Question

A typical kinetic energy for a nucleon in a middle-mass nucleus may be taken as 5.00MeV. To what effective nuclear temperature does this correspond, based on the assumptions of the collective model of nuclear structure?

Short answer

This energy corresponds to $3.87\times {10}^{10}\mathrm{K}$effective nuclear temperature based on the assumptions of the collective model of nuclear structure.

Step-by-step solution

Write the given data

Typical kinetic energy of the middle-mass nucleus,$K_{avg}=5\mathrm{Mev}$

Determine the concept of Equipartition theorem and the formulas

At thermal energy, an equal amount of energy is associated with each degree of freedom of the substance. Thus, the average kinetic energy is proportional to the equilibrium temperature. Based on the collective nuclear model, the structure attends to an equilibrium condition.

In thermal equilibrium, the average energy of each degree of freedom according to Equipartition theorem:

$K_{avg}=\frac{3kT}{2}$ …… (i)

Here,$k=\mathrm{Boltzmann}\text{'}\mathrm{s}\mathrm{constant},8.62\times {10}^{-5}\frac{\mathrm{eV}}{\mathrm{K}}$

Calculate the value of temperature

Using the given value of kinetic energy in equation (i), the value of the nuclear effective temperature can be given as follows:

$T=\frac{2K_{avg}}{3k}$

Substitute the values and solve as:

$$\begin{gathered} T=\frac{2\left(5\times {10}^6\mathrm{eV}\right)}{3\left(8.62\times {10}^{-5}{\displaystyle \frac{\mathrm{eV}}{\mathrm{K}}}\right)} \\ =3.87\times {10}^{10}\mathrm{K} \end{gathered}$$

Hence, the value of the temperature is $3.87\times {10}^{10}\mathrm{K}$.