91Ó°ÊÓ

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

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

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:

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

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