2.3 Solutions
121
deciding the model of the nucleus, that is discarding the electron–proton
hypothesis. Consider the nitrogen nucleus. The electron–proton hypothesis
implies 14 prorons+7 electrons. This means that it must have odd spin because
the total number of particles is odd (21) and Fermi statistics must be obeyed.
In the neutron–proton model the nitrogen nucleus has 7n + 7 p = 14 particles (even). Therefore Bose statistics must be obeyed. If the electronic wave
function for the molecules is symmetric it was shown that the interchange of
nuclei produces a factor (−1)
J (J = rotational quantum number) in the total
wave function of the molecule. Thus, if the nuclei obey Bose statistics symmetric nuclear spin function must be combined with even J rotational states
and antisymmetric with odd J . Because of the statistical weight attached to
spin states, the intensity of even rotational lines will be (I + 1)/I as great
as that of neighboring odd rotational lines where I is the nuclear spin. For
Fermi statistics of the nuclei the spin and rotational states combine in a manner
opposite to that stated previously, the odd rotational lines being more intense
in the ratio (I + 1)/I . The experimental ratio (I + 1)/I = 2 for even to odd
lines, giving I = 1, is consistent with the neutron–proton model.
2.60 The vibrational energy level is
E n =
n +
1
2
ω, n = 0, 1, 2 . . .
with ω =
√
(k/μ), k being the force constant and μ the reduced mass of the
oscillating atoms.
μ =
m c m 0
m c + m 0
=
12 × 16
12 + 16
= 6.857 amu
ω =
1908
6.857 × 1.67 × 10 −27
1/2
= 4.082 × 10
14 S
−1
Number of molecules in state E n is proportional to exp(−nω/kT ), k being
the Boltzmann constant and T the Kelvin temperature. The probability that
the molecule is in the first excited state is
P 1 =
exp(−ω/kT )
∞
0 exp(−nω/kT )
= exp(−ω/kT )[1 − exp(−ω/kT )]
ω
kT
=
1.054 × 10
−34
× 4.082 × 10
14
1.38 × 10 −23 × 1, 000
= 3.1177
Therefore, P 1 = exp(−3.117)[1 − exp(−3.1177)]
= 0.042
2.61 The rotational energy state is given by
E J = J
(J + 1)
2
2I
, J = 0, 1, 2 . . .
The state with quantum number J is proportional to (2J + 1) exp(−E J /kT )
The factor (2J + 1) arises from the J state.
N 0 /N 1 = (1/3) exp
2
/I o kT
121
deciding the model of the nucleus, that is discarding the electron–proton
hypothesis. Consider the nitrogen nucleus. The electron–proton hypothesis
implies 14 prorons+7 electrons. This means that it must have odd spin because
the total number of particles is odd (21) and Fermi statistics must be obeyed.
In the neutron–proton model the nitrogen nucleus has 7n + 7 p = 14 particles (even). Therefore Bose statistics must be obeyed. If the electronic wave
function for the molecules is symmetric it was shown that the interchange of
nuclei produces a factor (−1)
J (J = rotational quantum number) in the total
wave function of the molecule. Thus, if the nuclei obey Bose statistics symmetric nuclear spin function must be combined with even J rotational states
and antisymmetric with odd J . Because of the statistical weight attached to
spin states, the intensity of even rotational lines will be (I + 1)/I as great
as that of neighboring odd rotational lines where I is the nuclear spin. For
Fermi statistics of the nuclei the spin and rotational states combine in a manner
opposite to that stated previously, the odd rotational lines being more intense
in the ratio (I + 1)/I . The experimental ratio (I + 1)/I = 2 for even to odd
lines, giving I = 1, is consistent with the neutron–proton model.
2.60 The vibrational energy level is
E n =
n +
1
2
ω, n = 0, 1, 2 . . .
with ω =
√
(k/μ), k being the force constant and μ the reduced mass of the
oscillating atoms.
μ =
m c m 0
m c + m 0
=
12 × 16
12 + 16
= 6.857 amu
ω =
1908
6.857 × 1.67 × 10 −27
1/2
= 4.082 × 10
14 S
−1
Number of molecules in state E n is proportional to exp(−nω/kT ), k being
the Boltzmann constant and T the Kelvin temperature. The probability that
the molecule is in the first excited state is
P 1 =
exp(−ω/kT )
∞
0 exp(−nω/kT )
= exp(−ω/kT )[1 − exp(−ω/kT )]
ω
kT
=
1.054 × 10
−34
× 4.082 × 10
14
1.38 × 10 −23 × 1, 000
= 3.1177
Therefore, P 1 = exp(−3.117)[1 − exp(−3.1177)]
= 0.042
2.61 The rotational energy state is given by
E J = J
(J + 1)
2
2I
, J = 0, 1, 2 . . .
The state with quantum number J is proportional to (2J + 1) exp(−E J /kT )
The factor (2J + 1) arises from the J state.
N 0 /N 1 = (1/3) exp
2
/I o kT