Consider fusion occurring in a helium plasma. Determine the temperature at which the average thermal energy, given by 1.5 kT , becomes equal to the Coulomb potential energy at a separation of 2 fm .

Step-by-step solution

The Coulomb potential energy of two helium nuclei (each with charge q = 2e ) separated by a distance r is given by: U = 1 4 ₀ (2e)(2e) r Substituting the given values r = 2 fm = 2 10⁻¹⁵ m and e = 1.6 10⁻¹⁹ C : U = 9 10^9 4 (1.6 10⁻¹⁹)^2 2 10⁻¹⁵ U = 9 10^9 4 2.56 10⁻³⁸ 2 10⁻¹⁵ U = 46.08 10⁻¹⁴ 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 10⁻²³ J/K : 1.5 1.38 10⁻²³ T = 46.08 10⁻¹⁴ 2.07 10⁻²³ T = 46