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

Show by taking the derivative of the Maxwell–Boltzmann
probability distribution P(E) = {(m k / / p)(2 T) }3 Ee–E/kT with respect to
the energy E, and setting the derivative equal to 0, that the most probable
energy Ep of a particle in the distribution, which corresponds to the maximum
of the curve, is given by Ep = 1/2 kT.

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