Answer:
(a) E = [2[tex]N_A[/tex]ξ[tex]e^{\frac{-T}{kT}}[/tex]] / [1+2[tex]e^{\frac{-T}{kT}}[/tex]]]
(b) S = [tex]N_A[/tex]k.ln(1+2[tex]e^{\frac{-T}{kT}}[/tex]) + [[tex]N_A[/tex]ξ[tex]e^{\frac{-T}{kT}}[/tex]] / T[1+2[tex]e^{\frac{-T}{kT}}[/tex]]]
(c) S = [tex]N_A[/tex] k.ln 3 (These values directly obtained from the result of (b) when T⇒0 & T⇒∞.)
Explanation:
Given:
One of three quantum states labeled by the quantum numbers m,
Where m = -1 , 0 or 1
A nucleus has the same energy E= e in the state m = 1 & m = -1
Compared with an energy E = 0 in the state m = 0.