/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 73 Radioactive Iodine in Medicine. ... [FREE SOLUTION] | 91影视

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Chapter 43: Problem 73

Radioactive Iodine in Medicine. Iodine in the body is preferentially taken up by the thyroid gland. Therefore, radioactive iodine in small doses is used to image the thyroid and in large doses is used to kill thyroid cells to treat some types of cancer or thyroid disease. The iodine isotopes used have relatively short half-lives, so they must be produced in a nuclear reactor or accelerator. One isotope frequently used for imaging is \({ }^{123} \mathrm{I} ;\) it has a half-life of \(13.2 \mathrm{~h}\) and emits a \(0.16 \mathrm{MeV}\) gamma-ray photon. One method of producing \({ }^{123} \mathrm{I}\) is in the nuclear reaction \({ }^{123} \mathrm{Te}+\mathrm{p} \rightarrow{ }^{123} \mathrm{I}+\mathrm{n} .\) The atomic masses relevant to this reaction are \({ }^{123} \mathrm{Te}, 122.904270 \mathrm{u} ;{ }^{123} \mathrm{I}, 122.905589 \mathrm{u} ; \mathrm{n}, 1.008665 \mathrm{u} ;\) and \({ }^{1} \mathrm{H}, 1.007825 \mathrm{u}\) The iodine isotope commonly used for treatment of disease is \({ }^{131} \mathrm{I}\), which is produced by irradiating \({ }^{130}\) Te in a nuclear reactor to form \({ }^{131}\) Te. The \({ }^{131}\) Te then decays to \({ }^{131} \mathrm{I}\). \({ }^{131} \mathrm{I}\) undergoes \(\beta^{-}\) decay with a half-life of 8.04 d, emitting electrons with energies up to \(0.61 \mathrm{MeV}\) and gamma-ray photons of energy \(0.36 \mathrm{MeV}\). A typical thyroid cancer treatment might involve administration of \(3.7 \mathrm{GBq}\) of \({ }^{131} \mathrm{I}\).

Short Answer

Expert verified
To calculate the total energy released in the reactions involving iodine isotopes for medical use, first we need to calculate Q-value of the reaction using the conservation of mass-energy. Then, we determine the rate of decay of \({ }^{123} \mathrm{I}\) isotope using its given half-life values. Lastly, we compute the energy liberated during the \(\beta^{-}\) decay process of \({ }^{131} \mathrm{I}\) isotope. Each disintegration liberates 0.61 MeV of energy. Apply these steps to find total energy released.

Step by step solution

01

Calculating the Q-value (Energy release)

To find the energy released (\(Q\)) in the reaction鈥擻({ }^{123} \mathrm{Te}+\mathrm{p} \rightarrow{ }^{123} \mathrm{I}+\mathrm{n}\), one can use the principle of conservation of mass-energy. The Q-value can be obtained from the difference in total atomic masses before and after the reaction. It can be calculated using the formula: \[Q = (m_i - m_f)c^2\]where \(m_i\) and \(m_f\) are the initial and final total atomic masses respectively, and \(c\) is the speed of light.
02

Determine the half-life of \({ }^{123} \mathrm{I}\)

It's necessary to compute the Half Life (T) of \({ }^{123} \mathrm{I}\), which is provided in the exercise as 13.2 h. Understanding this will help gauge the rate at which the \({ }^{123} \mathrm{I}\) decays in the body.
03

Calculate energy release through beta decay

\({ }^{131} \mathrm{I}\) undergoes \(\beta^{-}\) decay with a half-life of 8.04 d. This means that half of the administered \({ }^{131} \mathrm{I}\) will decompose over an 8.04 days period into \({ }^{131} \mathrm{Xe}\) and an electron: \({ }^{131} \mathrm{I}\) \(\rightarrow { }^{131} \mathrm{Xe} + e^{-}\). Each disintegration liberates 0.61 MeV of energy. Use the principles from Step 1 to calculate the total energy release.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Radioactive Isotopes
Radioactive isotopes, also known as radioisotopes, are atoms with unstable nuclei that decay over time. They are known for emitting radiation in the form of alpha particles, beta particles, or gamma rays as they transform into more stable forms. In nuclear medicine, these isotopes are quite valuable because they can be used both for imaging and for therapeutic purposes.
An example of a radioisotope commonly used is Iodine-123. It is specifically taken up by the thyroid gland, making it ideal for imaging thyroid conditions. This isotope emits gamma rays, which can be detected to form images of the thyroid. Another isotope, Iodine-131, is utilized to treat thyroid diseases because it releases beta particles that can destroy thyroid cells in large doses. This selective uptake by the thyroid gland makes iodine isotopes particularly useful in medical settings.
Half-life
The concept of half-life is crucial to understanding how long a radioactive isotope will remain active. The half-life is the time it takes for half of the radioactive nuclei in a sample to decay.
  • Iodine-123 has a half-life of 13.2 hours, meaning it loses half its radioactive strength in that time period.
  • Iodine-131, on the other hand, has a longer half-life of 8.04 days.

This means Iodine-123 is better suited for diagnostic purposes since it decays quickly, allowing imaging procedures to be completed before it disappears. Iodine-131 is used in treatments where extended activity is needed to impact thyroid tissue over a longer period.
Beta Decay
Beta decay is a process by which a neutron in an unstable nucleus is transformed into a proton, emitting a beta particle (an electron in this context) and an anti-neutrino. This transformation helps the nucleus move towards a more stable state.
  • In the case of Iodine-131, it undergoes beta-minus decay, converting to Xenon-131, an electron (beta particle) and releasing energy.
  • This decay releases about 0.61 MeV of energy, which is enough to cause damage to thyroid cells in therapeutic uses.

Understanding beta decay helps us know how radioactive treatments affect the body and how radiation can be harnessed for both imaging and therapeutic purposes.
Q-value Calculation
The Q-value of a nuclear reaction quantifies the energy released during that reaction. It is calculated from the mass difference between the initial and final nuclear states multiplied by the speed of light squared, following Einstein鈥檚 famous equation, \[Q = (m_i - m_f)c^2\]where \[m_i\] is the initial mass and \[m_f\] is the final mass.
For example, during the reaction where Tellurium-123 absorbs a proton to become Iodine-123 and a neutron, we calculate the Q-value to show how energy is released. Understanding the Q-value helps in assessing the efficiency and feasibility of nuclear reactions for medical isotope production.

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Most popular questions from this chapter

The radioactive isotope \({ }_{7}^{12} \mathrm{~N}\) undergoes \(\beta^{+}\) decay. The energy released in the decay of one nucleus is \(16.316 \mathrm{MeV}\). What is the mass, in amu, of a neutral \({ }_{7}^{12} \mathrm{~N}\) atom? Give your answer to eight significant figures.

What nuclide is produced in the following radioactive decays? (a) \(\alpha\) decay of \({ }_{94}^{239} \mathrm{Pu} ;\) (b) \(\beta\) decay of \({ }_{11}^{24} \mathrm{Na} ;\) (c) \(\beta^{+}\) decay of \({ }_{8}^{15} \mathrm{O}\).

As a scientist in a nuclear physics research lab, you are conducting a photodisintegration experiment to verify the binding energy of a deuteron. A photon with wavelength \(\lambda\) in air is absorbed by a deuteron, which breaks apart into a neutron and a proton. The two fragments share the released kinetic energy equally, and the deuteron can be assumed to be initially at rest. You measure the speed of the proton after the disintegration as a function of the wavelength \(\lambda\) of the photon. Your experimental results are given in the table. $$ \begin{array}{l|ccccccc} \boldsymbol{\lambda}\left(\mathbf{1 0}^{-13} \mathbf{m}\right) & 3.50 & 3.75 & 4.00 & 4.25 & 4.50 & 4.75 & 5.00 \\ \hline \boldsymbol{v}\left(\mathbf{1 0}^{\mathbf{6}} \mathbf{~ m} / \mathbf{s}\right) & 11.3 & 10.2 & 9.1 & 8.1 & 7.2 & 6.1 & 4.9 \end{array} $$ (a) Graph the data as \(v^{2}\) versus \(1 / \lambda\). Explain why the data points, when graphed this way, should follow close to a straight line. Find the slope and \(y\) -intercept of the straight line that gives the best fit to the data. (b) Assume that \(h\) and \(c\) have their accepted values. Use your results from part (a) to calculate the mass of the proton and the binding energy (in \(\mathrm{MeV}\) ) of the deuteron.

Many radioactive decays occur within a sequence of decaysfor example, \({ }^{234} \mathrm{U} \rightarrow{ }_{98}^{230} \mathrm{Th} \rightarrow{ }_{84}^{226} \mathrm{Ra}\). The half-life for the \({ }_{92}^{234} \mathrm{U} \rightarrow{ }_{88}^{230} \mathrm{Th}\) decay is \(2.46 \times 10^{5} \mathrm{y},\) and the half-life for the \({ }_{88}^{230} \mathrm{Th} \rightarrow{ }_{84}^{226} \mathrm{Ra}\) decay is \(7.54 \times 10^{4} \mathrm{y}\). Let 1 refer to \({ }_{92}^{234} \mathrm{U}, 2\) to \({ }_{88}^{230} \mathrm{Th},\) and 3 to \({ }_{84}^{226} \mathrm{Ra} ;\) let \(\lambda_{1}\) be the decay constant for the \({ }_{92}^{234} \mathrm{U} \rightarrow{ }_{88}^{230} \mathrm{Th}\) decay and \(\lambda_{2}\) be the decay constant for the \({ }_{88}^{230} \mathrm{Th} \rightarrow{ }^{226} \mathrm{Ra}\) decay. The amount of \({ }_{88}^{230} \mathrm{Th}\) present at any time depends on the rate at which it is produced by the decay of \({ }_{92}^{234} \mathrm{U}\) and the rate by which it is depleted by its decay to \({ }_{84}^{226} \mathrm{Ra}\). Therefore, \(d N_{2}(t) / d t=\lambda_{1} N_{1}(t)-\lambda_{2} N_{2}(t) .\) If we start with a sample that contains \(N_{10}\) nuclei of \({ }_{92}^{234} \mathrm{U}\) and nothing else, then \(N(t)=N_{01} e^{-\lambda_{1} t}\) Thus \(d N_{2}(t) / d t=\lambda_{1} N_{10} e^{-\lambda_{1} t}-\lambda_{2} N_{2}(t) .\) This differential equation for \(N_{2}(t)\) can be solved as follows. Assume a trial solution of the form \(N_{2}(t)=N_{10}\left[h_{1} e^{-\lambda_{1} t}+h_{2} e^{-\lambda_{2} t}\right],\) where \(h_{1}\) and \(h_{2}\) are constants. (a) since \(N_{2}(0)=0,\) what must be the relationship between \(h_{1}\) and \(h_{2} ?\) (b) Use the trial solution to calculate \(d N_{2}(t) / d t,\) and substitute that into the differential equation for \(N_{2}(t) .\) Collect the coefficients of \(e^{-\lambda_{1} t}\) and \(e^{-\lambda_{2} t}\) since the equation must hold at all \(t,\) each of these coefficients must be zero. Use this requirement to solve for \(h_{1}\) and thereby complete the determination of \(N_{2}(t)\). (c) At time \(t=0\), you have a pure sample containing \(30.0 \mathrm{~g}\) of \({ }_{92}^{234} \mathrm{U}\) and nothing else. What mass of \({ }_{88}^{230} \mathrm{Th}\) is present at time \(t=2.46 \times 10^{5} \mathrm{y},\) the half-life for the \({ }_{92}^{234} \mathrm{U}\) decay?

In an industrial accident a \(65 \mathrm{~kg}\) person receives a lethal whole- body equivalent dose of \(5.4 \mathrm{~Sv}\) from \(\mathrm{x}\) rays. (a) \(\mathrm{What}\) is the equivalent dose in rem? (b) What is the absorbed dose in rad? (c) What is the total energy absorbed by the person's body? How does this amount of energy compare to the amount of energy required to raise the temperature of \(65 \mathrm{~kg}\) of water \(0.010 \mathrm{C}^{\circ} ?\)
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