Show by taking the derivative of the Maxwell–Boltzmann probability distribution P(E) = {(m k / /…


Show by vestibule the derivative of the Maxwell–Boltzmann probability disposal P(E) = {(m k / / p)(2 T) }3 Ee–E/kT after a while deference to the ghost E, and enhancement the derivative correspondent to 0, that the most probable ghost Ep of a tittle in the disposal, which corresponds to the maximum of the incurvation, is loving by Ep = 1/2 kT.