Question about the Boltzmann distribution

[dRic2]

I was reading about the Debye-Huckle theory for electrolytes solutions (https://en.wikipedia.org/wiki/Debye–Hückel_theory). In all the books, notes, and in the wikipedia age too, there is this statement that troubles me:

calculate the charge density by means of a Boltzmann distribution:
$$n(r) = \bar n exp( \frac {-z_i e \psi(r)} {k_B T})$$

Shouldn't I have the "normalization factor" (i.e $1/Z$) in the above equation?

Thanks Ric

 

[BvU]

No. It is not a probability distribution but a number density of ions.
Peter Atkins (Physical Chemistry) writes:

The difference in energy of an ion j of charge $z_je$ at a position where the central ion i is giving rise to a potential $\phi_i$ relative to its energy at infinity (where the potential is zero) is $\Delta E = z_je\phi_i$.
The Boltzmann distribution then gives the proportion of of ions at this location relative to the proportion in the bulk solution (effectively at infinity: $${{\mathcal N}_j\over {\mathcal N}_j^\circ}={ \text {
number of }{\sf j}\text{ ions per unit volume where the potential is }\phi_i \over \text {
number of }{\sf j}\text { ions per unit volume where the potential is zero}} \\ \mathstrut \\
=e^{-\Delta E/kT}\ , \ \ {\sf with} \quad \Delta E = z_je\phi_i\ .
$$This means that $$
{\mathcal N}_j/ {\mathcal N}_j^\circ =e^{- z_j e\phi_i/kT}\ . $$

 

[dRic2]

Ahhh. So basically
$$n(r) = \frac N Z e^{-\beta z_i e \psi(r)}$$
is the mean number of particles in the state with energy $z_i e \psi(r)$ (where $N$ is the total number of particles in the system). Also
$$\bar n( \infty) = \frac N Z e^{- \beta z_i e \psi( \infty )} = \frac N Z$$
since $\psi( \infty ) = 0$. Then:
$$ \frac {n(r)} {\bar n} = e^{-\beta z_i e \psi(r)}$$
The multiplying by $\frac V V$ I get the density distribution. Am I right ?

 

[BvU]

Remind us what your capital $Z$ is in this context. I don't see it appear in the analysis

 

[dRic2]

the partition function

 

[dRic2]

Do you think my reasoning is correct ?