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"N atoms are placed in a rigid grid, and each atom has 4 energi states, numbered 1-4, with the energy levels [tex]\epsilon_1=0 \epsilon_2=d \epsilon_3=\epsilon_4=2d[/tex]. The propability of an atom to be in the i'th state is given by [tex]n_i[/tex]. With a powerful elektromagnetic field, the system is made such that the propabilites are as follows: [tex]n_1=n_2=1/2 n_3=n_4=0[/tex]."

Then the first question.

"a) What is the inner energy [tex]U_o[/tex] and the entropy [tex]S_o[/tex] of this system?"

So i first i look at the energy, since [tex]U=N*<\epsilon>[/tex] i assumed that:

[tex]U=N*(\epsilon_1*P(\epsilon_1)+\epsilon_2*P(\epsilon_2)+\epsilon_3*P(\epsilon_3)+\epsilon_4*P(\epsilon_4))[/tex]

[tex]U=N*(0*1/2+d*1/2+2d*0+2d*0)=Nd*1/2[/tex]

Next the entropy, since we have no information about the spin excess, i assume i should use the approximation [tex]g=2^N[/tex], and hence [tex]S_o=N*ln(2)[/tex].

Then on to the B question:

b) "The system is now isolated from the enviroment and after a bit of time, it is in thermal equilibrium.

Write the propabilities [tex]n_i[/tex] expressed as functions of [tex]x=e^(-d/kT)[/tex] where T is the systems final temperature (and k the boltzman constant). Find the temperature T, the free energy F and entropy S for the thermal equilibrium, expressed as functions of N and d. Compare S with the start values of [tex]S_0[/tex] as well as the systems maximal entropy [tex]S_\infinity[/tex]."

So now i can see that the Z function must be:

[tex]Z=1+x+x^2[/tex]

And consequently the propabilities:

[tex]n_1=1/Z\ n_2=x/Z\ n_3=n_4=x^2/Z[/tex]

And using the same method as in a) to find the energy (with these new propabilities and old energy states) i get U to be:

[tex]U=Nd*(4x^2+X)/Z[/tex]

Now i'm not 100% sure if i can do this next step. But i assume that the energy before, is equal to the energy after, so [tex]U_0=U[/tex]. By doing that, i eliminate Nd from both equations and end up with a second order equation that i can easily solve. So:

[tex]Nd*(4x^2+X)/Z=Nd*1/2,\ =>\ 4x^2+x=1/2*Z:\ =>\ 3x^2+1/2*x-1/2=0[/tex]

Which has the solutions

[tex]x_1=1/3 x_2=-1/2[/tex]

Of which obviously 1/3 is the solution we can use, to find T from the x equation:

[tex]x=e^(-d/kT) x=1/3 => ln(1/3)=-d/kT => T=-d/k*ln(1/3) or \tau=-d/ln(1/3)[/tex]

Substituting x=1/3 into the Z function before, i get Z=14/9, and then i have tau and Z, so i can use the equation [tex]F=-\tau*ln(Z)[/tex] i get F to be:

[tex]F=-\tau*ln(Z)=-(-d/ln(1/3))*log(14/9)=d*ln(14/9)/ln(1/3)[/tex]

Then using the definiton of F, [tex]F=U-\tau*\sigma[/tex], where i have also substituted x=1/3 i U, i get:

[tex]F=U-\tau*\sigma \ =>\ \sigma=F-U/\tau=(d*ln(14/9)/ln(1/3)-1/2*nd)/(-d/ln(1/3) \ =>\ \sigma*k=k(ln(14/9)-1/2*N*ln(1/3)=S[/tex]

And this is where i'm stuck. Now i wasn´t sure if i could do the [tex]U_0=U[/tex] move, but i'm relatively sure that it was ok. but i have real problems with the last problem, where i am asked to compare the expressions for S. For the first, i can´t see what the answer should be, i could of course make obvious comments such as the entropy is bigger/smaller and what not, but i feel like they're looking for a more specific answer, and i'm not seeing the relation.

Also since they ask me to compare to the maximal S. So my question is, can i simply do [tex]S_\infinity=kN*ln(4)[/tex] or do i need to derive the equation from a g where there is an equal amount of particles in all states. I have done that, and ended up with the expression:

[tex]g=N!/((4*(N*1/4)!) \ and \ S=k*ln(g)=kN(ln(n)-4ln(N*1/4))[/tex]

Using the stirling approximation. Either way, i also have problems comparing this to the [tex]S_0[/tex], not sure what to say on that question.