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Chapter 19: Problem 15

Write an equation describing the radioactive decay of each of the following nuclides. (The particle produced is shown in parentheses, except for electron capture, where an electron is a reactant.) a. \({ }^{68} \mathrm{Ga}\) (electron capture) c. \({ }^{212} \mathrm{Fr}(\alpha)\) b. \({ }^{62} \mathrm{Cu}\) (positron) d. \({ }^{129} \mathrm{Sb}(\beta)\)

Short Answer

Expert verified
a. Electron Capture in \({ }^{68} \mathrm{Ga}\): \[{}^{68}_{31} \mathrm{Ga} + e^- \rightarrow {}^{68}_{30} \mathrm{Zn}\] b. Positron Decay in \({ }^{62} \mathrm{Cu}\): \[{}^{62}_{29} \mathrm{Cu} \rightarrow {}^{62}_{30} \mathrm{Zn} + e^+\] c. Alpha Decay in \({ }^{212} \mathrm{Fr}\): \[{}^{212}_{87} \mathrm{Fr} \rightarrow {}^{208}_{85} \mathrm{At} + {}^{4}_{2}\mathrm{He}\] d. Beta Decay in \({ }^{129} \mathrm{Sb}\): \[{}^{129}_{51} \mathrm{Sb} \rightarrow {}^{129}_{52} \mathrm{Te} + e^-\]

Step by step solution

01

a. Electron Capture in \({ }^{68} \mathrm{Ga}\)

Electron capture occurs when a proton in the nucleus captures an electron and converts into a neutron. The atomic number decreases by one, while the mass number remains the same. The general equation for electron capture is: \[{}^A_Z X + e^- \rightarrow {}^A_{Z-1} Y\] So for Ga-68, the equation will be: \[{}^{68}_{31} \mathrm{Ga} + e^- \rightarrow {}^{68}_{30} \mathrm{Zn}\]
02

b. Positron Decay in \({ }^{62} \mathrm{Cu}\)

Positron decay occurs when a neutron in the nucleus gets converted to a proton, and a positron is released. The atomic number increases by one, while the mass number remains the same. The general equation for positron decay is: \[{}^A_Z X \rightarrow {}^A_{Z+1} Y + e^+\] For Cu-62, the equation will be: \[{}^{62}_{29} \mathrm{Cu} \rightarrow {}^{62}_{30} \mathrm{Zn} + e^+\]
03

c. Alpha Decay in \({ }^{212} \mathrm{Fr}\)

Alpha decay occurs when a nucleus emits an alpha particle (consisting of 2 protons and 2 neutrons). The atomic number decreases by two, and the mass number decreases by four. The general equation for alpha decay is: \[{}^A_Z X \rightarrow {}^{A-4}_{Z-2} Y + {}^{4}_{2}\mathrm{He}\] For Fr-212, the equation will be: \[{}^{212}_{87} \mathrm{Fr} \rightarrow {}^{208}_{85} \mathrm{At} + {}^{4}_{2}\mathrm{He}\]
04

d. Beta Decay in \({ }^{129} \mathrm{Sb}\)

Beta decay occurs when a neutron in the nucleus gets converted to a proton, and a beta particle (an electron) is released. The atomic number increases by one, while the mass number remains the same. The general equation for beta decay is: \[{}^A_Z X \rightarrow {}^A_{Z+1} Y + e^-\] For Sb-129, the equation will be: \[{}^{129}_{51} \mathrm{Sb} \rightarrow {}^{129}_{52} \mathrm{Te} + e^-\]

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Key Concepts

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

Electron Capture
Electron capture is a form of radioactive decay where an atom's nucleus absorbs an orbiting electron, which causes a proton to convert into a neutron. This process results in the decrease of the atomic number by one, while the atomic mass number remains unchanged. It's important to note that during electron capture, no particles are emitted, differing it from other radioactive decay types where particles are released.

For instance, in the decay of gallium-68 gallium-68, which has an atomic number of 31, an electron is captured by the nucleus resulting in the production of zinc-68 with an atomic number of 30. The notation for this reaction is: gallium-68gallium-68: Zn
Positron Decay
Positron decay is a process where a neutron in an atom's nucleus is transformed into a proton and a positron (the antimatter counterpart of the electron). This reaction increases the atomic number by one but keeps the mass number stable. Positrons are positively charged particles that are identical to electrons except for their charge.

For example, when copper-62 undergoes positron decay, it changes into zinc-62. Here, a neutron becomes a proton, and a positron is expelled from the nucleus. The general equation with reference to copper-62 is as follows: Zn + e+.
Alpha Decay
In alpha decay, an atomic nucleus releases an alpha particle, which is composed of two protons and two neutrons, commonly represented as a helium-4 nucleus. This causes the original atom to lose both atomic mass and atomic number. Specifically, the atomic mass decreases by four units, and the atomic number decreases by two.

The case of francium-212 illustrates alpha decay. When francium-212 decays, it emits an alpha particle and transforms into astatine-208. The reaction can be written as follows: At + He.
Beta Decay
Beta decay occurs when a neutron in the nucleus of an atom is converted into a proton, resulting in the emission of a beta particle, which is essentially an electron. This transforms the original atom to a new element with an increased atomic number by one, while the mass number stays the same.

For antimony-129 (Sb-129), the process entails the conversion of a neutron to a proton and the ejection of an electron. As a result, the antimony atom becomes an atom of tellurium-129 (Te-129). The beta decay equation for Sb-129 is: Te + e-.

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Most popular questions from this chapter

Strontium-90 and radon-222 both pose serious health risks. \({ }^{90} \mathrm{Sr}\) decays by \(\beta\) -particle production and has a relatively long half-life (28.9 years). Radon-222 decays by \(\alpha\) -particle production and has a relatively short half-life (3.82 days). Explain why each decay process poses health risks.

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.

Which of the following statement(s) is(are) true? a. A radioactive nuclide that decays from \(1.00 \times 10^{10}\) atoms to \(2.5 \times 10^{9}\) atoms in 10 minutes has a half-life of 5.0 minutes. b. Nuclides with large \(Z\) values are observed to be \(\alpha\) -particle producers. c. As \(Z\) increases, nuclides need a greater proton-to-neutron ratio for stability. d. Those "light" nuclides that have twice as many neutrons as protons are expected to be stable.

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