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[QUOTE="tanaygupta2000, post: 6307613, member: 645573"] [B]Homework Statement:[/B] Three bosons are to be filled in two energy states with degeneracies 3 and 4 respectively. (1.) List all the macrostates. (2.) How many microstates does this 3-particle system has? (3.) Which macrostate is the most probable one? [B]Relevant Equations:[/B] Partition function, Z = ∑g(j)exp(-E(j)/kT) Upto now I've only dealt with the problems regarding non - degenerate energy states. Since bosons do not follow Pauli's Exclusion Principle, three bosons can be filled in two energy states (say E[SUB]1[/SUB] and E[SUB]2[/SUB]) as: [TABLE] [TR] [TD][CENTER][B]E[SUB]1[/SUB][/B][/CENTER][/TD] [TD][CENTER][B]E[SUB]2[/SUB][/B][/CENTER][/TD] [/TR] [TR] [TD][CENTER]1 boson[/CENTER][/TD] [TD][CENTER]2 bosons[/CENTER][/TD] [/TR] [TR] [TD][CENTER]2 bosons[/CENTER][/TD] [TD][CENTER]1 boson[/CENTER][/TD] [/TR] [TR] [TD][CENTER]3 bosons[/CENTER][/TD] [TD][CENTER]0 bosons[/CENTER][/TD] [/TR] [TR] [TD][CENTER]0 bosons[/CENTER][/TD] [TD][CENTER]3 bosons[/CENTER][/TD] [/TR] [/TABLE] so that there are 2 macrostates (corresponding to levels E[SUB]1[/SUB] and E[SUB]2[/SUB]) and 4 microstates (corresponding to 4 possibilities). Also the probability of occurrence of [LIST] [*]I Possibility = 3!/2! = 3 [*]II Possibility = 3!/2! = 3 [*]III Possibility = 3!/3! = 1 [*]IV Possibility = 3!/0! = 6 [/LIST] Is this correct way of dealing with this problem? I do not understand the meaning behind the given degeneracies of energy states. [/QUOTE]
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