- #1
TroyElliott
- 58
- 3
Originally posted in a technical forum section, so no HW template
The problem goes as follows... "
One model for the adsorption of gas atoms on a metal surface approximates the surface to be a corregated muffin-tin potential. A gas atom can lower its energy by sitting in one of the potential minima which are the adhesion sites on the surface each with a binding energy ∆. Ignore all other interactions. Show that the free energy of exactly N ≫ 1 atoms adsorbed to a metal surface with M > N adhesion sites is
F = −N∆ + MkBT[(1 − y) ln(1 − y) + y ln y],
where y = N/M."
My attempt at this problem goes as follows. This is a two level energy system, say with energy of 0 defined at the non-binding parts of the muffin-tin and ##-\Delta## in the binding wells. We can then write our partition function as follows,
$$Z = \binom{M}{0}+\binom{M}{1}e^{\beta \Delta}+\binom{M}{2}e^{\beta 2\Delta}+...+\binom{M}{N}e^{\beta N\Delta},$$
where ##\binom{M}{N}## is defined as ##\frac{M!}{N!(M-N)!}.##
This can be written as a series given by $$Z = \sum_{k=0}^{N}\binom{M}{k}e^{k\beta \Delta}.$$
When can be slightly further simplified to $$\ln{Z} = \ln{(M!)}+\ln{(\sum_{k=0}^{N}\frac{e^{k\beta \Delta}}{k!(M-k)!})}.$$
I know that the free energy is given by ##F = -k_{b}T\ln{(Z)}##, but I am not sure how to simplify the sum in order to get the correct free energy. Any ideas on where to go from here would be very much appreciated.
Thanks!
One model for the adsorption of gas atoms on a metal surface approximates the surface to be a corregated muffin-tin potential. A gas atom can lower its energy by sitting in one of the potential minima which are the adhesion sites on the surface each with a binding energy ∆. Ignore all other interactions. Show that the free energy of exactly N ≫ 1 atoms adsorbed to a metal surface with M > N adhesion sites is
F = −N∆ + MkBT[(1 − y) ln(1 − y) + y ln y],
where y = N/M."
My attempt at this problem goes as follows. This is a two level energy system, say with energy of 0 defined at the non-binding parts of the muffin-tin and ##-\Delta## in the binding wells. We can then write our partition function as follows,
$$Z = \binom{M}{0}+\binom{M}{1}e^{\beta \Delta}+\binom{M}{2}e^{\beta 2\Delta}+...+\binom{M}{N}e^{\beta N\Delta},$$
where ##\binom{M}{N}## is defined as ##\frac{M!}{N!(M-N)!}.##
This can be written as a series given by $$Z = \sum_{k=0}^{N}\binom{M}{k}e^{k\beta \Delta}.$$
When can be slightly further simplified to $$\ln{Z} = \ln{(M!)}+\ln{(\sum_{k=0}^{N}\frac{e^{k\beta \Delta}}{k!(M-k)!})}.$$
I know that the free energy is given by ##F = -k_{b}T\ln{(Z)}##, but I am not sure how to simplify the sum in order to get the correct free energy. Any ideas on where to go from here would be very much appreciated.
Thanks!
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