· Consider the recombination of hydrogen ions (and the ionization of hydrogen atoms) H* +e" ++ H or H – H* - e¯ = 0. Suppose that at some very low temperature, we start with a density no of pure hydrogen atoms in some fixed volume. The temperature is then raised to temperature T, at which some ionization may occur. Let K„(T) be the equilibrium constant for this reaction, and let IE noKn(T).
· Consider the recombination of hydrogen ions (and the ionization of hydrogen atoms) H* +e" ++ H or H – H* - e¯ = 0. Suppose that at some very low temperature, we start with a density no of pure hydrogen atoms in some fixed volume. The temperature is then raised to temperature T, at which some ionization may occur. Let K„(T) be the equilibrium constant for this reaction, and let IE noKn(T).
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Transcribed Image Text:. Consider the recombination of hydrogen ions (and the ionization of hydrogen atoms)
H+ +e + H
or
H – Ht - e = 0.
Suppose that at some very low temperature, we start with a density no of pure hydrogen
atoms in some fixed volume. The temperature is then raised to temperature T, at which
some ionization may occur. Let K,(T) be the equilibrium constant for this reaction, and let
* = noKn(T).
(a) Show that if r > 1, only one in every vI of the H atoms is ionized.
(b) Show that if r <1, the gas is almost entirely ionized.
(c) Show that if r = 1, ~62% of the H atoms are ionized.
(d) Treat the H atoms, and the H+ and e- ions, as classical "monatomic" ideal gases, but
let the H atoms have a constant negative potential energy -Eg (the binding energy).
Calculate K, (T) in terms of En; the masses of H, H+, and e-; T; and ho, the constant
that describes the counting of states in classical phase space.
(e) Explain why your answer to (d) shows that classical physics cannot really be valid here.
(f) Ignore (e) and plug in numerical values to get K„(T) at T = 104 K, T = 105 K, and
T = 106 K. Use Eo = 13.6 eV and ho = h (Planck's constant).
(g) Calculate the fraction of ionized H atoms at T = 10' K, T = 105 K, and T = 10 K. Use
no = 1023 /liter.
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