For two distinguishable, non-interacting particles in a box, Quantum Mechanics says the energy must be, for each individual particle,
E = (h^2)/(8mL^2)*[(n_x)^2 + (n_y)^2 + (n_z)^2)
where L is the length of one side of the cube, m is the mass of the particle, and the n's are independent integers that can take any value from 1 and up,
n_x = 1,2,3,4,K
n_y = 1,2,3,4,K
n_z = 1,2,3,4,K
In other words, the individual n's determine the microstates.
If the total energy is E = (h^2)/(8mL^2)(12) what is the multiplicity of that state?© BrainMass Inc. brainmass.com February 24, 2021, 2:24 pm ad1c9bdddf
First, you have to realize that because the two particles are symmetrical, there are a lot of equivalent states ie. for particle 1, P1(nx, ny, nz) we can have P1(1, 2, 1) and it will be equal to P1(2, 1, ...
The expert examines quantum models of non-interacting particles in a box. We have to solve P1(nx, ny, nz) + P2(nx, ny, nz) = 12.