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Use a computer to produce a table and graph, like those in this section, for two interacting two-state paramagnets, each containing 100 elementary magnetic dipoles. Take a "unit" of energy to be the amount needed to flip a single dipole from the "up" state (parallel to the external field) to the "down" state (antiparallel). Suppose that the total number of units of energy, relative to the state with all dipoles pointing up, is80; this energy can be shared in any way between the two paramagnets. 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 likely macrostate is when the energy units are evenly distributed,qA=qB=40 , with a probability of 0.07513. The least likely state is when all the energy units are in partition BorA,qA=40 , or when qB=40, with a chance of 7.8726×10-20.

Step by step solution

01

Expression for overall multiplicity

The probability of PqAis,

PqA=ΩtotalΩoverall

The overall multiplicity is,

ΩoverallNoverall,qoverall=qoverall+Noverall-1qoverall

The total multiplicity is ,

Ωtotal=ΩAΩB

ΩA=qA+NA-1qA

ΩB=qB+NB-1qB

02

Calculation for total multiplicity

Multiplicity is,

Ωoverall=qoverall+Noverall-1!qoverall!Noverall-1!

qoverall=qA+qB=80

Noverall=NA+NB=200

So,

role="math" localid="1650306409339" Ωoverall=(80+200-1)!80!(200-1)!

=2.1225×1071

Multiplicity of Ais,

ΩA=qA+99qA=qA+99!qA!(99)!

Multiplicity of Bis,

ΩB=qB+99qB

Substitute qB=80-qA

so,

role="math" localid="1650306390399" ΩB=179-qA80-qA

=179-qA!80-qA!(99)!

Probability is,

PqA=ΩAΩBΩoverall

=12.1225×1071qA+99!qA!(99)!179-qA!80-qA!(99)!

03

Python program for creation of graph

04

Graph for probability and energy

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