Show that by integrating the Maxwell–Boltzmann probability distribution N(E) = N {(m k / / p)(2 T)..

Show that by integrating the Maxwell–Boltzmann probability
distribution N(E) = N {(m k / / p)(2 T) }3 Ee– E/kT where the average
energy is dened by   = Nò (E)EdE/N,N is the total number of particles, and
the limits of integration are from E = 0 to E = ∞ that the average energy
of a particle that follows this distribution is = 3/2 kT.

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