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410 12MAGNETIC RESONANCE
E12A.4(b) �e energies of the nuclear spin states in a magnetic �eld are given by [12A.4d–
489], EmI = −gIµNB0mI where gI is the nuclear g-factor, µN is the nuclear
magneton, B0 is the magnitude of the magnetic �eld, and the component of
the angular momentum on a speci�ed axis is mIħ where mI = I, I − 1, ...,−I.
�erefore, since the possible values ofmI are 0,±1, the energies of nuclear spin
states are
EmI = −gIµNB0mI
= −(0.404) × (5.0508 × 10−27 J T−1) × (10.50 T) ×mI
= (−2.14... × 10−26 J) ×mI
Hence E0 = 0 and E±1 = ∓2.14 × 10−26 J .
E12A.5(b) �e energy level separation is ∆E = hν where ν = γNB0/2π, [12A.6–489].
Hence, in megahertz, the frequency separation is
ν = 10−6 × γNB0
2π
= 10−6 × (1.93 × 107 T−1 s−1) × (14.4 T)
2π
= 44.2 MHz
E12A.6(b) For nuclei the energy level separation is ∆E = hν where ν = γNB0/2π, [12A.6–
489]. For the samemagnetic �eld the Larmor frequencies, andhence the energy
separations, of the two nuclei are in the ratio of their magnetogyric ratios
ν14N
ν1H
= ∆E(
14N)
∆E(1H)
= γN(14N)
γN(1H)
hence
∆E(14N) = ∆E(1H) × γN(14N)
γN(1H)
= hν1H ×
γN(14N)
γN(1H)
With the values given
∆E(14N) = (6.6261 × 10−34 J s) × (600 × 106 Hz) × (1.93 × 107 T−1 s−1)
(26.75 × 107 T−1 s−1)
= 2.87 × 10−26 J
�e energies of the electron spin states in amagnetic �eld are given by [12A.11c–
492], Ems = geµBB0ms where ge = 2.0023 is the g-value of the free electron, B0
is the magnitude of the magnetic �eld,ms = ± 12 , and µB is the Bohr magneton.
�e energy separation is therefore
∆E = E+1/2 − E−1/2 = ( 12 geµBB0) − (− 12 geµBB0) = geµBB0
= (2.0023) × (9.2740 × 10−24 J T−1) × (0.300 T)
= 5.57 × 10−24 J
Hence the energy level separation of the electron is much greater than that of
the 14N nucleus under the conditions given.