In each of the following radioactive decay processes, supply the missing particle. a. \({ }^{73} \mathrm{Ga} \rightarrow{ }^{73} \mathrm{Ge}+\) ? b. \({ }^{192} \mathrm{Pt} \rightarrow{ }^{188} \mathrm{Os}+?\) c. \({ }^{205} \mathrm{Bi} \rightarrow{ }^{205} \mathrm{~Pb}+?\) d. \(^{241} \mathrm{Cm}+? \rightarrow{ }^{241} \mathrm{Am}\)

Short Answer

Expert verified
a. β¯ (electron) + νₑ (electron antineutrino) b. α (alpha particle) c. β⁺ (positron) + νₑ (neutrino) d. e⁻ (electron) + νₑ (neutrino)

Step by step solution

01

a. Finding the missing particle in the decay of \({ }^{73} \mathrm{Ga} \rightarrow{ }^{73} \mathrm{Ge}+\)

In this decay process, the mass number is conserved (73), while the atomic number increases by 1 (from 31 for Ga to 32 for Ge). This change in atomic number is characteristic of beta-minus decay, in which a neutron is transformed into a proton and emits an electron (known as a beta particle) and an electron antineutrino. So, the missing particle is a beta-minus particle (electron) and an electron antineutrino. Missing particle: β¯ (electron) + νₑ (electron antineutrino)
02

b. Finding the missing particle in the decay of \({ }^{192} \mathrm{Pt} \rightarrow{ }^{188} \mathrm{Os}+?\)

In this decay process, the mass number decreases by 4 (from 192 for Pt to 188 for Os) and the atomic number remains the same (from 78 for Pt to 76 for Os). This change in mass number is characteristic of alpha decay, in which two protons and two neutrons are emitted from the nucleus in the form of an alpha particle. Missing particle: α (alpha particle)
03

c. Finding the missing particle in the decay of \({ }^{205} \mathrm{Bi} \rightarrow{ }^{205} \mathrm{~Pb}+?\)

In this decay process, the mass number is conserved (205), while the atomic number decreases by 1 (from 83 for Bi to 82 for Pb). This change in atomic number is characteristic of beta-plus decay or positron emission, in which a proton is transformed into a neutron and emits a positron and a neutrino. So, the missing particle is a beta-plus particle (positron) and a neutrino. Missing particle: β⁺ (positron) + νₑ (neutrino)
04

d. Finding the missing particle in the reaction \(^{241} \mathrm{Cm}+? \rightarrow{ }^{241} \mathrm{Am}\)

In this reaction process, the mass number is conserved (241), while the atomic number decreases by 1 (from 96 for Cm to 95 for Am). This change in atomic number is characteristic of electron capture, in which a proton in the nucleus captures an electron from the inner electron shell and transforms into a neutron, emitting a neutrino. So, the missing particle is an electron (captured from the inner shell) and a neutrino in the reactants. Missing particle: e⁻ (electron) + νₑ (neutrino)

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Key Concepts

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

Beta-minus Decay
When a radioactive nucleus undergoes beta-minus decay, one of its neutrons is transformed into a proton. During this process, the nucleus emits two particles: an electron and an electron antineutrino (\(u_e\)). This type of decay results in an increase in the atomic number by one, while the mass number remains unchanged.

Imagine a crowded room where someone leaves (the neutron) and another person wearing a different shirt (the proton) enters immediately after; the room's population stays the same, but its composition changes. Similarly, in beta-minus decay, a neutron (\(n\)) becomes a proton (\(p\)), emitting an electron (\(e^-\))—referred to as a beta particle—and an antineutrino. This is what happens with \(^{73}_{31}Ga\) turning into \(^{73}_{32}Ge\).

Applications of beta-minus decay include carbon dating, where scientists can measure the age of archaeological finds by analyzing the decay of \(^{14}C\) into \(^{14}N\).
Alpha Decay
Alpha decay is another radioactive process whereby an unstable nucleus releases an alpha particle, which consists of two protons and two neutrons (\(^4_2He^{2+}\)), effectively decreasing the atom's mass by four units and the atomic number by two.

It's as if a family of four—two parents and two children—moves out of a building, leading to a notable drop in the building's occupancy. For example, when \(^{192}_{78}Pt\) undergoes alpha decay, it emits an alpha particle and transforms into \(^{188}_{76}Os\). This phenomenon is a common source of emitted radiation in heavy elements such as uranium and thorium, and it's utilized in smoke detectors that contain americium-241.
Beta-plus Decay
In contrast to beta-minus decay, beta-plus decay involves the transformation of a proton inside the nucleus into a neutron, accompanied by the release of a positron (\(e^+\)) and a neutrino (\(u_e\)). This ultimately leads to a decrease in the atomic number by one, leaving the mass number unaltered.

Using our previous analogy, this would be like someone putting on a different shirt, effectively changing their appearance. In beta-plus decay, a proton (\(p\)) changes into a neutron (\(n\)), releasing a positron—the positively charged counterpart to the electron—and a neutrino. For example, \(^{205}_{83}Bi\) decays into \(^{205}_{82}Pb\). Beta-plus decay is an essential aspect of the operation of Positron Emission Tomography (PET) scanners in medical imaging.
Electron Capture
The process of electron capture is somewhat akin to beta-plus decay. However, rather than emitting a positron, the nucleus draws an inner orbital electron into its midst. This captured electron then combines with a proton to form a neutron and a neutrino. The atomic number decreases by one, while the mass number stays the same.

Imagine a game of tag where, instead of running away, the target willingly joins the seeker. During electron capture, a proton in the nucleus 'tags' an electron, usually from the closest orbit, leading to the emission of a neutrino and the formation of a new neutron. When \(^{241}_{96}Cm\) captures an electron, it results in \(^{241}_{95}Am\). Electron capture is an important decay mode for isotopes that are unable to emit a positron due to energy considerations, and it also plays a crucial role in the stellar nucleosynthesis of elements inside stars.

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Most popular questions from this chapter

Radioactive copper-64 decays with a half-life of \(12.8\) days. a. What is the value of \(k\) in \(\mathrm{s}^{-1}\) ? b. A sample contains \(28.0 \mathrm{mg}^{64} \mathrm{Cu}\). How many decay events will be produced in the first second? Assume the atomic mass of \({ }^{64} \mathrm{Cu}\) is \(64.0 \mathrm{u}\). c. A chemist obtains a fresh sample of \({ }^{64} \mathrm{Cu}\) and measures its radioactivity. She then determines that to do an experiment, the radioactivity cannot fall below \(25 \%\) of the initial measured value. How long does she have to do the experiment?

A chemist wishing to do an experiment requiring \({ }^{47} \mathrm{Ca}^{2+}\) (half-life \(=4.5\) days) needs \(5.0 \mu \mathrm{g}\) of the nuclide. What mass of \({ }^{47} \mathrm{CaCO}_{3}\) must be ordered if it takes \(48 \mathrm{~h}\) for delivery from the supplier? Assume that the atomic mass of \({ }^{47} \mathrm{Ca}\) is \(47.0 \mathrm{u}\).

Uranium- 235 undergoes many different fission reactions. For one such reaction, when \({ }^{235} \mathrm{U}\) is struck with a neutron, \({ }^{144} \mathrm{Ce}\) and \({ }^{90} \mathrm{Sr}\) are produced along with some neutrons and electrons. How many neutrons and \(\beta\) -particles are produced in this fission reaction?

The rate constant for a certain radioactive nuclide is \(1.0 \mathrm{X}\) \(10^{-3} \mathrm{~h}^{-1}\). What is the half-life of this nuclide?

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.

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