Statistical Mechanics 299
fMB()AekT (9.2)
The value of Adepends on the number of particles in the system and plays a role here
analogous to that of the normalization constant of a wave function. As usual, kis Boltz-
mann’s constant, whose value is
k1.381 10 ^23 J/K8.617 10 ^5 eV/K
Combining Eqs. (9.1) and (9.2) gives us the number n() of identical, distinguish-
able particles in an assembly at the temperature Tthat have the energy :
Maxwell-Boltzmann n()Ag()ekT (9.3)
Example 9.1
A cubic meter of atomic hydrogen at 0°C and at atmospheric pressure contains about 2.7 1025
atoms. Find the number of these atoms in their first excited states (n2) at 0°C and at 10,000°C.
Solution
(a) The constant Ain Eq. (9.3) is the same for atoms in both states, so the ratio between the
numbers of atoms in the n1 and n2 states is
e(^2 ^1 )kT
From Eq. (7.14) we know that the number of possible states that correspond to the quantum
number nis 2n^2. Thus the number of states of energy 1 is g( 1 ) 2; a 1selectron has
l 0 and ml0 but mscan be ^12 or ^12 . The number of states of energy 2 is g( 2 ) 8; a
2 s(l0) electron can have ms^12 and a 2p(l1) electron can have ml0, 1, in each
case with ms^12 . Since the ground-state energy is 1 13.6 eV, 2 1 /n^2 3.4 eV and
1 2 10.2 eV. Here T0°C 273 K, so
434
The result is
e^434 1.3 10 ^188
Thus about 1 atom in every 10^188 is in its first excited state at 0°C. With only 2.7 1025 atoms
in our sample, we can be confident that all are in their ground states. (If all the known matter
in the universe were in the form of hydrogen atoms, there would be about 10^78 of them, and if
they were at 0°C the same conclusion would still hold.)
(b) When T10,000°C 10,273 K,
11.5
and e11.54.0 10 ^5
Now the number of excited atoms is about 10^21 , a substantial number even though only a small
fraction of the total.
8
2
n( 2 )
n( 1 )
2 1
kT
8
2
n( 2 )
n( 1 )
10.2 eV
(8.617 10 ^5 eV/K)(273 K)
2 1
kT
g( 2 )
g( 1 )
n( 2 )
n( 1 )
Boltzmann’s
constant
Maxwell-Boltzmann
distribution function
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