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SOLUTIONSMANUAL TO ACCOMPANY ATKINS' PHYSICAL CHEMISTRY 285
P8C.12 For a ‘red star’, with a surface temperature in the range 3 000 K to 4 000 K, the
thermal energy is insu�cient to cause the excitation of the electron in hydrogen
from its ground state: no emission spectra from hydrogen is therefore seen. In
contrast, for a ‘blue star’, with a surface temperature in the range 15 000 K to
20 000 K, the thermal energy is su�cient to cause ionization of the electron.
�erefore, neither absorption not emission from hydrogen is seen simply be-
cause there are no hydrogen atoms present – they have all been ionized. If
the with a surface temperature is in the range 8 000 K to 10 000 K the thermal
energy is su�cient to cause excitation of hydrogen atoms, but not su�cient to
cause extensive ionization. Intense hydrogen emission lines are seen from such
stars.
�is explanation can be explored in amore quantitative way be examining how
the frequency spectrum of black body radiation changes with temperature,
and examining what part of the radiation is at high enough frequencies to
cause excitation or ionization.�e Planck distribution is given by [11A.5–420],
ρ(ν) = 8πhν3c−3(ehν/kT − 1)−1, and in Fig. 8.1 this is plotted against hν/I,
where I is the ionization energy of hydrogen, for three di�erent temperatures.
At 25000 K a signi�cant fraction of the radiation is at frequencies above that
needed to ionize hydrogen (hν/I > 1).
0.0 0.5 1.0 1.5 2.0
0.000
0.002
0.004
0.006
hν/I
ρ(
ν)
h2
c3
/8
πI
3
25 000 K
20 000 K
15 000 K
Figure 8.1
Answers to integrated activities
I8.2 (a) �e energy levels of a Hydrogen atom (with Z = 1) are given by [8A.13–
308], En = −hcR̃Hn−2, where R̃H = 109 677 cm−1 is the Rydberg constant
for the Hydrogen atom. Hence, the separation between between the en-
ergy levels n and n + 1 is
∆E = En+1 − En = hcR̃H ( 1
n2
− 1
(n + 1)2
) = hcR̃H
(n + 1)2 − n2
n2(n + 1)2
= hcR̃H(2n + 1)/[n2(n + 1)2] ≈ 2hcR̃H/n3