E
R= P
2
/2m = E
2
/2MC
2
= (E
t-E
R)
2
/2MC
2
≈ E
t/2MC
2
Since E
Ris small
•IfMincreasestherecoilenergycanbestill
lower.Sothesourceandtheabsorberare
fixedonalargerlatticetoincreasemass
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What happens because of recoil effect ?
9SANTHANAM SCSVMV
E
ϒ E
ϒ
E
R
E
R
E
tE
t
E.S E.S
G.S G.S
Emission [Eγ= E
t–E
R] Absorption [Eγ= E
t+ E
R]
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Why recording MB spectrum is difficult?
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What is there in X axis? λor νor E?
•Itisalreadyknownthat,forspectroscopic
techniqueweneedamonochromatic
radiation.
•Butgammaraysarecomingoutbecauseof
theenergydifferencebetweenthenuclear
levels.
•Wecannotalterit,tochangethegamma
photon’senergy.
•SoweuseDopplereffecttochangethe
energyofthephoton.
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Doppler Effect
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Tuning the energy by using Doppler Effect
Source moves towards absorber, νincreases
Source moves away from absorber, νdecreases 14SANTHANAM SCSVMV
57
Co
270
days
57
Fe*
127.90 k.eV/ 91%
57
Fe* (99.3 nS)
14.41 k.eV
57
Fe
136.32 k.eV /
9 %
Why Fe is the most studied?
20SANTHANAM SCSVMV
Why Fe is the most studied?
•Theprecursorof
57
Feis
57
Cowhichdecaysto
57
Fe*withahalflifeof270days(highly
stableprecursor)
•9%of
57
Fe*decaystogroundstatedirectly
withemissionofϒphotonofenergy136.32
keV.
•91%of
57
Fe*decaystoanotherexcitedstate
withemissionof121.91keVenergy
21SANTHANAM SCSVMV
•The lower excited state is having a life time of 99.3nS
(more stable)
•This decays to ground state with emission of 14.41
keV.
•This transition satisfies all the conditions for MB
spectra except 2
nd
condition.
•But it is compensated by larger absorption cross
section.
•Other elements that can be studied-are
119
Sn,
121
Sb,
125
Te,
129
I,
129
Xe and
197
Au
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Isomer shift
•The change in electrostatic energy on decay is
given by (chemical shift / Isomer shift)
δ= (ε
0/5) (Ze
2
R
2
)(ΔR/R)[|ψ
s(abs)|
2
-|ψ
s(source)|
2
]
where
ε
0 –Permittivity of free space
Z -atomic number of the nucleus
e -electronic charge
ψ
s(abs) -s orbital wave function of absorber
ψ
s(source)-s orbital wave function of source
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SPHERICAL Q IS 0 PROLATE Q IS +IVEOBLATE Q IS -IVE
41SANTHANAM SCSVMV
Oblate and Prolate nuclei
42SANTHANAM SCSVMV
•In the case of an isotope with a I=3/2 excited
state, such as
57
Fe or
119
Sn, the excited state
is split into two sub states m
I=±1/2 and
m
I=±3/2.
•This gives a two line spectrum or 'doublet'.
•The magnitude of splitting, Delta, is related
to the nuclear quadrupole moment, Q, and
the principle component of the EFG, V
zz, by
the relation Δ=eQV
zz/2
43SANTHANAM SCSVMV
Quadruple Splitting for
57
Fe and
119
Sn
44SANTHANAM SCSVMV
5/2
3/2
1/2
7/2
5/2
3/2
1/2
I
e= 5/2
I
g= 7/2
Selection Rule
for MB spectra
m
I= 0,1
45SANTHANAM SCSVMV
Stereochemical activity of lone pair
•Consider Te(IV) compounds with six ligands
•If the lone pair of electron of Te is
stereochemically active , it will occupy one
vertex of a pentagonal bipyramide(pbp)
•This will create an efg at the nucleus,
resulting in q.s
•But no q.s in these class of compounds.
•So the lone pair is stereochemically inactive
•It occupies the 5s orbital
48SANTHANAM SCSVMV
Magnetic interactions
•:Inthepresenceofamagneticfieldthe
nuclearspinmomentexperiencesadipolar
interactionwiththemagneticfieldieZeeman
splitting
•Therearemanysourcesofmagneticfields
thatcanbeexperiencedbythenucleus.
•Thetotaleffectivemagneticfieldatthe
nucleus,B
effisgivenby
B
eff= (B
contact+ B
orbital+ B
dipolar) + B
applied
50SANTHANAM SCSVMV
•Thefirstthreetermsbeingduetotheatom's
ownpartiallyfilledelectronshells.
•B
contactisduetothespinonthoseelectrons
polarisingthespindensityatthenucleus
•B
orbitalisduetotheorbitalmomentonthose
electrons,andB
dipolaristhedipolarfielddue
tothespinofthoseelectrons.
Energy of the nuclear levels E
m= -gμ
NBm
I
51SANTHANAM SCSVMV