Spring 2010 Solutions - FSU Physics Department

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Problem 11The ground level of the neutral lithium atom is doubly degenerate (that is, d0 = 2). Thefirst excited level is 6-fold degenerate (d1 = 6) and is at an energy 1.2 eV above theground level.(a) In the outer atmosphere of the Sun, which is at a temperature of about 6000 K, whatfraction of the neutral lithium is in the excited level? Since all the other levels of Li are ata much higher energy, it is safe to assume that they are not significantly occupied.(b) Find the average energy of Li atom at temperature T(again, consider only the ground state and the first excited level).(c) Find the contribution of these levels to the specific heat per mole, CV, and sketch CVas a function of T.Solution(a) If the ground level energy is defined as zero and E is the energy ofexcited level:Z = Σ i d i exp(-βε i ) = 2 + 6 exp (−βΕ)The probability that the atom is in its excited level:P(E) = 6 exp(−βΕ) / Ζ= exp(−βΕ)]/ [2 + 6 exp(−βΕ)]= 3 / (3 + exp(βΕ)Since E = 1.2 eV, T = 6000K (~0.5 eV), βE = 2.32, exp(βE ) ≈ 10we get: P(Ε ) = [3/ (3+ 10)] = 0.23(b) The average energy per atom is: = (1/Ζ)(δΖ/δβ) = [(ε) 6 exp(−βε )] / [2+6 exp(−βε)] = [3ε /(exp(βε) + 3)](c) The specific heat is:C v = [δ/δΤ] v = [-3ε x ε exp(βε)( δβ/δT)] / [exp(βε) + 3] 2= [3k B (βε) 2 exp(βε)] / [exp(βε) +3] 2
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Spring 2010 Solutions - FSU Physics Department

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