Question

(a) What temperature gas would have atoms moving fast enough to bring two \(^{\rm{3}}{\rm{He}}\) nuclei into contact? Note that, because both are moving, the average kinetic energy only needs to be half the electric potential energy of these doubly charged nuclei when just in contact with one another.

(b) Does this high temperature imply practical difficulties for doing this in controlled fusion?

Short answer

(a) The gas with temperature\(T = 6.43 \times {10^9}{\rm{K}}\)would have atoms moving fast enough to bring two\({}^{\rm{3}}{\rm{He}}\)nuclei into contact.

(b) Yes, this high temperature implies practical difficulties for doing this in controlled fusion as materials which can withstand high-temperatures are not available.

Step-by-step solution

Concept Introduction

The entire work done by an external agent in transporting a charge or system of charges from infinity to the current configuration without incurring any acceleration is referred to as the electric potential energy of that charge or system of charges.

Forming the equations

The average kinetic energy only needs to be half the electric potential energy of these doubly charged nuclei when just in contact with one another.

The kinetic energy is defined as –

\(E = \frac{1}{2}m{v^2}\)

From thermodynamics the kinetic energy is also a measure of temperature –

\({E_k} = \frac{{3{k_b}T}}{2}\)

And average (root mean square velocity) is equal to the –

\(\frac{1}{2}m{\left( {{v^2}} \right)_{av}} = \frac{3}{2}{k_b}T\)

Where average (root mean square speed is equal to the –

\(\begin{array}{c}{v_{rms}} = \sqrt {{{\left( {{v^2}} \right)}_{av}}} \\ = \sqrt {\frac{{3{k_b}T}}{m}} \\ = \sqrt {\frac{{3RT}}{M}} \end{array}\)

Two same nuclei are repelled by the Coulomb force and the potential energy between them is –

\({E_p} = qV\)

Where\({\rm{q}}\)is the charge of the nuclei and\({\rm{V}}\)represents the electric potential between two nuclei, which is –

\(V = \frac{{kq}}{r}\)

Calculation for Temperature

The electric potential is dictated by the separation distance from the source charge, which has dependence (knowing from the droplet model) –

\(r = {r_0}{A^{\frac{1}{3}}},\,\,{r_0} = 1.2 \times {10^{ - 15}}\;{\rm{m}}\)

The kinetic energy is only half of the potential kinetic energy, so it can be written –

\({E_p} = \frac{{{E_{{k_b}}}}}{2}\)

Plugging in all the relations:

\(\begin{array}{c}qV = \frac{m}{2} \times \frac{1}{2}{\left( {\sqrt {\frac{{3{k_b}T}}{m}} } \right)^2}\\qV = \frac{3}{4}{k_b}T\end{array}\)

Now putting in the potential and the radius:

\(\frac{{k{q^2}}}{{2{r_o}{A^{\frac{1}{3}}}}} = \frac{3}{4}{k_b}T\)

And expressing the temperature from the previous relation:

\(T = \frac{{2{k_b}{q^2}}}{{3k{r_o}{A^{\frac{1}{3}}}}}\)

Plugging in all the values:

\(\begin{array}{c}T = \frac{{2\left( {9 \times {{10}^9}{\rm{N}}{{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right){{\left( {1.60 \times {{10}^{ - 19}}{\rm{C}}} \right)}^2}}}{{3\left( {1.38 \times {{10}^{ - 23}}\;{\rm{J/K}}} \right)\left( {1.2 \times {{10}^{ - 15}}\;{\rm{m}}} \right) \times {3^{\frac{1}{3}}}}}\\T = 6.43 \times {10^9}{\rm{K}}\end{array}\)

Therefore, the value for temperature is obtained as \(T = 6.43 \times {10^9}\,{\rm{K}}\).

Implications of high temperature

Achieving a temperature like this is extremely difficult. Temperatures like this are much higher than the ones are present on the sun, due to this making a controllable fusion is very hard, as one need to have materials that can stand a high-level temperature.

Therefore, there will be difficulties for the fusion to carry out.