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    3. Consider again the lattice of N spin-1/2 particles in an external homogeneous magnetic field, where each particle has two possible states: spin "down" with energy e = 0 and spin "up" with energy e = 1/2". The microstate of the system is specified by the energy states of all the particles, i.e. the list (e1, e2...

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    https://brainmass.com/physics/beta/problems-statistical-mechanics-428804

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    (a) We have S = k log W(n) = k log NCn = k log (N!/n!(N-n)!). Using Sterling's approximation to the factorial, we have

    S is approximately equal to k log (N^N e^-N/((n^n e^-n)((N-n)^(N-n) e^-(N-n)))
    = k (N log N - n log n - (N-n) log(N-n))

    Since E(n) = n/2, we have

    S(E) = k (N log N - 2E log 2E - (N-2E) log(N-2E))

    where the above equality is approximate in the limit of large N and n.

    (b) We have

    1/T = dS/dE = k(-2E/E - 2 log 2E + 2 log(N-2E) - 2(N-2E)/(N-2E))
    = 2k log((N-2E)/2E)
    whence

    T(E) = 1/(2k log(N/2E - 1)).

    (c) Inveting the above formula, we find

    E(T) = N/(2(1 + e^(1/(2kT))).

    We have

    C(T) = dE/dT

    = N/2 (-1/(2kT^2)) (-1/(1 + e^(1/(2kT)))^2
    = N/(4kT^2) (1 + e^(1/(2kT)))^-2.

    (d) We have

    Z = sum_{n=0}^infty(NCn e^(-beta E(n)))
    = sum_{n=0}^infty(NCn e^(-beta n/2))

    where beta = 1/(kT).

    (e) We have

    <E> = -1/Z dZ/dbeta
    = 1/(2Z) sum_{n=0}^infty(n NCn e^(-beta n/2)).

    (f) We have

    C = d<E>/dT
    = d<E>/dbeta dbeta/dT
    = -1/(kT^2) d<E>/dbeta
    = 1/(kT^2) [(1/(2Z^2) dZ/dbeta) sum_{n=0}^infty(n NCn e^(-beta n/2))
    + 1/(4Z) sum_{n=0}^infty(n^2 NCn e^(-beta n/2))]

    = 1/(kT^2) [1/(4Z) sum_{n=0}^infty(n^2 NCn e^(-beta n/2)) - E^2/2].

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com December 24, 2021, 9:57 pm ad1c9bdddf>
    https://brainmass.com/physics/beta/problems-statistical-mechanics-428804

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