Consider fusion occurring in a helium plasma. Determine the...

The Coulomb potential energy of two helium nuclei (each with charge $q = 2e$) separated by a distance $r$ is given by:

$U = \dfrac{1}{4\pi\epsilon_0} \dfrac{(2e)(2e)}{r}$

Substituting the given values $r = 2 \text{ fm} = 2 \times 10^{-15} \text{ m}$ and $e = 1.6 \times 10^{-19} \text{ C}$:

$U = \dfrac{9 \times 10^9 \times 4 \times (1.6 \times 10^{-19})^2}{2 \times 10^{-15}}$

$U = \dfrac{9 \times 10^9 \times 4 \times 2.56 \times 10^{-38}}{2 \times 10^{-15}}$

$U = 46.08 \times 10^{-14} \text{ J}$

The average thermal energy is given as $1.5 kT$. Equating the thermal energy to the Coulomb potential energy:

$1.5 kT = U$

Substituting the Boltzmann constant $k = 1.38 \times 10^{-23} \text{ J/K}$:

$1.5 \times 1.38 \times 10^{-23} \times T = 46.08 \times 10^{-14}$

$2.07 \times 10^{-23} \times T = 46.08 \times 10^{-14}$

$T = \dfrac{46.08 \times 10^{-14}}{2.07 \times 10^{-23}}$

$T \approx 22.26 \times 10^9 \text{ K} = 2.23 \times 10^{10} \text{ K}$