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SOLUTIONSMANUAL TO ACCOMPANY ATKINS' PHYSICAL CHEMISTRY 411
E12A.7(b) �e ground state has mI = + 12 (α spin) and population Nα , and the upper
state has mI = − 12 (β spin) and population Nβ . �e total population N is N =
Nα+Nβ , and the population di�erence is Nα−Nβ .�e Boltzmann distribution
gives Nβ/Nα = e−∆E/kT , where ∆E is the energy di�erence between the two
states: ∆E = γNħB0. It follows that Nβ = Nαe−∆E/kT . With these results
Nα − Nβ
N
=
Nα − Nβ
Nα + Nβ
= Nα(1 − e−∆E/kT)
Nα(1 + e−∆E/kT)
= 1 − e
−∆E/kT
1 + e−∆E/kT
Because ∆E ≪ kT the exponential e−∆E/kT is approximated as 1 − ∆E/kT to
give
Nα − Nβ
N
≈ 1 − (1 − ∆E/kT)
1 + (1 − ∆E/kT)
= ∆E/kT
2 + ∆E/kT
= ∆E
2kT
= γNħB0
2kT
For a 13C nucleus and at 298 K
Nα − Nβ
N
= γNħB0
2kT
= (6.73 × 107 T−1 s−1) × (1.0546 × 10−34 J s) × B0
2 × (1.3806 × 10−23 JK−1) × (298 K)
= 3.42... × 10−6 × (B0/T)
For B0 = 0.50 T, (Nα − Nβ)/N = 4.3 × 10−7 ; for B0 = 2.5 T, the ratio is
2.2 × 10−6 ; for B0 = 15.5 T, the ratio is 1.34 × 10−5 .
E12A.8(b) �e population di�erence for a collection ofN spin- 12 nuclei is given by [12A.8b–
491], (Nα − Nβ) ≈ NγNħB0/2kT , where Nα is the number of spins in the
lower energy state and Nβ is the number of spins in the higher energy state.
At constant magnetic �eld, (Nα − Nβ)/N ∝ T−1. Hence, for the relative
population di�erence to be increased by a factor of 5, the temperature must
decreased by this same factor – that is the temperature must by multiplied by
factor of 0.2 .
For many samples this is simply not a feasible way of increasing the sensitivity
as the solvent would freeze long before the temperature had been lowered by
this amount. High-resolution spectra require the sample to remain in a liquid
solution.
E12A.9(b) �e resonance frequency νNMR in an NMR spectrometer is given by [12A.6–
489], hνNMR = γNħB0, where γN is the nuclear magnetogyric ratio and B0 is
the magnetic �eld strength. �e EPR resonance frequency νEPR is given by
[12A.12b–492], hνEPR = geµBB0, where ge is the magnetogyric ratio of the
electron.
�e EPR resonance frequency in amagnetic �eld forwhich theNMR frequency
for 1H nuclei is νNMR is calculated by taking the ratio of the two resonance
conditions, assuming B0 is equal for both systems. Hence,
hνEPR
hνNMR
= geµBB0
γNħB0
hence νEPR
νNMR
= geµB
γNħ