Chapter 7: Q.7.4 (page 261)

Repeat the previous problem, taking into account the two independent spin states of the electron. Now the system has two "occupied" states, one with the electron in each spin configuration. However, the chemical potential of the electron gas is also slightly different. Show that the ratio of probabilities is the same as before: The spin degeneracy cancels out of the saha equation.

Short Answer

Expert verified

A system of a single Hydrogen atom/ion, which has two possible states: Unoccupied and occupied. Taking into account the two independent spin states of the electron, now the system has two occupied states one with the electron in each spin configuration.

Step by step solution

Unoccupied state:

Energy $(ε) = 0$

Occupation number $(n) = 0$

Gibbs Factor $= e x p [- \frac{1}{K T} (ε - n μ)] = e x p [- \frac{(0 - 0 . μ)}{k T}] = e x p (0) = 1$

Singly occupied states:

Energy

Occupation

Degeneracy

Gibbs Factor

The partition function is

(ε) = - 1

The probability that the hydrogen atom is ionized is:

$P_{i o n i z e d} = \frac{1}{1 + 2 e x p (\frac{(1 + μ)}{k T})}$

The probability that the hydrogen atom is neutral

$P_{n e u t r a l} = \frac{2 e x p (\frac{(I + μ)}{k T})}{1 + 2 e x p (\frac{(I + μ)}{k T})}$

The chemical potential of the electron gas can be written as:

$μ = - k T \ln (\frac{V Z_{i n t}}{N v_{Q}})$ Where $Z_{i n t} = 2$ for the two spin states.

role="math" $\frac{P_{i o n i z e d}}{P_{n e u t r a l}} = \frac{1}{2 e x p (\frac{(I + μ)}{k T})} = \frac{e x p (\frac{- I}{k T})}{2 e x p (\frac{μ}{k T})} μ = - k T \ln (\frac{V Z_{i n t}}{N v_{Q}}) = k T \ln (\frac{N v_{Q}}{2 V}) e x p (\frac{μ}{k T}) = \frac{N v_{Q}}{2 V} \frac{P_{i o n i z e d}}{P_{n e u t r a l}} = \frac{e x p (\frac{- I}{k T})}{2 (\frac{N v_{Q}}{2 V})} \frac{P_{i o n i z e d}}{P_{n e u t r a l}} = \frac{V}{N v_{Q}} e x p (\frac{- I}{k T})$

As $T \to 0, v_{Q} \to \infty, e x p (\frac{- I}{k T}) \to \infty$

And $\frac{P_{i o n i z e d}}{P_{n e u t r a l}} \to 0,$ which is what you would expect, if none of the hydrogen atoms are ionized.

role="math" $T \to \infty, v_{Q} \to 0, e x p (\frac{- I}{k T}) \to 1 \frac{P_{i o n i z e d}}{P_{n e u t r a l}} \to \infty$

i.e., all the hydrogen atoms are ionized and $P_{n e u t r a l} \to 0$

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Short Answer

Step by step solution

Unoccupied state:

Singly occupied states:

The probability that the hydrogen atom is ionized is:

The chemical potential of the electron gas can be written as:

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