   # Estimate the temperature needed to achieve the fusion of deuterium to make an α particle. The energy required can be estimated from Coulomb’s Jaw [use the form E = 9.0 × 10 9 ( Q 1 Q 2 / r ), using Q = 1.6 × 10 −19 C for a proton, and r = 2 × 10 −15 m for the helium nucleus; the unit for the proportionality constant in Coloumb’s law is J ∙ m/C 2 ]. ### Chemistry: An Atoms First Approach

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

#### Solutions

Chapter
Section ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 18, Problem 86CP
Textbook Problem
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## Estimate the temperature needed to achieve the fusion of deuterium to make an α particle. The energy required can be estimated from Coulomb’s Jaw [use the form E = 9.0 × 109 (Q1Q2/r), using Q = 1.6 × 10−19 C for a proton, and r = 2 × 10−15 m for the helium nucleus; the unit for the proportionality constant in Coloumb’s law is J ∙ m/C2].

Interpretation Introduction

Interpretation: The temperature needed to achieve the fusion of deuterium to make an alpha particle is to be calculated.

Concept introduction: The coulombs law expresses the electrostatic force between the two charged particles.

Coulombs energy barrier is the energy required to overcome a barrier by the two nuclei to come in close contact with each other to make a nuclear reaction possible.

To determine: The temperature needed to achieve the fusion of deuterium to make an alpha particle.

### Explanation of Solution

Explanation

The energy of coulombs barrier is calculated by the formula,

E=K×Q1×Q2r

Where,

• K is the proportionality constant.
• r is the nuclear separation.

The charge of the particles Q1=Q2=1.6×1019C .

The value of r is 2×1015m .

The value of K is 9×109JmC-2 .

Substitute the value of Q1,Q2 , r and K in the above equation.

E=9×109JmC-2×(1.6×1019C)22×1015mE=1.152×1013J/nucleus

The conversion of J/nucleus to J/mole is done as,

1mole=6.023×1023nucleus

Therefore, the conversion of 1.152×1013J/nucleus into J/mole is,

1

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