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MICROCOPY RESOLUTION TEST CHART
NATIONAL BUREAU OF STANDARDS -1963
ORN P1194
Conf-65HELA
rack presented at the Meeting of Southeastern Section of the American
Physical Society, Charlottesville, Virginia, November 1~3, 1965

- MASTER
23: NN
RECOILLESS GAMMA RAY EMISSION FOLLOWING COULOMB EXCITATION*
G. Czjzek
Metals and Ceramics Division, Oak Ridge National Laboratory
Oak Ridge, Tennessee
37830

RELEASED FOR ANNOUNCEMENT
IN NUCLEAR SOLANCE ABSTRACTS
*Research sponsored by the U.S. Atomic Energy Commission under contract
with the Union Carbide corporation,
LEGAL NOTICE
This report was preparad u an account of Government sponsored work. Noither the United
Statos, por the Commission, nor any person soting on behall of the Commission:
A. Makes any warranty or ropresentation, expressed or implied, with respect to the accu-
racy, completeneII, or usotaloon of the taformation contained in this roport, or that the ne
of any information, apparatus, method, or process dieclosed in this roport may not latring
prinatoly owned rights; or
B. Asnumss any liabilues with respect to the use of, or for damages resulting from the
urs of any information, apparatus, method, or procon disolound in this report
As und in the abovo, "person noting on behall of the Commission includes way om-
ploys or contractor of the Commission, or employs of such contractor, to the extent that
such employo. or contractor of the Commission, or omploym of such contractor properts,
disnominatos, or provides acconto, any information pursuant to his employment or contract
with the commission, or hin omployment with much contractor.
Introduction
(a) Mössbauer effect
(b) Coulomb excitation
(c) Slowing-down process
wowote na
(2) Experiments performed
(a) General description of apparatus
(b) 6Ni
(c) 736
mi
V
-
veure
.
.
.
KW
.
1. Introduction
(la) Mossbauer effect
Any Y-ray emitted from a nucleus carries with it a momentum.
The conservation of momentum requires that the nucleus recoils.
This
recoil of course takes kinetic energy and so the actual energy of the
7-ray emitted differs from the energy of the nuclear transition.
In the more precise mathematical language:
0 = p + B
momentum conservation
energy conservation
-
.
-
-
•
Recoil energy: R =
2MC2
2Mc2
Ey = E. - R
Now: cross section for resonance absorpotion (Breit-Wigner formula)
0 =
0.
N.
(E-E.)2 +(1 2)/4
I is the so called natural width of the level, is connected with the
mean life time I by Heisenberg's uncertainty relation:
for = h .
|
Comparison of orders of magnitude for y-rays:
E ~ 104 - 105 er
Then: R ~ 10-2 ev
•
q10-9 - 10-7 sec, then: I ~ 10-8 - 10-6 ev
-
-
:-
-.
.
-
That is, the width of the resonance is many orders of magnitude smaller
-
than the misfit of energies as given by R. The situation is actually worse,
since for the absorption the reverse of the energy-momentum equations applies,
the gamma ray would have to give the nucleus the additional energy R.
Therefore the misfit is actually 2R. In general, no resonance absorption can be
observed.
R. Mössbauer showed in an experimental work with Iryt (1957, published
1958) that for a nucleus bound in a crystal lattice the situation is different.
There the nucleus cannot recoil.
It is bound to the lattice with an energy
of several ev. The recoil momentum is given to the entire crystal. Energy
can only be lost by the generation of lattice vibrations (phonons). The
probability for the generation of phonons is smaller than one: Short wavelength
phonons have a high energy for the generation of long-wavelength phonons with
smaller energies (w = 27Ck) the in-phase movement of several atoms is required -
again the probability for the occurrence of this is small. So there is a
finite probability that no vibrations in the crystal are excited; that is,
-
- -
no recoil energy is lost. This probability depends on the dynamics of the
lattice, that is essentially on the binding energy of the atoms in the
lattice, and on the temperature.
..
If one observes a large nuwber of events, this means that for a certain
fraction of them the emission takes place without loss of recoil energy. This
so-called recoilless fraction is given by:
* = exple to pe < >> 14?) = explainen,
The same arguments hold for the resonance absorption. The cross-section o
has to be multiplied by f.
Requirements for isotopes to be suitable for M. E. measurements:
(a) small R →E. 5 150 keV
(3) 20-10 árs 206
(a) atoms must exist in some chemical environment in solid foru
The importance of the Mössbauer effect lies in the extremely small line width.
For this reason, small changes in the energy levels caused by the interaction
between the nucleus and its surroundings can be detected. I want to list the
most important interactions:
(a) Electrostatic interaction between the nuclear charge distribution
and the electronic charge density (i.e. monopole interaction). causes
a shift of the line, given by:
S = ZR&R {\\(o)l source - lvolabal
(slides 20,21)
......
.......
...........
(B) Magnetic interaction between nuclear magnetic moment and the electronic
spins. The interaction can be expressed as being due to an effective
magnetic field at the nucleus. The m-degeneracy of the nuclear levels
is listed, a state of spin J splits into 2J + 1 levels, the distance
between the levels is given by.
4 = $ Help
Here = 2<5,+ Hext
Interaction between the quadrupole moment of the nucleus and electric
fieid gradients in the material.
This interaction also lifts the
m-degeneracy, but only in part, Levels with opposite sign of
m, but same im become not separated. This gives J + 1 levels for J integer,
J +
levels if J =(2n + 1)2. The separation between the levels is
usually written as:
8 = ega
(e:atomic charge, q: nuclear quadrupole moment, Q:field gradient)
Actually the interaction term in the Hamiltonian is the product of two tensors
of second rank, the details of the spectrum depend on the symmetry of the physical
situation,
.
E
.
.
...........................
5
other effects: temperature-dependent second order Doppler-shift proportional
to <3<>, gravitational shift (Pound and Rebka ), shift due to acceleration:
Higher multipole interaction, effects are smaller by orders of magnitude, have
never been observed.
The experimental arrangement for observing resonance absorption is schematically
shown in
SLIDE 1
Source, absorber, detector. To scan the energy spectrum: the absorber is moved
relative to the source to impose a Doppler shift. The relation between
energy and velocity is:
(
somewhat peculiar dimension m/ sec for energies used by Mössbauerists)
Usually applied velocities: 1 mm/sec < Umax < 10 cm/sec
The next
SLIDE 2
shows the experimental arrangement as we mostly use it in more detail. In this
case the source is driven in a sine wave, - all velocities between 0 and av
max
are scanned. A pickup signal is either used to modulate the pulse height, or
fed directly into the MCA
Background signals are used to compensate for the
....
..
...
I
..
.....
......................
...
inte lindo vivimitesinif
he
d
fact that the sine waves spends different times at different velocities. :
e
The next
o monte
SLIDE 3
(L. D, Roberts) shows a typical one-line Mössbauer spectrum, here of 19 Au.
Note isomer shift.
In the source: The excited state of the nucleus has to be populated by
some means. The old-fashioned (but not outdated!) method: Radioactive decay
of a parent isotope. This imposes an additionai, pure technical restriction on
the possibility to do M.E. measurements with an isotope :
(8) The excited state must be reached by some radioactive decay, and the
mean lifetime of the parent isotope must be long enough to allow source
preparation and measurement.
R. Mössbauer suggested in 1960 (first international conference on the M.E.,
Allerton Park, Illinois) to use Coulomb excitation.
The first attempt by
:
E. Cotton, et al., published 1961 in J. Phys. et le Radium 22, 538) with "Fe
failed. Since 1964, a series of papers appeared, describing experiments with various
new methods:
247Au * 23TNV
2374 237NVO
J. A. Stone and W. L. Pillinger
Savannah River Lab
P.R.L. 13, 200 (1964)
3°K(n,x)40K:
D.W. Hafemeister and E. Brooks Shera
Los Alamos National Laboratory
P.R.L. 14, 593 (1965)
39K(d, py)40K:
S.L. Ruby and R.E. Holland
Argonne National Laboratory
P.R.L. 14, 591 (1965)
Coulomb Excitation:
6Ixi:
D. Seyboth et al.
Oak Ridge National Laboratory
P.R.L. 14, 954 (1965)
57 Fe:
Y. K. Lee et al.
Johns Hopkins University
P.R.L. 14, 957 (1965)
In this talk I want to report on the method using Coulomb excitation. Before
describing the Oak Ridge experiments I want to say a few words about Coulomo
excitation as far as we have to know how it works for these experiments.
(10) Coulomb Excitation
· Definition: Transition of a nucleus into an excited state induced by the
electromagnetic interaction with another nucleus. This process occurs only at
high relative velocities between the two nuclei: In accordance with the
:
experimental procedure I will speak about the target nucleus (the one being
excited) and the projectile nucleus inducing the interaction).
The total cross section is given by:
o(XL) = (
LD -? B(XL, Jų – Jp) ERICE)
X:M, E;L: multipole order of interaction
vuru
Z, e: charge of projectile
04: initial velocity of projectile
hall distance of closest approach
op: final velocity of projectile
m: reduced mass
Zine; charge of target nucleus
b
spins of nuclear states of transition
aAE
un
where AE: energy difference between nuclear staies of transition.
B(...) reduced XL-transition probability
: an integral taken over the orbits of the projectile,
S
.
NL. VAD..
.
.
This is the particular form for a classical treatment of the projectile
Trajectory. In our case: good approximation i'or o tot. A quantum meccanical
treatment changes only f, introduces the additional parameter
2.2.62
12
Kve
(class. nu → )
i
In most cases: E2 - transition has the largest cross section.
SLIDE 4
shows the dependence of the cross section on energy for the
excitation of the 67.4 keV-level in Odri by various projectiles. For our
purpose only energies below the Coulomb barrier are useful. C. B. is the
energy for which the distance of closest approach becomes smaller than the
· range of nuclear forces, roughly a = R + Rs. Above this energy: other
nuclear reactions - y-spectra become messed up. C.B. for
Ni ~ 5 MeV for
protons, ~ 10 MeV for a-s, ~ 40 MeV for 100.
Obviously heavy ions are much better to use.
For the y-yield we also have to take into account the energy loss
of the projectiles in the target. The actual yield of a target of thickness
d is given by :
O(E) E
I = const.
ſ
Elpa) dE/ (px)
DE/a(px): stopping power of target, that is, rate of energy loss of projectile.
"thick target": Elcdt) = 0, also: range of projectiles = d
10
For a > 04: I independent of d. Results of calculation of I(a) for 40
ini
Ni in next
SLIDE 5
shows that actually a thin layer of Ni is a "thick' target. Good for M.E...
because then the line-broadening due to self-absorption in the target is
negligible.
SLIDE 6
shows the distribution of the activity in the target.
E
11
ni
(1c) Slowing down process
During the interaction with the projectile: kinetic energy is
transferred to the excited nucleus. The average kinetic energy of bi,
Ge after interaction with 100: ~ 7 MeV. Since recoil-free emission
of y-rays is only possible for nuclei bound in a crystal lattice, it is
necessary that the nuclei lose their energy in a time short compared
to the mean lifetime of the excited state.
Very little is known
about the details of the slowing down process. From the known ranges of
charged particles in solids, stopping power (= dE/(ox)) and the average
initial energy: we have estimated that the nuclei will come to rest within
about 10.-13-10-12 sec.
Another unknown quantity is the time necessary for the dissipation
of the heat generated locally in the slowing down process.
A crude theoretical consideration on the problem of heat dissipation
has been published by J. G. Mullen, Phys. Letters 15, 15 (1965).
Also, the question arises what the final position of the Coulomb
excited nucleus will be - whether it comes to rest at a regular lattice
site or in a disturbed region of the lattice.
12
P. H. Deterichs et al, have done a calculation for the probability
of replacement collisions – that is, collisions by which the excited
atom will replace another atom in its lattice position, the other one being
kicked away. Their result, that about 50–70% of the nuclei should end in
lattice positions by this process is not in disagreement with our experimental
results (phys, stat, sol. 8, 213 (1965).
It can be hoped that these experiments combining the Coulomb
excitation process and M.E. may help to clarify these questions.
Experimental work at Oak Ridge carried out by
Felix Obenshain
Dietrich Seyboth (Erlangen, Germany)
Jin Ford
G. Czjzek
(2a) General description of equipment.
Used for work on 61V1 and 73Ge.
100** ions accelerated to 25 MeV in ORNL-tandem Van de Graaff accelerator.
(Highest beam-current for this charge-state and energy)
SLIDE 7
shows the principle of the accelerator.
Next
SLIDE 8
gives schematic sketch of our chamber. Liquid No, target on solid copper
block in direct contact with N, bath. Absorber in well also at liquid Ng.
Detector:
3 in. x 3 in. NaI Harshaw integral-line detector.
SLIDE 9
Cutside view of chamber under working condition.
SLIDE 10
Detail: vibrator. Symmetric arrangement of drive-coil, springs, Absorber in tube
....
.........................
r
14.
eine
in center. Driven in sine wavé. Pickup coil. Signal directly into multi-
channel analyzer.
tion
minimaaniline transak
-
-
-
-
-
.
Attir
Yr
(20) 62v1
Parameters of the nuclear states:
----
G.S.:
J = 3/2
...
-
Tul - (0.746 + 0.007) mra
E.S.:
E = 67.4 kev
T = 7.6 * 10-9 sec
J = 5/2
HI = (0.35 + 0.06) nm (He/Hg < 0)
B(E2) = 0.00071
in nickel metal: Hefe = (–70 15 )KG.
Target: 2.8 mg/cm?
-Ni, electroplated on a copper disk.
Absorber: Disk of natural nickel, 0.020 in, thick.
Hence: We have a split spectrum both in target and absorber.
Expected Mossbauer spectrum: 12 lines both for target and absorber. Folded
spectruim: 35 unresolved lines, if Hep the same in target and absorber.
SLIDE 11
shows the observed y-spectrum and Mössbauer spectrum.
-
-
-
-
SLIDE 12
shows Mössbauer spectrum, fitted by least squares method with 35 lines.
In this fit: only four free parameters: two positions determining the
overall center of the spectrum and the splitting, one intensity and one
width. The other 101 parameters were related to these four by the theoretical
linear relations. The results were compared with the spectrum obtained
.
by H. Wegener and F. E. Obenshain (Phys. Rev. 121, 1344 (1965)) who used
the radioactive decay of "fco. The individual lines are broadened by a
meiwuweni
factor of 2; the intensity of each line is reduced by a factor of 2.
Therefore, the total absorption, defined by the area of the line (proportional
to the product of intensity and width), is unchanged. That means the recoilless
fraction is not affected by the processes occurring in the target. But
Obviously many of the nuclei are in final positions with disturbed surroundings,
either interstitial, or lattice position disturbed by radiation damage.
We observed no isomer shift.
From the splitting of the fitted lines: average value of.
Here = (74 = 10)kg.
.
17
These two results indicate that the electronic surrounding of the nuclei
is on the average the same as in an undisturbed crystal.
We have also measured the maximum absorption at zero velocity
versus irradiation time.
SLIDE 13
shows the result. No change can be seen for an irradiation time up to 100 hr.
The average particle flux during this time was ~ 5.1045 particles/ar cm?.
18
(2c) 73 Ge
The Mössbauer effect in TSGe has not been observed previously. The
reason can best be demonstrated with a look at
SLIDE 14
showing the level scheme of
Ge. There is a low-lying level at 13.5 kev
which should give a tremendous M.E. 1f the lifetime were not very long:
4.6 x 20° sec. This makes Mössbauer experiments extremely difficult.
Also the conversion coefficient for this transition is extremely high, which
makes it even worse. The next level, at 67.0 keV is not reached in any
radioactive decay, from the level at 67.4 kev, no transitions to the G.S.
have been observed, also the lifetime of this level is ~ 55 sec. But the
.
level at 67.0 keV can be populated by Coulomb excitation.
Nuclear parameters:
G.S.: J = 9/2
H = - 0.877 nm
E.S.:
E = 67.0 keV
T = (2.33 + 0.20).10-9 sec
.
J = (7/2, 9/2, 11/2)?
YS
= ?
T
ro!.
B(E2) = 0.057
.."-
Crystal: belongs to cubic system, the atomic sites have tetrahedral
symmetry
Target: 2.3 mg/cm?
Soe, evaporated on nickel disk.
Absorber: Pieces cut from single cryst
ane, assembled in
mosaic.
Thickness: 0.010 in.
SLIDE 15
shows the y-spectrum obtained by Coulomb excitation. The B(E2 )-value
is 80 times that for
ni. The counting rate in the photo peak was by a
factor ~ 10 higher than for
Ni.
SLIDE 16
shows the Mössbauer spectrum. Since the effect, was very small we had to
accumulate a million counts per data point. The continuous line is the result
of a least-squares fit with a single line of Lorentz shape.
The parameters
of this line are:
Position: (0.11 $ 0.14) mm/sec ->20 isomer shift.
Maximum absorption: (0.64 $ 0.04)%
Line wiđth: (4.4 £ 0.6) mm/sec
The recoilless fraction, calculated from the Debye-model, 00 = 360°K (at
liquid N2-teraperature) is: fth = 0.125. We assume that this f-value holds
initio conicom
imation
20
t
for the absorber. Then we deduce from the effective thickness of the
ne
es
absorber and the measured absorption: Itarget = (0.009 1 0.002) (the
passi
error comes from the statistical error in the measurement, and from
uncertainties in the effec.ve thickness: Jexe., Ore are not well known.)
in
intenderezovica
Under the same assumption for the effective absorber thickness we
Weitere
derive from the width of the line that the width of the emission spectrum
ALINA
of the target is Iterset = (1.3 + 0.4) nomi/sec.
This is in agreement with the natural width r = (1.27 $ 0.11) mm/sec
calculated from the lifetime of the state:
In contrast to
Ni the recoilless fraction in the target is reduced
by a factor greater than 10 compared to the expected value. We can give
two possible explanations for this result:
(a) The binding energy of the Coulomb excited nuclei in their final
positions can be very small – the germanium lattice has large holes compared
to the dense-packed structure of nickel. This cov.ld also be expressed as
a reduction in the effective Debye-temperature for these nuclei to about 220°K.
in
,
.
21
(B) The heat generated in the slowing down process of the Coulomb excited
nuclei may not be dissipated fast enough. The effective temperature of these
nuclei would have to be ~ 300°K to explain the experimental p-value.
We cannot exclude a third possibility: That the recoilless fraction of
the absorber is smaller than the computed value. The observed broadening
would then be due to quadrupole splitting in target or absorber or in
both, and to radiation damage effects in the target. Assuming that the temperature
of the emitting nuclei is 78°K and that the recoilless fractions in target .
and absorber are equal, we obtain a Mössbauer-Debye-temperature ex ñ 250°K.
In this connection it is o2 interest to mention a paper presented at
the APS meeting in Chicago by E. T. Ritter et al. from Johns Hopkins University...
They have observed the Mossbauer effect following Coulomb excitation of
Fe
in both metallic targets and in Fe og targets. In the first case they obtain
a recoilless fraction equal to that observed otherwise, but a reduction
by a factor of about 3 with the oxide target. On the other hand they find no change
in.the hyperfine splitting with the oxide target. From this they conclude
that the reduction in f cannot be due to a lack of heat dissipation.
The
temperature that they would have to assume to explain the reduction of f would
be higüer than the Néel-temperature of Fe og.
.
......

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25.75 :*
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ISO1ER SHIFT (mm/s.:)
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20 40
X, ATOMIC PERCENT NOBLE METAL
60 80
+ 8€ (FINITE NUCLEAR SIZE)
Enuclear + E&I (POINT NUCLEUS) + Eldnival
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