Question

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

Short answer

The least separation any two protons can have is 1pm.

Step-by-step solution

Write the given data

1. Kinetic energy of each proton, K = kT
2. Temperature of the system, T =$1\times {10}^{7}K$

Determine the concept of least separation

For the least separation between any two protons, the energy considered should be the maximum for the given system. Again due to the conservation of energy within the system the total maximum energy is converted to the potential energy of the system at the distance of the least separation. Thus, using this concept, the required separation value is calculated.

Formula:

The potential energy of the two charged system is given as follows:

$U=\frac{q_1{q}_2}{4\pi {\epsilon }_{0}r}$ ….. (i)

Calculate the least separation between any two protons

As per the given data, the kinetic energy of each proton is given as:

$K=k_{B}T$

Substitute the values and solve as:

$$\begin{gathered} \mathrm{K}=\left(1.38\times {10}^{-23}\frac{\mathrm{J}}{\mathrm{K}}\right)\left(1\times {10}^7\mathrm{K}\right) \\ =1.38\times {10}^{16}\mathrm{J} \end{gathered}$$

At the closest separation, all the kinetic energy is converted to potential energy. Thus, using the value in equation (i), the total kinetic energy of the two-proton system can be given that is used to calculate the least separation between the two protons as follows:

2K = U

$$\begin{gathered} 2\mathrm{K}=\frac{{\mathrm{e}}^2}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}_{\min }} \\ {\mathrm{r}}_{\min }=\frac{{\mathrm{e}}^2}{4{\mathrm{\pi \epsilon }}_0\left(2\mathrm{K}\right)} \end{gathered}$$

Substitute the values and solve as:

$$\begin{gathered} {\mathrm{r}}_{\min }=\frac{\left(9\times {10}^9{\displaystyle \frac{\mathrm{N}.{\mathrm{m}}^2}{{\mathrm{C}}^2}}\right){\left(1.6\times {10}^{-19}\mathrm{C}\right)}^2}{2\left(1.38\times {10}^{-16}\mathrm{J}\right)} \\ =8.33\times {10}^{-13}\mathrm{m} \\ \approx 1\mathrm{pm} \end{gathered}$$

Hence, the value of the least separation is 1 pm .