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. In the given decay process, the missing particle is a beta-minus particle (an electron and an electron antineutrino): \(^{0}_{-1} \mathrm{e}^- + \bar{\nu}_e\). b. In the given decay process, the missing particle is an alpha particle: \(^{4}_{2} \mathrm{He}\). c. In the given decay process, the missing particle is a beta-plus particle (a positron and an electron neutrino): \(^{0}_{1} \mathrm{e}^+ + \nu_e\). d. In the given decay process, the missing particle is an electron in electron capture: \(^{0}_{-1} \mathrm{e}^- + \nu_e\).

Step by step solution

01

Identify the elements and their atomic numbers

Ge (Germanium) has an atomic number of 32, and Ga (Gallium) has an atomic number of 31. We can now write the decay process with the atomic numbers included: \[^{73}_{31} \mathrm{Ga} \rightarrow^{73}_{32} \mathrm{Ge} + ?\]
02

Determine the type of decay and missing particle

Since the atomic number has increased by 1, it is a beta-minus decay (\(\beta^-\)), and the missing particle is an electron (e\(^-\)) and an electron antineutrino (\(\bar{\nu}_e\)). So the reaction is: \[^{73}_{31} \mathrm{Ga} \rightarrow^{73}_{32} \mathrm{Ge} + ^{0}_{-1} \mathrm{e}^- + \bar{\nu}_e\] b. \(^{192} \mathrm{Pt} \rightarrow^{188} \mathrm{Os}+?\)
03

Identify the elements and their atomic numbers

Pt (Platinum) has an atomic number of 78, and Os (Osmium) has an atomic number of 76. We can now write the decay process with the atomic numbers included: \[^{192}_{78} \mathrm{Pt} \rightarrow^{188}_{76} \mathrm{Os} + ?\]
04

Determine the type of decay and missing particle

Since the atomic number has decreased by 2, it is an alpha decay, and the missing particle is an alpha particle, which consists of 2 protons and 2 neutrons. So the reaction is: \[^{192}_{78} \mathrm{Pt} \rightarrow^{188}_{76} \mathrm{Os} + ^{4}_{2} \mathrm{He}\] c. \(^{205} \mathrm{Bi} \rightarrow^{205} \mathrm{Pb}+?\)
05

Identify the elements and their atomic numbers

Bi (Bismuth) has an atomic number of 83, and Pb (Lead) has an atomic number of 82. We can now write the decay process with the atomic numbers included: \[^{205}_{83} \mathrm{Bi} \rightarrow^{205}_{82} \mathrm{Pb} + ?\]
06

Determine the type of decay and missing particle

Since the atomic number has decreased by 1, it is a beta-plus decay (\(\beta^+\)), and the missing particle is a positron (e\(^+\)) and an electron neutrino (\(\nu_e\)). So the reaction is: \[^{205}_{83} \mathrm{Bi} \rightarrow^{205}_{82} \mathrm{Pb}+^{0}_{1} \mathrm{e}^+ + \nu_e\] d. \(^{241} \mathrm{Cm}+? \rightarrow^{241} \mathrm{Am}\)
07

Identify the elements and their atomic numbers

Cm (Curium) has an atomic number of 96, and Am (Americium) has an atomic number of 95. We can now write the decay process with the atomic numbers included: \[^{241}_{96} \mathrm{Cm} + ?\rightarrow^{241}_{95} \mathrm{Am}\]
08

Determine the type of decay and missing particle

Since the atomic number has decreased by 1, it is a capture process. The missing particle will be an electron from an inner shell of the atom. This is called electron capture (EC). So the reaction is: \[^{241}_{96} \mathrm{Cm} + ^{0}_{-1} \mathrm{e}^- \rightarrow^{241}_{95} \mathrm{Am} + \nu_e\]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Describe the relative penetrating powers of alpha, beta, and gamma radiation.

The first atomic explosion was detonated in the desert north of Alamogordo, New Mexico, on July \(16,1945 .\) What percentage of the strontium- 90\(\left(t_{1 / 2}=28.9 \text { years) originally produced }\right.\) by that explosion still remains as of July \(16,2017 ?\)

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 Ga (electron capture) b. 62 Cu (positron) c. 212 \(\mathrm{Fr}(\alpha)\) d. 129 \(\mathrm{Sb}(\beta)\)

When using a Geiger-Müller counter to measure radioactivity, it is necessary to maintain the same geometrical orientation between the sample and the Geiger-Muller tube to compare different measurements. Why?

A chemist studied the reaction mechanism for the reaction $$ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g) $$ by reacting \(\mathrm{N}^{16} \mathrm{O}\) with \(^{18} \mathrm{O}_{2}\) . If the reaction mechanism is $$ \begin{aligned} \mathrm{NO}+\mathrm{O}_{2} & \rightleftharpoons \mathrm{NO}_{3}(\text { fast equilibrium }) \\ \mathrm{NO}_{3}+\mathrm{NO} & \longrightarrow 2 \mathrm{NO}_{2}(\text { slow }) \end{aligned} $$ what distribution of \(^{18} \mathrm{O}\) would you expect in the NO \(_{2} ?\) Assume that \(\mathrm{N}\) is the central atom in \(\mathrm{NO}_{3},\) assume only \(\mathrm{N}^{16} \mathrm{O}^{16} \mathrm{O}_{2}\) forms, and assume stoichiometric amounts of reactants are combined.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free