18 per second. Now the last part of this question is asking us to take that and figure out what that the dosage isn't is in Curie's. So we will take this value this 1.97 times, turning to the 17th her second. Multiply that by one Curie emitted Um, divided by 3.7 times, turned to the 10th stuck in per second, and we get that that is 2.7 times 10 to this six Curie's admit it.

Hello. So the decay constant K. Is you got to uh 0.693 over half life. So half life was given in days, which is uh talk my name and it has to be converted to second since he took my name times 24 hours times 3600 seconds. Okay, that's going to give you that for the Education council. Now. We are dealing with copper 64 copper 64 contains avocado with a number of molecules. So we want to know what 2.8 mg. So you need to prevent that. two g. That's gonna be 0.28 Okay. It's going to contain this amount of molecules. All right now, the rate of uh decay or the number of decay produced per second uh is given by lambda times and lambda is what we found the same as the K. Ok, so longer times. And it's gonna give us the number of the Cape recycling. Okay, So that's gonna be done for the last part. This is the explanation to that. Okay, thank you very much.


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