Q10PE - College Physics (Urone)

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^-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?

Step-by-Step Solution

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 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 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,

Step 2: Principle and Formula

Electric Potential Energy: 

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

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.

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:

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

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

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:

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

 

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.