Question

Assume that the protons in a hot ball of protons each have a kinetic energy equal to $kT$ where $k$ is Boltzmann constant and $T$ is absolute temperature. If $T = 1 \times {10^7}K$ what (approximately) is the least separation any two protons can have?

Answer 1

Hint:In order to solve this question we need to understand the distance of closest approach. Distance of closest approach is the minimum distance up to which two same charged particle can interact with each other. We know same charges repel each other so when they bring closer to each other than due to high force of repulsion between them a large potential energy is stored in the configuration. So due to this high repulsive force those two particles upon releasing go far away from each other and thereby converting the stored potential energy into kinetic energy.

Complete step by step answer:
So according to energy conservation rules upon releasing two particles they try to move far away from each other so all of potential energy is converted into kinetic energy. Let the two protons have charges $e$ where $e$ is an electronic charge and it is equal to $e = 1.67 \times {10^{ - 19}}C$.Let the minimum distance between two charges be ${r_{\min }}$.

Then the potential energy stored in configuration is given by
$U = \dfrac{{{e^2}}}{{4\pi {\varepsilon _0}{r_{\min }}^2}}$
Where ${\varepsilon _0} = 8.854 \times {10^{ - 12}}{C^2}{N^{ - 1}}{m^{ - 2}}$.
And Kinetic energy is given by $E = kT$.
Where $k = 1.38 \times {10^{ - 23}}J{K^{ - 1}}$ is Boltzmann constant and $T = 1 \times {10^7}K$ is temperature
So Kinetic energy is $E = (1.38 \times {10^{ - 23}} \times {10^7})J$
$E = 1.38 \times {10^{ - 16}}J$
So Using Closest distance approach relation we Know $E = U$

Putting values we get,
$1.38 \times {10^{ - 16}} = \dfrac{{{e^2}}}{{4\pi {\varepsilon _0}{r_{\min }}^2}}$
$\Rightarrow {r_{\min }}^2 = \dfrac{{{e^2}}}{{4\pi {\varepsilon _0}(1.38 \times {{10}^{ - 16}})}}$
$\Rightarrow {r_{\min }}^2 = \dfrac{{(9 \times {{10}^9}) \times {{(1.67 \times {{10}^{ - 19}})}^2}}}{{1.38 \times {{10}^{ - 16}}}}$
Here, $\dfrac{1}{{4\pi {\varepsilon _0}}} = 9 \times {10^9}$
$\Rightarrow {r_{\min }}^2 = 1.8188 \times {10^{ - 12}}{m^2}$
$\Rightarrow {r_{\min }} = 1.34 \times {10^{ - 6}}m$
$\therefore {r_{\min }} = 1.34\mu m$

So the minimum distance is ${r_{\min }} = 1.34\mu m$.

Note: It should be remembered that the value of kinetic energy is calculated when the configuration of protons is considered similar to gases configuration. Also hot ball of photons could be considered as blackbody radiation having all radiation trapped inside it so energy is directly proportional to temperature of black body as explained by plank and $1\mu m = {10^{ - 6}}m.$