Page 1 of 1

stat mech question

Posted: Fri Dec 05, 2008 4:26 pm
by rohit
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

Re: stat mech question

Posted: Fri Dec 05, 2008 5:39 pm
by matonski
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

Re: stat mech question

Posted: Sat Dec 06, 2008 9:41 am
by rohit
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

Re: stat mech question

Posted: Sat Dec 06, 2008 11:14 am
by WakkaDojo
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)

Re: stat mech question

Posted: Sat Dec 06, 2008 12:49 pm
by matonski
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.

Re: stat mech question

Posted: Sat Dec 06, 2008 1:05 pm
by rohit
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

Re: stat mech question

Posted: Mon Aug 24, 2009 9:45 am
by blackcat007
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

Re: stat mech question

Posted: Mon Aug 24, 2009 7:16 pm
by WakkaDojo
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.