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.

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) atomswhich can absorb them to become excited themselves.

NTQE

ppN fi EEQ

Nm

pQE N

2

2

2

2

2 cm

EQ

N

If this change is large enough, the will not be absorbed by an identical nucleus. 2

2

2 cm

pQE

N

emitted

In fact, for absorption, actually need to exceed the step between energylevels by enough to provide the nucleus with the needed recoil:

p=E/cTN =

pN2

2mN

= p2

2mN

The photon energy is mismatched by

2

2

2 cm

EQE

N

absorbed

2

2

2

2

22

cm

E

cm

E

NN

For atomic resonance experiments, note Q ~ few eV (for visible transitions) and mN ~ Au ~ A1000 MeV

MeV

eVTN 1000)10010(

)101( 2

1012 – 1010eV

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 ~108 sec

The energy uncertainty eV10 7 E

Notice the uncertainty E >> 2TN

Q

2TN

with an enormous amount of overlap allowing resonance fluorescence

For NUCLEAR resonance experiments

Q ~ few MeV (for -ray emissions)

with mN ~ Au ~ A1000 MeV

MeV

MeVkeVTN 1000)10010(

)10100( 2

0.1 – 104eV

For nuclear transitions, the typical lifetime is ~1010 sec

The energy uncertainty eV10 5 E

This time the uncertainty E << 2TN

Q

2TN

which provides no overlap allowing resonance fluorescence

57Co7/2

5/2

EC

=270d

57Fe1/23/2 14.4keV

136keV

=10-7s

As an example consider the distinctive 14.4 keV from 57Fe.

The recoil energy of the iron-57 nucleus is

this is 5 orders of magnitude greater than the natural linewidth of the iron transition which produced the photon!

eVGeV

keV

cm

EE

N

recoil

002.0)022.53(2

)4.14(

22

2

2

With = 107 s, =108 eV

~90% of the 57Fe* decays are through this intermediate level produce 14.4 keV s.

1958 Rudolf MössbauerWorking with 129-keV ray of 191Ir

Discovered by imbedding theradioactive samples in crystals, and cooling them, their tightly held crystal positionsprevented 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?

)022.53(2

)4.14(10

28

GeVN

keVeVErecoil

000,200N

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

c

csourceobserved /1

/1

v

v

v is positive

for an approaching

source

)1(

)/1(

)/1)(/1(

)/1)(/1(22 c

cv

cc

ccvv s

so v

v

vv

vv

)/1( cvvso

vIf v/c <<1 this simplifies to

This can be written as

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

                                               

   

                                       This was in fact demonstrated with the source in a centrifuge

m/sec 42 v

vv

ckeV

chvvvheV

ss4.14)(002.0

0

Cool an embedded sample to produce recoilless emissionand drive the source or absorber to scan the resonance.

vibrator

servo-motorcontrols

dataacquisition

radioactivesource

absorbingsample detector

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:

cvkeVeV /4.1410 8

gives:

0.0002 meter/sec = 2 mm/sec

Setting

Mossbauer Absorption of 191Ir129-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.

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.

Source

Absorber

Detector

191Irv

191Ir

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 SpectroscopyObserved Effect Illustration Observed Spectrum

 

Isomer ShiftInteraction 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 SplittingInteraction of the nuclear electric quadrupole moment with the EFG and the nucleus

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              , so its total energy E, as measured by an observer at the foot of the tower is

ghv 2

).(

)(2

1

42

422

O

O

mghmc

mvmcEbottom

Let the particle rebound elastically at the bottom and return.

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.

).(

42

2

OmghmcE

mcE

bottom

top

Should mtop=mbottom? Or is the mass now greater than it began?

What must be true, even for the photon is

.)(

42

2

Omghmc

mc

E

E

bottom

top

)(

42

2

Omghmc

mc

E

E

bottom

top

242 /)(/1

1

mccghE

E

bottom

top

O

.1 2c

gh

E

E

bottom

top

Since for photons we have Etop = htop

2/1 cghvbottomtop

This implies a photon climbing in the earths gravitational field

will lose energy and consequently be redshifted.

                                                                           

2

2

2

/

]/11[

/1

cghv

cghv

cghvvv

bottom

bottom

bottombottomtopbottom

The redshift is:

2/ cghv

v

bottom

topbottom

In just 22.6 meters, the fractional redshift

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.

2

0/1 cgh

The gain in energy for a photon which falls distance h = 22.6 m is

Comparing the energy shifts on the upward and downward paths gives a predicted difference

mgc

keVgh

c

EmghE 6.22

4.1422

eVE 11105.3

1511

109.44.14

)105.3(2

keV

eV

E

E

E

E

updown

1510)5.01.5(

updown E

E

E

EThe measured difference was