Consider two deuterons moving towards each other with equal speeds in a deuteron gas. What should be their kinetic energies (when they are widely separated) so that the closest separation between them becomes `2 fm?` Assume that the nuclear force is not effective for separations greater than `2 fm`. At what temperature will the deuterons have this kinetic energy on an average?

Explanation

As the deuterons move, the Coulomb repulsion will slow them down. The loss in kinetic energy will be equal to the gain in Coulomb potential energy. At the closest separation, the kinetic energy is zero and the potential energy is `(e^2)/((4piepsilon_0) (2 fm))`
`= ((1.6 xx 10^(-19) C)^2 xx (9 xx 10^9 N m^2 C^(-2))`/(2 xx 10^(-15) m`
or, `K = 5.7 xx 10^(-14) J`.
If the temperature of the gas is `T`, the average kinetic energy of random motion of each nucleus will be `1.5 kT`. The temperature needed for the deuterons to have the average kinetic energy of `5.7 xx 10^(-14) J` will be given by
`1.5 kT = 5.7 xx 10^(-14) J`
or, `T = (5.7 xx 10^(-14) J)/(1.5 xx 1.38 xx 10^(23) J K`
`= 2.8 xx 10^9 K`.