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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.

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Chemistry: An Atoms First Approach

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

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Chapter
Section
BuyFindarrow_forward

Chemistry: An Atoms First Approach

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

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Chapter 18 Solutions

Chemistry: An Atoms First Approach
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