Statistical mechanics: multiplicity

[SoggyBottoms]

Homework Statement

We have a surface that can adsorb identical atoms. There are N possible adsorption positions on this surface and only 1 atom can adsorb on each of those. An adsorbed atom is bound to the surface with negative energy -\epsilon (so \epsilon > 0). The adsorption positions are far enough away to not influence each other.

a) Give the multiplicity of this system for n adsorbed atoms, with 0 \leq n \leq N.

b) Calculate the entropy of the macrostate of n adsorbed atoms. Simplify this expression by assuming N >> 1 and n >> 1.

c) If the temperature of the system is T, calculate the average number of adsorbed atoms.

The Attempt at a Solution

a) \Omega(n) = \frac{N!}{n! (N - n)!}

b) S = k_b \ln \Omega(n) = k_b \ln \left(\frac{N!}{n! (N - n)!}\right)

Using Stirling's approximation: S \approx k_B ( N \ln N - N - n \ln n - n - (N - n) \ln (N - n) - (N - n) = k_B ( N \ln N - n \ln n - (N - n) \ln (N - n)

A Taylor expansion around n = 0 then gives: S \approx k_B (- \frac{n^2}{2N} + ...)\approx -\frac{k_b n^2}{2N}

c) I'm not even sure if the previous stuff is correct, but I have no idea how to do this one. Any hints?

 

[Mute]

Homework Statement

We have a surface that can adsorb identical atoms. There are N possible adsorption positions on this surface and only 1 atom can adsorb on each of those. An adsorbed atom is bound to the surface with negative energy -\epsilon (so \epsilon > 0). The adsorption positions are far enough away to not influence each other.

a) Give the multiplicity of this system for n adsorbed atoms, with 0 \leq n \leq N.

b) Calculate the entropy of the macrostate of n adsorbed atoms. Simplify this expression by assuming N >> 1 and n >> 1.

c) If the temperature of the system is T, calculate the average number of adsorbed atoms.

The Attempt at a Solution

a) \Omega(n) = \frac{N!}{n! (N - n)!}

b) S = k_b \ln \Omega(n) = k_b \ln \left(\frac{N!}{n! (N - n)!}\right)

Using Stirling's approximation: S \approx k_B ( N \ln N - N - n \ln n - n - (N - n) \ln (N - n) - (N - n) = k_B ( N \ln N - n \ln n - (N - n) \ln (N - n)

A Taylor expansion around n = 0 then gives: S \approx k_B (- \frac{n^2}{2N} + ...)\approx -\frac{k_b n^2}{2N}

c) I'm not even sure if the previous stuff is correct, but I have no idea how to do this one. Any hints?

Part (a) looks good, but you made a couple of sign mistakes in part (b). In the terms in the denominator you forgot to distribute the negative sign to the second term in n ln n - n and (N-n) ln(N-n). I've corrected the signs here:

S \approx k_B ( N \ln N - N - n \ln n + n - (N - n) \ln (N - n) + (N - n) = k_B ( N \ln N - n \ln n - (N - n) \ln (N - n)

This change will result in some cancellations that help simplify your expression. Rewriting

(N-n)\ln(N-n) = \left(1-\frac{n}{N}\right)N\ln N + (N-n)\ln\left(1-\frac{n}{N}\right)

will help you make some more cancellations.

Another mistake you made in your original attempt was that you expanded around n = 0, but you are told n is much greater than 1, so you can't do that expansion. What you can do, however, is assume that while n is much greater than 1, it is still much less that N, such that n/N is small, and you can expand the above logarithms in n/N.

This will give you a simple expression for the entropy. To get the temperature, you need to write the entropy as a function of the total energy. Right now your entropy is a function of number. However, you are told how much energy there is per site, so you can figure out what the total energy is for n adsorbed atoms. Use this to rewrite the entropy in terms of the total energy.