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 deutrons 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 closet separation, the kinetic energy is zero and the potential energy is `(e^(2))/(4piepsilon_(0)r)`. If the initial kinetic energy of each deuteron is `K` and the closest separation is `2 fm`, we shall have
`2K=(e^(2))/(4pi epsilon_(0)(2fm))=((1.6xx10^(-19)C)^(2)xx(9xx10^(9)N-m^(2)//C^(2)))/(2xx10^(-15)m)`
or, `K=5.7xx10^(-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 `5.7xx10^(-14)J`. will be given by
`1.5kT=5.7xx10^(-14)J`
where `K=` Botzmann constant
or, `T=(5.7xx10^(-14)J)/(1.5xx1.38xx10^(-23)J//K)`