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 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
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
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)
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.
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
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