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Chapter 19: Problem 55

Write balanced equations for each of the processes described below. a. Chromium- 51 , which targets the spleen and is used as a tracer in studies of red blood cells, decays by electron capture. b. Iodine-131, used to treat hyperactive thyroid glands, decays by producing a \(\beta\) particle. c. Phosphorus- 32, which accumulates in the liver, decays by \(\beta\) particle production.

Short Answer

Expert verified
a. The balanced equation for the electron capture decay of Chromium-51 is: \[ ^{51}_{24}Cr + e^- \rightarrow ^{51}_{23}V \] b. The balanced equation for the beta decay of Iodine-131 is: \[ ^{131}_{53}I \rightarrow ^{131}_{54}Xe + \beta^- \] c. The balanced equation for the beta decay of Phosphorus-32 is: \[ ^{32}_{15}P \rightarrow ^{32}_{16}S + \beta^- \]

Step by step solution

01

a. Chromium-51 (electron capture)

In electron capture, a proton in the nucleus captures an electron and becomes a neutron. The atomic number decreases by 1, and the mass number remains the same. The general formula for an electron capture reaction is: \[ ^A_Z X + e^- \rightarrow ^A_{Z-1}Y \] So for Chromium-51: 1. Identify the isotope and its atomic number (Z) and mass number (A). Chromium-51 (Cr): Z = 24, A = 51. 2. Write the balanced nuclear equation: \[ ^{51}_{24}Cr + e^- \rightarrow ^{51}_{23}V \] Thus, the balanced equation for the electron capture decay of Chromium-51 is: \[ ^{51}_{24}Cr + e^- \rightarrow ^{51}_{23}V \]
02

b. Iodine-131 (beta decay)

In beta decay, a neutron in the nucleus converts into a proton and releases a beta particle (electron). The atomic number increases by 1, and the mass number remains the same. The general formula for a beta decay reaction is: \[ ^A_Z X \rightarrow ^A_{Z+1} Y + \beta^- \] So for Iodine-131: 1. Identify the isotope and its atomic number (Z) and mass number (A). Iodine-131 (I): Z = 53, A = 131. 2. Write the balanced nuclear equation: \[ ^{131}_{53}I \rightarrow ^{131}_{54}Xe + \beta^- \] Thus, the balanced equation for the beta decay of Iodine-131 is: \[ ^{131}_{53}I \rightarrow ^{131}_{54}Xe + \beta^- \]
03

c. Phosphorus-32 (beta decay)

For Phosphorus-32, we have another beta decay: 1. Identify the isotope and its atomic number (Z) and mass number (A). Phosphorus-32 (P): Z = 15, A = 32. 2. Write the balanced nuclear equation: \[ ^{32}_{15}P \rightarrow ^{32}_{16}S + \beta^- \] Thus, the balanced equation for the beta decay of Phosphorus-32 is: \[ ^{32}_{15}P \rightarrow ^{32}_{16}S + \beta^- \]

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Most popular questions from this chapter

Iodine- 131 is used in the diagnosis and treatment of thyroid disease and has a half-life of \(8.0\) days. If a patient with thyroid disease consumes a sample of \(\mathrm{Na}^{131}\) I containing \(10 . \mu \mathrm{g}{ }^{131} \mathrm{I}\), how long will it take for the amount of \({ }^{131} \mathrm{I}\) to decrease to \(1 / 100\) of the original amount?

In 1994 it was proposed (and eventually accepted) that element 106 be named seaborgium, \(\mathrm{Sg}\), in honor of Glenn \(\mathrm{T}\). Seaborg, discoverer of the transuranium elements. a. \({ }^{263} \mathrm{Sg}\) was produced by the bombardment of \({ }^{249} \mathrm{Cf}\) with a beam of \({ }^{18} \mathrm{O}\) nuclei. Complete and balance an equation for this reaction. b. \({ }^{263} \mathrm{Sg}\) decays by \(\alpha\) emission. What is the other product resulting from the \(\alpha\) decay of \({ }^{263} \mathrm{Sg}\) ?

A rock contains \(0.688 \mathrm{mg}^{206} \mathrm{~Pb}\) for every \(1.000 \mathrm{mg}{ }^{238} \mathrm{U}\) present. Assuming that no lead was originally present, that all the \({ }^{206} \mathrm{~Pb}\) formed over the years has remained in the rock, and that the number of nuclides in intermediate stages of decay between \({ }^{238} \mathrm{U}\) and \({ }^{206} \mathrm{~Pb}\) is negligible, calculate the age of the rock. (For \({ }^{238} \mathrm{U}\), \(t_{1 / 2}=4.5 \times 10^{9}\) years. \()\)

Why are the observed energy changes for nuclear processes so much larger than the energy changes for chemical and physical processes?

A positron and an electron can annihilate each other on colliding, producing energy as photons: $${ }_{-1}^{0} \mathrm{e}+{ }_{+1}^{0} \mathrm{e} \longrightarrow 2{ }_{0}^{0} \gamma$$ Assuming that both \(\gamma\) rays have the same energy, calculate the wavelength of the electromagnetic radiation produced.
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