Resonance fluorescence absorption (& re-emission)
of
s emitted by nuclei of the same type
Radiation from a sample of excited atoms
illuminates a collection of identical (ground state) atoms
which can absorb them to become excited themselves.
When an nucleus releases the transition energy Q (say 14.4 keV)
in a -decay, the  does not carry the full 14.4 keV.
Conservation of momentum requires the nucleus recoil.
pN  p
Q  Ei  E f
E  Q  TN
2
E2
p
Q
E  Q 
2mN c 2
2mN
N
If this change is large enough, the  will
not be absorbed by an identical nucleus.
E
emitted
Q
p2
2mN c 2
In fact, for absorption, actually need to exceed the step between energy
levels by enough to provide the nucleus with the needed recoil:
2
2
p
p

N
TN =
=
2mN 2mN
p=E /c
E
absorbed
Q
E2
2mN c 2
The photon energy is mismatched by
2
E2
2mN c
2

E2
mN c 2
For atomic resonance experiments, note
Q ~ few eV (for visible transitions)
and mN ~ Au ~ A1000 MeV
(1  10eV )
TN 
 1012 – 1010eV
(10  100)1000 MeV
2
But how precisely fixed is the emitted energy anyway?
Recall:



there is a “natural width” to the energy, related
to how stable the initial energy state was.
For atomic transitions, the typical lifetime is ~108 sec
The energy uncertainty
  E  10 eV
7
Notice the uncertainty E >> 2TN
2TN

Q
with an enormous amount of overlap allowing resonance fluorescence
For NUCLEAR resonance experiments
Q ~ few MeV (for -ray emissions)
with mN ~ Au ~ A1000 MeV
(100keV  10MeV )
TN 
 0.1 – 104eV
(10  100)1000 MeV
2
For nuclear transitions, the typical lifetime is ~1010 sec
The energy uncertainty
  E  10 eV
5
This time the uncertainty E << 2TN
2TN

Q
which provides no overlap allowing resonance fluorescence
As an example consider the distinctive 14.4 keV
 from 57Fe.
~90% of the 57Fe* decays are through this intermediate level produce 14.4 keV s.
=270d
57Co
7/2
The recoil energy of the iron-57 nucleus
is
EC
5/2 136keV
Erecoil
=10-7s
57Fe
3/2 14.4keV
1/2
E2

2 mN c 2
(14.4keV ) 2

 0.002eV
2(53.022GeV )
With  = 107 s,  =108 eV
this is 5 orders of magnitude greater than the natural linewidth
of the iron transition which produced the photon!
1958 Rudolf Mössbauer
Working with 129-keV  ray of 191Ir
Discovered by imbedding the
radioactive samples in crystals,
and cooling them,
their tightly held crystal positions
prevented them from recoiling.
The energy of recoil had been
absorbed by the lattice as a whole.
To keep the  within its natural linewidth how many iron nuclei
would have to recoil together in our example of 57Fe?
2
Erecoil
(14.4keV )
 10 eV 
2( N  53.022GeV )
8
N  200,000
Very small compared to Avogadro's number!
(In fact a speck too small to be seen in a microscope).
Any tiny crystal within a 57Cobalt-containing piece of iron would meet
the conditions for resonance absorption if cooled sufficiently.
You can also destroy that resonance by moving the source relative
to the absorber and Doppler shifting the photons off resonance.
The Doppler shift of a photon is a relativistic shift given by
 observed   source
1 v / c
1 v / c
v is positive
for an
approaching
source
This can be written as
vo  vs
(1  v / c)(1  v / c) vs (1  v / c)

2
2
(1  v / c)(1  v / c)
(1  v c )
If v/c <<1 this simplifies to
vo  vs (1  v / c)
shift recoiling emissions to resonance by moving
the source relative to the absorber and Doppler shifting
the photons to the necessary energy for absorption.
Continuing our 57Fe example: The source velocity necessary to shift the
photon to resonance absorption energy is
v
v
0.002eV  h ( v0  vs )  hvs  14.4keV  
c
c
v  42 m/sec
This was in fact demonstrated with the source in a centrifuge
Cool an embedded sample to produce recoilless emission
and drive the source or absorber to scan the resonance.
radioactive
source
absorbing
sample
detector
vibrator
servo-motor
controls
data
acquisition
Continuing with our example of 57Fe :
Using the uncertainty in energy given by 
as a measure of how far you need to
Doppler shift frequencies to be off resonance:
Setting
  108 eV  14.4keV v / c 
gives:
0.0002 meter/sec = 2 mm/sec
Mossbauer Absorption of 191Ir
129-keV gamma rays from iridium-191 were measured as a function of source velocity.
A velocity of only about 1.5 cm/s was enough to drop the absorption to half its peak value.
Sample and absorber were cooled to 88K.
191Ir
v
191Ir
Detector
Source
Absorber
A half-width of only about 0.65 x 10-5 electron volts makes this absorption an extremely
sensitive test of any influence which would shift the frequency. It is sensitive enough to
measure the Zeeman splittings from the magnetic field of the nucleus.
The incredibly high resolution of the Mössbauer effect in 57Fe
makes possible the measurement of the nuclear Zeeman effect .
O. C. Kistmer and A. W. Sunyar, Physical Review Letters, 4, 412(1960).
The splittings are 11 orders of magnitude smaller than the nuclear transition energy!
Nuclear Hyperfine Interactions Observable with Mossbauer Spectroscopy
Observed Effect
Isomer Shift
Interaction of the nuclear charge
distribution with the electron cloud
surrounding the nuclei in both the
absorber and source
Zeeman Effect(Dipole Interaction)
Interaction of the nuclear magnetic
dipole moment with the external
applied magnetic field on the
nucleus.
Quadrupole Splitting
Interaction of the nuclear electric
quadrupole moment with the EFG
and the nucleus
Illustration
Observed Spectrum
http://www.fastcomtec.com/fwww/moss.htm
http://www.cryoindustries.com/moss.htm
http://www.dwiarda.com/scientific/Moessbauer.html
Gravitational redshift
A ``gedänken'' experiment first suggested by Einstein:
A particle of rest mass m is dropped
to fall freely with an acceleration g
from a tower of height h.
It reaches the ground with a velocity v  2 gh ,
so its total energy E, as measured by an observer
at the foot of the tower is
1 2
4
Ebottom  mc  mv  O ( )
2
2
4
 mc  mgh  O ( ).
2
Let the particle rebound elastically at the bottom
and return.
Etop
 mc
2
Ebottom  mc  mgh  O ( ).
2
4
Suppose the rebounding particle could be converted to a photon of
energy Ebottom & upon its arrival at the top changed back into a
particle of rest mass m = E/c2.
Should mtop=mbottom?
Or is the mass now
greater than it began?
What must be true, even for the photon is
Etop
Ebottom
2
mc
 2
.
4
mc  mgh  O ( )
Etop
Ebottom
mc
 2
4
mc  mgh  O ( )
Etop
Ebottom
1

2
4
2
1  gh / c  O ( ) / mc
2
Etop
Ebottom
gh
1 2 .
c
Since for photons we have Etop = htop
top  vbottom1  gh / c
2

This implies a photon climbing in the earths gravitational field
will lose energy
and consequently be redshifted.
vbottom   top  vbottom  vbottom1  gh / c
 vbottom[1  1  gh / c ]
2
 vbottomgh / c
2
2

The redshift is:
vbottom   top
2
 gh / c
vbottom
In just 22.6 meters, the fractional
redshift
   0 1  gh / c
2

is only 4.92  10-15 but using the
Mössbauer effect on the 14.4 keV
gamma ray from 57Fe should
provide high enough resolution to
detect that difference!
In the early 60's physicists Pound,
Rebka,and Snyder at the Jefferson
Physical Laboratory at Harvard
measured the shift to within 1%
of this predicted shift.
Pound, R. V. and Rebka, G. A. Jr. "Gravitational Red-Shift in Nuclear Resonance." Phys. Rev. Lett. 3, 439-441, 1959.
The gain in energy for a photon which falls distance h = 22.6 m is
E
14.4keV
E  mgh  2 gh 
g  22.6m
2
c
c
E  3.5 10 eV
11
Comparing the energy shifts on the upward and downward paths gives a
predicted difference
2(3.5  10 eV )
 E 
 E 
15
 4.9  10

 
 
14.4keV
 E down  E up
11
The measured difference was
 E 
 E 
15

 
  (5.1  0.5)  10
 E down  E up