Ch.19 - Nuclear ChemistrySee 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: Calculate the energy change when one 238U nucleus undergoes the reaction:238U + n → 239Pu + 2βThe masses needed are 238U, 238.0508 amu; 239Pu, 239.0522 amu; n, 1.0087 amu. a) -1.1 x 10-12 Jb) -1.5 x 10-10 Jc) -4.5 x 10-10 Jd) +1.5 x 10-12 Je) +3.0 x 10-10 J

Problem

Calculate the energy change when one 238U nucleus undergoes the reaction:

238U + n → 239Pu + 2β

The masses needed are 238U, 238.0508 amu; 239Pu, 239.0522 amu; n, 1.0087 amu.

 

a) -1.1 x 10-12 J

b) -1.5 x 10-10 J

c) -4.5 x 10-10 J

d) +1.5 x 10-12 J

e) +3.0 x 10-10 J