## stat mech question

rohit
Posts: 104
Joined: Tue Nov 04, 2008 9:12 am

### stat mech question

N particles are distributed amongst three levels having energies 0, kT and 2kT. If
the total equilibrium energy of the system is approximately 425kT, what is the value
of N?

Thnx

matonski
Posts: 121
Joined: Thu Dec 04, 2008 5:03 pm

### Re: stat mech question

Let me try:

Total Energy = Sum [ (Number of particles in a state i) * (energy of state i) ]

425kT = N exp(0)*0 + N exp(-1)*kT + N exp(-2)*2kT
= NkT [ exp(-1) + 2 exp(-2) ]
= .503 NkT

N = 845

rohit
Posts: 104
Joined: Tue Nov 04, 2008 9:12 am

### Re: stat mech question

matonski wrote:Let me try:

Total Energy = Sum [ (Number of particles in a state i) * (energy of state i) ]

425kT = N exp(0)*0 + N exp(-1)*kT + N exp(-2)*2kT
= NkT [ exp(-1) + 2 exp(-2) ]
= .503 NkT

N = 845

er.. i think each of the 3 levels does not have N particles - its more like n1+n2+n3 = N
And the choices in the answer are
1001
335
425
390
181

Posts: 41
Joined: Sat Dec 06, 2008 11:10 am

### Re: stat mech question

n1*E1 + n2*E2 + n3*E3 = 425kT

where n1+n2+n3 = N

P(n1)/P(n2) = n1/n2 = exp(-0*kt/kT)/exp(-kt/kT) = e

Similarly, n1/n3 = e^2

Hence 0*exp(0)kT + (n2*exp(-1))kT + (n3*exp(-2))*2kT = 425kT, which has n1 = 665. Then we use the above relative n relations to get n2 = 244, n3 = 90.

These numbers have N = 999 (numerical error suggests it should be the first option)
Last edited by WakkaDojo on Mon Aug 24, 2009 7:54 pm, edited 1 time in total.

matonski
Posts: 121
Joined: Thu Dec 04, 2008 5:03 pm

### Re: stat mech question

rohit wrote:er.. i think each of the 3 levels does not have N particles - its more like n1+n2+n3 = N
And the choices in the answer are

My thought was that N x the probability being in state i = ni. My mistake was that exp(-E/kT) is not the probability of being in a state because it's not normalized. Oh well. Shows that I need to review.

rohit
Posts: 104
Joined: Tue Nov 04, 2008 9:12 am

### Re: stat mech question

WakkaDojo wrote:n1*E1 + n2*E2 + n3*E3 = 425kT

where n1+n2+n3 = N

P(n1)/P(n2) = n1/n2 = exp(-0kT)/exp(-kT) = e

Similarly, n1/n3 = e^2

Hence 0*exp(0)kT + (n1*exp(-1))kT + (n2*exp(-2))*2kT = 425kT, which has n1 = 665. Then we use the above relative n relations to get n2 = 244, n3 = 90.

These numbers have N = 999 (numerical error suggests it should be the first option)

thats it! thnx

blackcat007
Posts: 378
Joined: Wed Mar 26, 2008 9:14 am

### Re: stat mech question

WakkaDojo wrote:Hence 0*exp(0)kT + (n1*exp(-1))kT + (n2*exp(-2))*2kT = 425kT, which has n1 = 665. Then we use the above relative n relations to get n2 = 244, n3 = 90.

shouldn't this be 0*exp(0)kT + (n2*exp(-1))kT + (n3*exp(-2))*2kT = 425kT ??
and i am still not getting n1=665

Posts: 41
Joined: Sat Dec 06, 2008 11:10 am

### Re: stat mech question

blackcat007 wrote:
WakkaDojo wrote:Hence 0*exp(0)kT + (n1*exp(-1))kT + (n2*exp(-2))*2kT = 425kT, which has n1 = 665. Then we use the above relative n relations to get n2 = 244, n3 = 90.

shouldn't this be 0*exp(0)kT + (n2*exp(-1))kT + (n3*exp(-2))*2kT = 425kT ??
and i am still not getting n1=665

Yes, sorry for the typo (at the time I was quite used to indexing things n0, n1, ... instead of n1, n2, ... since I was doing a lot of computational work). But my calculations are still correct:

n1/n2 = e --> n2 = n1 / e
n1/n2 = e^2 --> n3 = n1 / e^2

So we have

n1 (e^{-1} + 2*e^{-2}) = 425, and we get n1 = 665. Then just plug in n2 = n1 / e, and n3 = n2 / e.