Ch.19 - Nuclear ChemistryWorksheetSee all chapters
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Ch.1 - Intro to General Chemistry
Ch.2 - Atoms & Elements
Ch.3 - Chemical Reactions
BONUS: Lab Techniques and Procedures
BONUS: Mathematical Operations and Functions
Ch.4 - Chemical Quantities & Aqueous Reactions
Ch.5 - Gases
Ch.6 - Thermochemistry
Ch.7 - Quantum Mechanics
Ch.8 - Periodic Properties of the Elements
Ch.9 - Bonding & Molecular Structure
Ch.10 - Molecular Shapes & Valence Bond Theory
Ch.11 - Liquids, Solids & Intermolecular Forces
Ch.12 - Solutions
Ch.13 - Chemical Kinetics
Ch.14 - Chemical Equilibrium
Ch.15 - Acid and Base Equilibrium
Ch.16 - Aqueous Equilibrium
Ch. 17 - Chemical Thermodynamics
Ch.18 - Electrochemistry
Ch.19 - Nuclear Chemistry
Ch.20 - Organic Chemistry
Ch.22 - Chemistry of the Nonmetals
Ch.23 - Transition Metals and Coordination Compounds

Solution: Write a nuclear equation for the indicated decay of each of the following nuclides. Express your answer as a nuclear equation.Ra−210 (alpha) Sn−126 (beta) Th−234 (beta) Mn−49 (positron emission) Ar−37 (electron capture)

Solution: Write a nuclear equation for the indicated decay of each of the following nuclides. Express your answer as a nuclear equation.Ra−210 (alpha)Sn−126 (beta)Th−234 (beta)Mn−49 (positron emission)Ar−37 (el

Problem

Write a nuclear equation for the indicated decay of each of the following nuclides. 

Express your answer as a nuclear equation.

Ra−210 (alpha)


Sn−126 (beta)


Th−234 (beta)


Mn−49 (positron emission)


Ar−37 (electron capture)

Solution

We’re being asked to write a nuclear equation for the indicated decay of each nuclide.


Recall that in a nuclear reaction, the number of protons and neutrons is affected and the identity of the element changes


The different types of radioactive decay are:

• Alpha decay: forms an alpha particle (42α, atomic mass = 4, atomic number = 2)

• Beta decay: forms a beta particle (0–1β, atomic mass = 0, atomic number = –1). The beta particle appears in the product side.

• Gamma emission: forms a gamma particle (00γ, atomic mass = 0, atomic number = 0)

• Positron emission: forms a positron particle (01e, atomic mass = 0, atomic number = 1)

• Electron capture: the initial nuclide captures an electron (0–1e, atomic mass = 0, atomic number = –1). The electron appears in the reactant side.


The nuclear equation must be balanced with the same total atomic mass and atomic number on both sides.


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