Assume that the reaction H ⇌p+e occurs in thermal equilibrium at T=4000 K in a very low density

Assume that the reaction H ⇌p+e occurs in thermal...

Assume that the reaction $\mathrm{H} \rightleftharpoons \mathrm{p}+\mathrm{e}$ occurs in thermal equilibrium at $T=4000 \mathrm{~K}$ in a very low density gas (no degeneracy) of each species with overall charge neutrality. (a) Write the chemical potential of each gas in terms of its number density $[\mathrm{H}],[\mathrm{p}]$, or [e]. For simplicity you may ignore the spectrum of excited bound states of H and consider only the ground state. Justify this assumption. (b) Give the condition for thermal equilibrium and calculate the equilibrium value of $[\mathrm{e}]$ as a function of $[\mathrm{H}]$ and $T$. (c) Estimate the nucleon density for which the gas is half-ionized at $T$ $=4000 \mathrm{~K}$. (Note that this is an approximate picture of the universe at a redshift $z=10^3$.)

Assume that the reaction $\mathrm{H} \rightleftharpoons \mathrm{p}+\mathrm{e}$ occurs in thermal equilibrium at $T=4000 \mathrm{~K}$ in a very low density gas (no degeneracy) of each species with overall charge neutrality.
(a) Write the chemical potential of each gas in terms of its number density $[\mathrm{H}],[\mathrm{p}]$, or [e]. For simplicity you may ignore the spectrum of excited bound states of H and consider only the ground state. Justify this assumption.
(b) Give the condition for thermal equilibrium and calculate the equilibrium value of $[\mathrm{e}]$ as a function of $[\mathrm{H}]$ and $T$.
(c) Estimate the nucleon density for which the gas is half-ionized at $T$ $=4000 \mathrm{~K}$. (Note that this is an approximate picture of the universe at a redshift $z=10^3$.)

Key Concepts

Chemical Potential in an Ideal Gas

The chemical potential of an ideal gas relates to the logarithm of its number density and partition function. It quantifies the change in free energy when particles are added or removed from the system. In the low density (non-degenerate) limit, it can be expressed in terms of the number density, temperature, and mass of the particles, which provides a clear picture of the thermodynamic properties crucial for analyzing reactions under equilibrium conditions.

Thermal Equilibrium and Chemical Reaction Equilibrium

Thermal equilibrium in the context of chemical reactions means that the chemical potentials of the reactants and products balance out according to the reaction stoichiometry. For a reaction where particles convert one type to another, the condition ?_A = ?_B + ?_C (for a reaction A ? B + C) assures that there is no net change in the free energy, thereby maintaining the equilibrium state. This is a fundamental concept in thermodynamics that underlies the derivation of equilibrium concentration relationships.

Saha Equation

The Saha equation is used to calculate the ionization state of a gas in thermal equilibrium by relating the number densities of ions, electrons, and neutral atoms. It incorporates factors such as temperature, ionization energy, and statistical weights (degeneracies) of the species. This relation is especially important in astrophysics and cosmology, where it helps predict the ionization fraction of gases under different thermal conditions, as seen in the early universe.

Ground State Approximation

When dealing with the chemical potentials of atoms, one may sometimes ignore excited bound states and restrict the analysis to the ground state. This approximation is justified if the temperature is such that the majority of atoms remain in the ground state, which simplifies the calculation of the partition function. The assumption is particularly valid in low density environments or when the excitation energy far exceeds the thermal energy, making excited state occupancy negligible.

Transcript

00:01 In the first item we have to estimate the fraction of hydrogen atoms in the n -equal 2 and n -equal 3 levels relative to the ground state.

00:09 So we have to evaluate n2 divided by n1 and n3 divided by n1.

00:19 We begin by n2.

00:21 This is for the first item.

00:23 N2 divided by n1 is equal to e to minus e2 minus e1.

00:35 This is divided by k times t.

00:39 But the temperature on the first item is 300 kelvin.

00:44 So k times 300.

00:47 Now we use this equation to calculate e2 and e1.

00:52 So we got n2 divided by n1 is equals to e to the minus minus 13 .6 times 1 divided by 2 squared minus 1 and this is divided by k which is 1 .38 times 10 to minus 23 times 300 but note that the energy difference we are evaluating here using this equation is in electron volts we have to convert it to jules and we do that by multiplying it by the electron charge so we multiply it by 1 .6 times 10 to minus 19 this results in e to 13 .6 times 1 over 4 minus 1 is equal to minus 3 over 4 times 1 .6 times 10 to minus 19 this is divided by 1 .38 times 10 to minus 23 times 300 and finally, to account for the degeneracy of the n equals 2 level, we have to multiply it by a factor.

02:09 So, the degeneracy of n1 is equal to 2 because there are two possible quantum states which has m equals to 1.

02:19 The degeneracy of the second level is equal to 2 times n squared.

02:25 So 2 times 2 squared, which is equal to 8.

02:28 Then we have to multiply this result by 8 divided by 2.

02:33 Therefore, the final result is equal to 8 divided by 2 times e to minus 13 .6 times 1 .6 times 10 to minus 19 divided by 4 times 1 .38 times 10 to minus 23 times 300.

02:55 And this is approximately 3 times 10 to minus 171.

03:03 Now, we evaluate the same quantity but for n3 divided by n1.

03:10 N3 divided by n1 is equal to e to minus minus 13 .6 times 1 divided by 3 squared, which is 9 minus 1 .6 times 1 .6 times 1 .6 times 1 .6 times 10 to minus 1 .6 times 1 .6 times 10 to minus 1.

03:26 6 times 1 .6 times 1 .6 times 10 to 19 to convert the energy to joules and this is divided by 1 .38 times 10 to minus 23 times 300.

03:39 This is equals to we to minus 13 .6 times 8 times 1 .6 times 10 to minus 19 divided by 1 .38 times 10 to minus 233 times times 9 because this is equal to minus 8 divided by 9.

04:03 And note that g3 is equal to 2 times 3 squared, which is 2 times 9, 18.

04:12 Therefore, we have to multiply this result by 18 divided by 2.

04:20 And 18 divided by 2 is equal to 9.

04:22 So finally, we have n3 divided by n1 being equals to 18, divided by 2 times z to minus 13 .6 times 8 times 1 .6 times 10 to minus 19 divided by 1 .38 times 10 to minus 23 times times 300.

04:48 I forgot the 300 on the previous line and this is approximately 10 to minus 202.

04:57 On the next item we have to do the same calculations but using t equals to 6 ,000 kelvin instead...

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