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422 12MAGNETIC RESONANCE
magnetogyric ratio and, all other things being equal, the relaxation caused by
these �elds will therefore be faster the greater the magnetogyric ratio.
�e magnetogyric ratio for 1H is about four times that of 13C, hence the inter-
actions between 1H and local �elds is stronger leading to faster relaxation and
shorter relaxation times. A somewhat separate issue is that in typical molecules
H atoms tend to bemore accessible thanC atoms, and so if the local �eld is from
outside themolecule it may be felt more strongly by themore exposedH atoms.
D12C.4 �is is discussed in Section 12C.4 on page 516.
Solutions to exercises
E12C.1(b) From the discussion in Section 12C.1(a) on page 510, the �ip angle ϕ of a pulse
of duration ∆τ is given by ϕ = γNB1∆τ, where B1 is the strength of the applied
�eld. It follows that B1 = ϕ/γN∆τ. With the data given
B1 =
ϕ
γN∆τ
= π/2
(26.752 × 107 T−1 s−1) × (5 × 10−6 s)
= 1.2 mT
�e corresponding 180○ pulse has twice the �ip angle so, for the same B1, ∆τ
is doubled: ∆τ = 2 × (5 µs) = 10 µs .
E12C.2(b) �e e�ective transverse relaxation time, T∗2 , is given by [12C.6–514], T∗2 =
1/π∆ν1/2, where ∆ν1/2 is the width of the (assumed) Lorentzian signal mea-
sured at half the peak height. Hence
T∗2 = 1
π∆ν1/2
= 1
π × (12 Hz)
= 0.027 s
E12C.3(b) �e amplitude of the free induction decay, S(t), is proportional to e−t/T2 , where
t is time and T2 is the transverse relaxation time. If S(t2) = 1
2 S(t1) then
S(t2)
S(t1)
= 1
2
= e
−t2/T2
e−t1/T2
= e−(t2−t1)/T2 = e−∆t/T2
taking logarithms gives
ln 12 = −∆t/T2 therefore ∆t = −T2 ln 12 = T2 ln 2
With the data given
∆t = (0.050 s) × ln 2 = 0.035 s