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Radioactive tracers. Radioactive isotopes are often intro- duced into the body through the bloodstream. Their spread through the body can then be monitored by detecting the appearance of radiation in different organs. $^{131} \mathrm{I}, \mathrm{a} \beta^{-}$ emitter with a half-life of 8.0 $\mathrm{d}$ is one such tracer. Suppose a scientist introduces a sample with an activity of 375 $\mathrm{Bq}$ and watches it spread to the organs. (a) Assuming that the sample all went to the thyroid gland, what will be the decay rate in that gland 24 $\mathrm{d}$ (about 2$\frac{1}{2}$ weeks) later? (b) If the decay rate in the thyroid 24 d later is actually measured to be 17.0 $\mathrm{Bq}$ , what percent of the tracer went to that gland? (c) What isotope remains after the I-131 decays?

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a. 46.9 $\mathrm{Bq}$b. 36.25$\%$c. $\frac{131}{54} X e$ (Xenon) remains after lodine decay.

Physics 103

Chapter 30

Nuclear and High-Energy Physics

Atomic Physics

Nuclear Physics

Particle Physics

University of Michigan - Ann Arbor

Hope College

McMaster University

Lectures

02:42

Atomic physics is the field of physics that studies atoms as an isolated system of electrons and an atomic nucleus. It is primarily concerned with the arrangement of electrons around the nucleus and the processes by which these arrangements change. The theory of quantum mechanics, a set of mathematical rules that describe the behaviour of matter and its interactions, provides a good model for the description of atomic structure and properties.

02:26

In physics, nuclear physics is the field of physics that studies the constituents and interactions of atomic nuclei. The most commonly known applications of nuclear physics are nuclear power generation and nuclear weapons technology, but the research has provided application in many fields, including those in nuclear medicine and magnetic resonance imaging, ion implantation in materials engineering, and radiocarbon dating in geology and archaeology.

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Radioactive Tracers. Radio…

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BIO Radioactive Tracers. R…

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43.24. Radioactive Tracers…

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A sample of a radioactive…

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The greatest concentration…

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Radioactive Fallout. One o…

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Radioactive fallout. One o…

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A sample of the radioactiv…

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In one common type of ho…

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BIO Radioactive Fallout. O…

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A by-product of some fissi…

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$\textbf{Radioactive Fallo…

for this question were told that we have a radioactive isotope of iodine. Um, with 1/2 life t 1/2 of eight days, which I then convert into seconds of the morass of supposed scientists introduces a sample with an activity of 375 becquerels. So Delta and over Delta T, which I also call a is 325 becquerels, Uh, for part A. It wants to assume that sample went through the thyroid grand and will decay at a rate of 24 days, or about two and 1/2 weeks. So I convert that Decatur that time t from 24 days into second. So 2.74 times 10 to the six seconds Okay, so to begin, in part A, the first thing we did find is the number of atoms that we have in the beginning. So we'll call this, uh and not. And in order to do this, we're gonna consider the equation that says the decay rate a is equal to the care a constant Lambda Times. And not for another word solving for and not since we know a and we know Lambda we find in not is one over Lambda Time's A well, lambda 0.693 divided by the half life t 1/2 or one over Lambda is t 1/2 divided by 0.69 three. This gets multiplied by a so plugging those values into this expression, we find that the answer here is equal to 3.74 times 10 88 nuclear ons. Okay, so then from that, we can get in since in here is equal to and not times the exponential e to the minus lambda t again Lambda is t 1/2 divided by 0.693 and T is the value of 24 days which be converted into seconds. So plugging those values and we find that the new nuclear on number is equal to 4.68 times tend to the seven nuclear eons after 24 days. Okay, so therefore, now we can get the new do the new decay rate constant. Um hey, so a here, we'll call this one a prime since it's different is equal Lambda Times in In this case, we have the same value for Lambda of Tea 1/2 divided by 0.6 night works. Excuse me, would be 0.693 divided by t, 1/2 multiplied by our new value of in that we just calculated previously, 4.68 times 10 to the seven. This comes out to equal 46 0.9 becquerels or B Q Lincoln box that in as our solution for a okay heart be says if the decay rate of the thyroid 24 days later is actually measured to be 17 becquerels. What percent of the tracer went into the gland so we can say percent tracer, And this is gonna be equal to the amount that we actually measured. 17 becquerels divided by the amount that we should have had, which was a prime times 100 to put it into percents. This is times 100%. Otherwise, it would just be a fraction clunking those values and we find that this is equal to 36.25% Quentin to the gland go. I'm thinking box it in as the solution to be and lastly, uh, part See, it says what isotope remains after the I won 31 2 case So I won. 31 is decaying via baited a case. So we have 1 31 is our nuclear on number and there are 53 protons and this is going to decay. Two if it's beta decay we saw at 1 31 But now we have a 54. So this is Xena. And since it was beta decay, we have an electron left which has a nuclear on number of zero, but a charge of negative one proton. So we can say that scene on remains. Let's go ahead and type that out, making boxing and is our solution for C, and that's the final part of our question.

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