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### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

# Fusion probability is greatly enhanced when appropriate nuclei are brought close together, but mutual Coulomb repulsion must be overcome. This can be done using the kinetic energy of high-temperature gas ions or by accelerating the nuclei toward one another. (a) Calculate the potential energy of two singly charged nuclei separated by 1.00 x 10-12 m by finding the voltage of one at that distance and multiplying by the charge of the other. (b) At what temperature will atoms of a gas have an average kinetic energy equal to this needed electrical potential energy?

(a) The potential energy is 2.30 x 10-16 J.

(b) At temperature 1.11 x 107 K atoms of a gas has an average kinetic energy equal to this needed electrical potential energy.

See the step by step solution

## Step 1: Definition of electric potential

Electric potential due to a single charge: To determine the electric potential due to a single source charge Q at a specific location, place a test charge q at that location and determine the electric potential energy UQq of a system containing the test charge and the source charge that creates the field.

The electric potential at that location

$\begin{array}{rcl}{\mathbf{V}}& {\mathbf{=}}& \frac{{\mathbf{U}}_{\mathbf{Qq}}}{\mathbf{q}}\\ & {\mathbf{=}}& \frac{\mathbf{kQ}}{\mathbf{r}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\\ & & \end{array}$

where ${\mathrm{k}}{=}{8}{.}{99}{×}{{10}}^{{9}}{}{\mathrm{N}}{·}{{\mathrm{m}}}^{{2}}{/}{\mathrm{C}}$is Coulomb's constant,

## Step 2: Principle and Formula

Electric Potential Energy:

${{\mathbf{U}}}_{{\mathbf{e}}}{\mathbf{=}}{\mathbf{qV}}{\mathbf{}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{\left(2\right)}}$

where q is the charge and V is the potential.

The average translational kinetic energy per molecule of a gas

${\mathbf{K}}{\mathbf{.}}{\mathbf{E}}{\mathbf{=}}\frac{\mathbf{3}}{\mathbf{2}}{{\mathbf{k}}}_{{\mathbf{B}}}{\mathbf{T}}{\mathbf{}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{\left(3\right)}}$

where ${{\mathbf{k}}}_{{\mathbf{B}}}{\mathbf{=}}{\mathbf{1}}{\mathbf{.}}{\mathbf{38}}{\mathbf{×}}{{\mathbf{10}}}^{\mathbf{-}\mathbf{23}}{\mathbf{}}{{\mathbf{m}}}^{{\mathbf{2}}}{\mathbf{.}}{\mathbf{}}{\mathbf{kg}}{\mathbf{/}}{{\mathbf{s}}}^{{\mathbf{2}}}{\mathbf{.}}{\mathbf{}}{\mathbf{K}}$is the Boltzmann's constant and T is the temperature measured in Kelvin.

## Step 3: Calculation of potential energy

(a)

Potential energy due to two charge e and distance r between them:

from eq.(1) and(2)

${\mathbf{U}}{\mathbf{=}}\frac{{\mathbf{ke}}^{\mathbf{2}}}{\mathbf{r}}$

Substitute values:

$\begin{array}{rcl}{\mathbf{U}}& {\mathbf{=}}& \frac{\left(8.99×{10}^{9}N·{m}^{2}/C\right){\left(1.60×{10}^{-19}\text{}C\right)}^{\mathbf{2}}}{\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{12}}\mathbf{}\mathbf{m}}\\ & {\mathbf{=}}& {\mathbf{2}}{\mathbf{.}}{\mathbf{30}}{\mathbf{×}}{{\mathbf{10}}}^{\mathbf{-}\mathbf{16}}{\mathbf{}}{\mathbf{J}}\\ & & \end{array}$

Therefore, potential energy is 2.30 x 10-16 J.

## Step 4: Calculation of the temperature

(b)

Equating KE from eq.(3) with potential energy, U

${\mathbf{U}}{\mathbf{=}}\frac{\mathbf{3}}{\mathbf{2}}{{\mathbf{k}}}_{{\mathbf{B}}}{\mathbf{T}}$

Solve for T:

${\mathbf{T}}{\mathbf{=}}\frac{\mathbf{2}\mathbf{U}}{\mathbf{3}{\mathbf{k}}_{\mathbf{B}}}$

Substitute given values:

$\begin{array}{rcl}{\mathbf{T}}& {\mathbf{=}}& \frac{\mathbf{2}\left(2.30×{10}^{-16}J\right)}{\mathbf{3}\left(1.38×{10}^{-23}{m}^{2}\\mathrm{cdotkg}/{s}^{2}\\mathrm{cdotK}\right)}\\ & {\mathbf{=}}& {\mathbf{1}}{\mathbf{.}}{\mathbf{11}}{\mathbf{×}}{{\mathbf{10}}}^{{\mathbf{7}}}{\mathbf{}}{\mathbf{K}}\end{array}$

Therefore, the atoms of a gas has an average kinetic energy equal to this needed electrical potential energy at temperature 1.11 x 10+ K.