Chapter 2: Q. 2.9 (page 60)

Use a computer to reproduce the table and graph in Figure $2.4$ : two Einstein solids, each containing three harmonic oscillators, with a total of six units of energy. Then modify the table and graph to show the case where one Einstein solid contains six harmonic oscillators and the other contains four harmonic oscillators (with the total number of energy units still equal to six). Assuming that all microstates are equally likely, what is the most probable macrostate, and what is its probability? What is the least probable macrostate, and what is its probability?

Short Answer

Expert verified

The most probable macrostate in the first solids with $N_{A} = N_{B} = 3$ is when the energy units are evenly distributed, $q_{A} = q_{B}$ , with a probability of $0.21645$ ; the least probable macrostate is when all the energy units are in one of the partitions, $q_{A} = 6$ or when $q_{B} = 6$ , with a probability of $0.06060$ .

The most probable macrostate in the second solids with $N_{A} = 6$ and $N_{B} = 4$ is when all energy units are in solid $B, q_{B} = 6$ , and its probability is $0.016783$ , while the least probable macrostate is when four energy units are in solid one and the rest in solid $B$ , $q_{A} = 4$ and $q_{B} = 2$ .

Step by step solution

Overall multiplicity and probability

Multiplicity is,

$Ω_{overall} (N_{overall}, q_{overall}) = (\begin{matrix} q_{overall} + N_{overall} - 1 \\ q_{overall} \end{matrix})$

$= \frac{(q_{overall} + N_{overall} - 1)!}{q_{overall}! (N_{overall} - 1)!}$

$q_{overall} = q_{A} + q_{B} = 6,$

$N_{overall} = N_{A} + N_{B} = 6$

So,

$Ω_{overall} = (\begin{matrix} 6 + 6 - 1 \\ 6 \end{matrix})$

$= \frac{(6 + 6 - 1)!}{6! (6 - 1)!}$

$= 462$

Total multiplicity is,

$Ω_{total} = Ω_{A} Ω_{B}$

$Ω_{A} = (\begin{matrix} q_{A} + N_{A} - 1 \\ q_{A} \end{matrix})$

For $N_{A} = 3$

$Ω_{A} = (\begin{matrix} q_{A} + 2 \\ q_{A} \end{matrix})$

$= \frac{(q_{A} + 2)!}{q_{A}! (2!)}$

$Ω_{B} = (\begin{matrix} q_{B} + N_{B} - 1 \\ q_{B} \end{matrix})$

For $N_{B} = 3$ , $q_{B} = 6 - q_{A}$

So,

$Ω_{B} = (\begin{matrix} 6 - q_{A} + 3 - 1 \\ 6 - q_{A} \end{matrix})$

$= \frac{(8 - q_{A})!}{(6 - q_{A})! (2)!}$

Probability is,

$P (q_{A}) = \frac{Ω_{A} Ω_{B}}{Ω_{overall}}$

$= \frac{1}{462} \frac{(q_{A} + 2)!}{q_{A}! (2)!} \frac{(8 - q_{A})!}{(6 - q_{A})! (2)!}$

Result of table when both energy is equal

Graph for equal energy

Multiplicity for different energy and probability

Multiplicity,

$Ω_{overall} (N_{overall}, q_{overall}) = (\begin{matrix} q_{overall} + N_{overall} - 1 \\ q_{overall} \end{matrix})$

$Ω_{overall} = (\begin{matrix} 6 + 10 - 1 \\ 6 \end{matrix})$

$= 5005$

$Ω_{A} = (\begin{matrix} q_{A} + N_{A} - 1 \\ q_{A} \end{matrix})$

For $N_{A} = 6$

$Ω_{A} = (\begin{matrix} q_{A} + 5 \\ q_{A} \end{matrix})$

$= \frac{(q_{A} + 5)!}{q_{A}! (5!)}$

$Ω_{B} = (\begin{matrix} q_{B} + N_{B} - 1 \\ q_{B} \end{matrix})$

For $N_{B} = 4$ , $q_{B} = 6 - q_{A}$

$Ω_{B} = (\begin{matrix} 6 - q_{A} + 4 - 1 \\ 6 - q_{A} \end{matrix})$

$= \frac{(9 - q_{A})!}{(6 - q_{A})! (3)!}$

Probability is,

$P (q_{A}) = \frac{Ω_{A} Ω_{B}}{Ω_{overall}}$

$= \frac{1}{5005} \frac{(q_{A} + 5)!}{q_{A}! (5!)} \frac{(9 - q_{A})!}{(6 - q_{A})! (3)!}$

Result of table for different energy level

Graph for different energy values

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Short Answer

Step by step solution

Overall multiplicity and probability

Result of table when both energy is equal

Graph for equal energy

Multiplicity for different energy and probability

Result of table for different energy level

Graph for different energy values

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