GREPhysics.NET: GR9277 #71

Alternate Solutions

moonrazor
2006-03-27 13:37:23

(GR9277-71) Alternate Solution with less assumptions:

Using P(E) = (1/Z)exp[-E/kT] we have

P(E_1) = exp[-E_1/kT] / (exp[-E_1/kT] + exp[-E_2/kT])

= 1 / (1 + exp[-(E_1-E_2)/kT])

= 1 / (1 + exp[-\epsilon /kT])

LaTeX

\begin{eqnarray*}
P(E_{1}) &=& \frac{e^{-E_{1}/kT}}{e^{-E_{1}/kT} + e^{-E_{2}/kT}}\\

&=& \frac{1}{1 + e^{-(E_{2}-E_{1})/kT}}\\

&=& \frac{1}{1 + e^{-\epsilon/kT}}
\end{eqnarray*}

radicaltyro
2006-10-31 14:05:37

Hi moonrazor. Wrap your latex in single dollar signs.

$\begin{eqnarray*} P(E_{1}) &=& \frac{e^{-E_{1}/kT}}{e^{-E_{1}/kT} + e^{-E_{2}/kT}}\\&=& \frac{1}{1 + e^{-(E_{2}-E_{1})/kT}}\\&=& \frac{1}{1 + e^{-\epsilon/kT}} \end{eqnarray*}$

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Comments

Fv4
2008-10-04 09:43:14

as epsilon -> $\infty$ , number of subsystems at E1 should approach $N_{0}$ . Thus, only B) can be the answer.

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hassanctech
2007-09-30 22:02:11

Correct me if I'm wrong but if E2-E1 = e>0 then lets look at the limit as e --> infinity. This means that E2 is of infinite energy and you would expect all the sybsystems to be in the E1 state. Choice B is the only one for which e--> infinity reduces to N0.

Blue Quark
2007-10-31 19:22:45

Am I missing something?
As T approaches infinity all the particles should be in the highest energy state, E2.
For (b), as T approaches infinity the average number of particles in the lower energy state, E1, is equal to No. So as the temperature increases more particles are dropping to lower energy states?

Jeremy
2007-11-01 16:58:13

Blue Quark,

Check your math on the limit. The high temperature limit is in fact $E \rightarrow N_{0}/2$ . Here's a qualitative idea of why it's true. At low temperatures, most of the subsystems are in the lowest energy state because there is not enough energy for a transition (taking $k T$ to represent the amount of energy available to the system). Of course, there may be a few subsystems in excited states, but I'm speaking about the typical case - what the "big picture" looks like. Well, as $T$ increases, the available energy increases, and more and more subsystems will be in their excited states. For $k T \gg \epsilon$ , the limitation imposed by energy is completely removed, so that it is equally likely to be in any of its available states. This is why we expect to find half of the subsystems in the lower state and half in the higher state. Also note that this macrostate maximizes entropy, whereas putting all subsystems in one state (for example the one with higher energy) will minimize entropy. And one last thing, this understanding allows you to answer #73 (from this test) immediately - no equations needed!

Jeremy
2007-11-13 11:14:08

I finally found the other problem that my previous post churns out an instant answer for! It's GR8677 #67. So you can see everything one page, here's what it says:
__________
67. A large isolated system of $N$ weakly interacting particles is in thermal equilibrium. Each particle has only 3 possible nondegenerate states of energies $0$ , $\epsilon$ , and $3\epsilon$ . When the system is at an absolute temperature $T \gg \epsilon/k$ , where $k$ is Boltzmann's constant, the average energy of each particle is

(A) $0$
(B) $\epsilon$
(C) $\frac{4}{3}\epsilon$
(D) $2\epsilon$
(E) $3\epsilon$
__________

Using the reasoning in my previous post, we know that the 3 states are equally occupied, so the average energy for any individual particle is just the average energy of the possible states: $\frac{0+\epsilon+3\epsilon}{3}=\frac{4}{3}\epsilon$ .

Here " target="_blank">http://grephysics.net/ans/8677/67">Here is a link to the page discussing that problem, where others have already mentioned this shortcut.

Jeremy
2007-11-13 11:23:27

Why can I never get the link to work the first time? I must not be doing it right... but it works the second time. Anyone care to share a linking syntax that works?

Link

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Richard
2007-09-13 23:48:09

I too wanted to say that the posted solution is rather poor. This is a Boltzman stat. problem: The probability for a state is the Boltzman factor for that state divided by the Partition Function. What's with the, "let's assume that $E_1=\epsilon$ ?"

evanb
2008-06-25 12:56:18

It turns out that it doesn't matter what $E_1$ is equal to... it adds a factor of $e^{E_1/kT}$ to the numerator and to all the factors in the denominator. Better off setting it to zero, in my opinion, but in terms of the physics it doesn't make one lick of difference: it's like setting your h in mgh potential energy to 0 on the ground.

See radicaltyro's comment on moonrazor's solution for evidence.

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georgi
2007-08-26 22:07:45

the posted solution is actually just purely incorrect. a fermi dirac distribution assumes that the occupancy of a particular quantum state is either 0 or 1 in accordance with the pauli-exclusion principle. instead we are actually interested in a maxwell boltzmann distribution in which the particles (in our case subsystems) are distinguishable and any number can occupy either energy state. then we apply the partition function as posted by radicaltyro to obtain our solution.

FortranMan
2008-10-02 15:33:49

So because a system only has two states doesn't alone mean that it's Fermi-Dirac?

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moonrazor
2006-03-27 13:37:23

radicaltyro
2006-10-31 14:05:37

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