Ch.19 - Nuclear ChemistryWorksheetSee all chapters
All Chapters
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: Fill in the blanks in the partial decay series. Express your answers as a chemical expression.Part A24194Pu → 24195Am +                 Part B24195Am → 23793Np +                  

Solution: Fill in the blanks in the partial decay series. Express your answers as a chemical expression.Part A24194Pu → 24195Am +                 Part B24195Am → 23793Np +                  

Problem

Fill in the blanks in the partial decay series. Express your answers as a chemical expression.

Part A

24194Pu → 24195Am +                 


Part B

24195Am → 23793Np +                  

Solution

We’re being asked to complete the given partial decay series


To do so, we need to determine the product that forms during the decay of the initial 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.

Step 1: Balance the atomic mass on both sides.

Step 2: Balance the atomic number on both sides.

Step 3: Identify the particle formed.


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