Suppose a patient is given 1.55 mg of I-131, a beta emitter with a half-life of 8.0 days. Assuming that none of the I-131 is eliminated from the person's body in the first 4.0 hours of treatment, what is the exposure (in Ci) during those first four hours?

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Hello everyone. So in this video we're trying to calculate for the exposure for the first six hours of treatment, assuming that none of the P 32 excited the patient's body. So we're dealing with here the radioactive and nuclear decay of isotopes. And whenever dealing with that, we're gonna follow the first order kinetics. So the integrated rate law for the first order reactions is going to be as far as well have L. N. Times the concentration of N. At time T. Equaling two negative K. Which is going to be the decay constant times time plus the L. N. Of again the concentration of N. But the spine has at the initial concentration so but did not. All right. So we go and also recall that the half life is the time needed for the amount of a reacting to decrease by 50% or by one half. So the half life of a first order reaction is given by this reaction or this equation. So we have T. Half equals two L. N. Of two over cave. So we first need to calculate the decay constant using the given half life which is going to be 14.3 days. All right over here. So then our K. Is equal to R. L. N. Of two divided by T. Of half. So basically what we did here is we want to head to isolate our K. And by rearranging that equation then we get this new equation. So let's put in the numerical value. So we have Ln of two divided by 14.3 days. So putting that into my calculator my value will be 0.04847 days to the negative one power. Because our days unit was at the bottom of fraction. So we just put this All right. So since no concentration is given we need or we can use the given mass instead of concentration. So what we're given, let's go ahead and trade into color. So we're given we're not giving the concentration of time or at a certain time. So we still don't know what that is. We'll put a big question mark. As for the initial concentration, That's going to be 0.015 mg we calculated for our K. Which is 0.04847 days. If we have our tea at four or four hours then two days six hours, One day has 24 hours Which will give us 0.25 days. Alright now just scrolling down then. So now we are solving for time. So the L. N. Of the concentration at tea time Is equal to negative 0.04847 days. 2 -1 times 0. days. And we're gonna go ahead and add this with the natural log of 93. Again solving for the natural log of the concentration of end at time. That's equal to if it's just simplify what we have at top. We have negative 0. Plus 4.532, 60. So it's going to go ahead and equal 4.5- notice solving and isolating the concentration. Here we go ahead and raise it to the power of E, which will go ahead and get this isolated. Okay, that's going to be e to the 4.5- power, Putting that into my calculator. My numerical value is going to be 91.8797. Course the unit is in Milligrams. So finally calculate for the milligrams that is decayed, It's equal to 93.0 mg minus what we just saw for the concentration of 91.8797 mg. Then we'll get the value of 1. milligrams. Alright, now, last part here, we're gonna go ahead and see the beta decades how we do this. Well we have the milligrams. So let's go ahead and start the dimensional analysis about that. We have the 1. mg of RP 32. We're going to first convert our milligrams into grams. So tend to negril third grams for one g or one mg. Then we'll go ahead and have the graham's going to moles. So everyone mole RP We have 32 g of p 32, then we go ahead and convert the moles of P 32 debated the case. So it's similar to Aligarh summer. So every one mole of P 32 will have 6.22 times 10 to the 23rd beta decays. Alright, as you can see Milgram's council grams cancel, moles cancel. So numerical value we are left with is 2.1083 times 10 to the 19th beta two case as for our tea. So we start off with these six hours converting the hours in two minutes. Every one hour is 60 minutes, then minutes, two seconds. So every one minute is equal to seconds. Okay, so you can see hours cancel minutes cancel. Perfect. So that my value is going to be 2.16 times 10-4 seconds. Alright, now finally software what we've been waiting for. So let's go ahead and start with our beta decays of 2.1083 times 10 to the better. Tks Per the time is calculated for 2.16 Times 10 to the four seconds. So putting that into my calculator for my new micro, values we get 9.7606 times 10 to the 14 decays per second. Then we go ahead and multiply this by for everyone. See I There is 3.7 times 10 to 10 decades per second. So then my final answer for this question is going to be 2.64 times 10 to the four power of R. C. I. And that is going to be my final answer for this problem. Thank you all so much for watching.

Suppose a patient is given 1.55 mg of I-131, a beta emitter with a half-life of 8.0 days. Assuming that none of the I-131 is eliminated from the person's body in the first 4.0 hours of treatment, what is the exposure (in Ci) during those first four hours?

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Key Concepts

Radioactive Decay

Half-Life

Curie (Ci)