The easiest fusion reaction to initiate is H 1 2 + H 1 3 → H 2 4 e + n 0 1 Calculate the energy released per H 2 4 e nucleus produced and per mole of H 2 4 e produced. The atomic masses are H 1 2 , 2.01410 u; H 1 3 , 3.01605 u; and H 2 4 e , 4.00260 u. The masses of the electron and neutron are 5.4858 × 10 −4 u and 1.00866 u, respectively.

Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

Chapter
Section

Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 18, Problem 50E
Textbook Problem
1 views

The easiest fusion reaction to initiate is H 1 2   +   H 1 3   →   H 2 4 e   +   n 0 1 Calculate the energy released per H 2 4 e nucleus produced and per mole of H 2 4 e produced. The atomic masses are H 1 2 , 2.01410 u; H 1 3 , 3.01605 u; and H 2 4 e , 4.00260 u. The masses of the electron and neutron are 5.4858 × 10−4 u and 1.00866 u, respectively.

Interpretation Introduction

Interpretation: The amount of energy released per 24He nucleus and per mole of 24He produced in the given reaction is to be calculated.

Concept introduction: The sum of masses of the component nucleons and the actual mass of a nucleus is known as the mass defect and it can be used to calculate the nuclear binding energy.

To determine: The amount of energy released per 24He nucleus and per mole of 24He produced in the given reaction.

Explanation of Solution

Explanation

The atomic mass of 24He is 4.00260amu .

The atomic mass of 12He is 2.01410amu .

The atomic mass of 13He is 3.01605amu

The mass of neutron is 1.0087amu

Number of protons in 24He=2

Number of neutron in 24He=2

Number of electrons in 24He=2

The given reaction for which the energy is to be calculated is,

12H+13H24He+10n

The mass defect is calculated by the formula,

Δm=([Mass of 24Henucleus+Massofneutron][Massof12Hnucleus+Massof13Hnucleus])

Substitute the value of the atomic mass 12H , atomic mass of 24He , atomic mass of 13He , the number of protons and and that of the neutron in the above equation.

Δm=[4.00260+1.00866][2.01403.01605]Δm=-0.01899amu_

Explanation

The conversion of amu/nucleus to Kg/nucleus is done as,

1amu=1.66×1027Kg

Therefore, the conversion of 0.589amu/nucleus into kg/nucleus is,

0.01899amu/nucleus=(0.01899×1.66×1027)kg/nucleus=3.14×1029kg/nucleus

The binding energy per nucleon is calculated by Einstein’s mass energy equation, that is,

ΔE=ΔmC2

Where,

• ΔE is the change in energy

Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started

Find more solutions based on key concepts
Most people who successfully maintain weight loss do all of the following except continue to employ many of the...

Nutrition: Concepts and Controversies - Standalone book (MindTap Course List)

What is the defining feature of a synovial joint?

Human Biology (MindTap Course List)

Why does helium fusion require a higher temperature than hydrogen fusion?

Horizons: Exploring the Universe (MindTap Course List)

Four resistors are connected to a battery as shown in Figure P27.15. (a) Determine the potential difference acr...

Physics for Scientists and Engineers, Technology Update (No access codes included)

Which is the largest marine community?

Oceanography: An Invitation To Marine Science, Loose-leaf Versin