Question

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^-12m 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?

Short answer

(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.

Step-by-step solution

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 U_Qq 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 $\mathrm{k}=8.99\times {10}^9\mathrm{N}·{\mathrm{m}}^2/\mathrm{C}$is Coulomb's constant,

Principle and Formula

Electric Potential Energy:

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

where q is the charge and V is the potential.

The average translational kinetic energy per molecule of a gas

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

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

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:

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

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

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:

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

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.