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No measurement on a single decay reveals its angular momentum (identifying the multi-pole nature of the g ). . source. Need to take many measurements on a sample counting the occurrence of E g -decays as a function of angle. - PowerPoint PPT Presentation
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No measurement on a single decay reveals its angular momentum
(identifying the multi-pole nature of the ).
Need to take many measurements on a sample
counting the occurrence of E -decays
as a function of angle
source
We know, for fixed mj the radiation is given by the Poynting vector
)(4
1 *jmjmjm HE
cS
Re
where ijm
ijm
ijm
ijm
AH
ikAE
2
21
21
2
)]1()1(
)]1()1([
)1(2
1
mjj
mjjj
mjjj
jm
Ym
Ymmjj
Ymmjj
jjS
Same for E and M multipoles
60Co
I = 0+
60Ni GROUND
1333 keV
2506 keV
I = 2+
I = ?
60Co’s -decay to 60Ni is accompanied by two rapid gamma emissionsin succession (lifetimes of ~10-12 sec, make them seem simultaneous)
½ = 5.26 years
Obviously cascading to its ground state
1g7/2
1g9/2
2p1/2
1f 5/2
2p3/2
1f 7/2 81d3/2 42s1/2 21d5/2 6 281p1/2 21p3/2 41s1/2 2
4
60Co
I = 0+
60Ni GROUND
1333 keV
2506 keV
I = 2+
I = ?
60Co’s -decay to 60Ni is accompanied by two rapid gamma emissionsin succession (lifetimes of ~10-12 sec, make them seem simultaneous)
½ = 5.26 years
Assuming (trying) different values for I1
would lead you to expect
42 cos10cos2
15
2
5)( W
I1 = 0
4cos24
9
10
11)( W
I1 = 3
42 cos24
1cos
8
11)( W
I1 = 4
Jintial
Jfinal
mj1 = -j1, -j1+1, …, j1-1, j1
mj2 = -j2, -j2+1, …, j2-1, j2
unless special steps taken to orient nucleithe various mj states are equally populated
and mj Sjmj
= constant
Jintial = 1
Jfinal = 0
mj1 = -1, 0, +1
mj2 = 0
0 0
1 0
minitial mfinal
2sin
)cos1(2
1 2
With the three mj1 states equally populated:
)cos1(2
1
3
2 2 2sin3
13
2 Isotropic!
independentof .
Ji = 1
Jf = 0
mi = -1, 0, +1
mf = 0
mi = 0
mi = 1
mi = 1
Jf = 0 mf = 0
E+E
EEE
B-field
E = B but to compare atomic to nuclear splittinglook at one Bohr magneton and the nuclear magneton:
eN m
e
m
e
22
Howeverapplying a strong magnetic field
to a nuclear system at low temperature
can exploitthe Boltzmann distribution
kTEmem /)( PAt kT<<B
the nuclear spins are aligned with the external B-field
minitial = +1
/2 3/2 2
At T << B nuclear spins tend to be aligned with the external B
Yielding, for example for mi=+1
At high temperatures ( 1 K !) occupation of the nuclear energy levels
are equal.
Low temperature Nuclear Orientation
For this a 3He:4He dilution refrigerator is widely used.
At lower temperatures (100 mK) the lower energy levels
are preferentially occupied.
Measurement of temperatureResistance thermometry
3He Melting curve thermometryNuclear orientation thermometry
Nuclear orientation is detected by the temperature-dependent change in
the pattern of emitted radiation from appropriate nuclei.
most easily measured for the radiation that exits the cryostat walls of the cooling chamber
(for low energy and radiation the detectors need to be inside the cryostat).
For 60Co, even at 1 K radiation, the pattern of radiation
is uniform in all directions.
At lower temperatures the radiation pattern becomes distorted
An external magnetic field may be needed to sweep out magnetic domains, ensuringall of the target nuclei are correctly aligned.
This is most easily detected by a -detector aligned with the sample axis
though frequently an azimuthal detector is also used.
For work below 0.3 K3He/4He dilution refrigerator
Capable of reaching ultra low temperatures down to 10 mK.
Ultra Low Temperature Thermometry
The equilibrium concentration of 3He/4He is temperature and pressure dependent.
The vapor pressure of 3He is higher than 4He
Manipulating the 3He/4He concentration(by pumping) can control the temperature.
Similar to evaporation techniques in your refrigerator.
The production of low temperaturesEvaporation refrigeration using liquid 3He (T~0.3 to 0.5 Kelvin)
Dilution refrigerationcontinuous refrigeration to low temperatures
but low cooling power
Phase Separation Driven by osmotic pressure differences
when temperature is low enough~0.8 K
3He
4He
3He
4He
Dilution and CoolingDriving temperatures lower
T0.01 K
3He + 4He mixtureat T>0.8 K
Useful when there is a cascade of
successive radiations
source
Angular Correlation Technique
Ei-counter
Ef -counter
Ji
Jf
J1
2
After the 1st transition the orientation of atoms is no longer randomi.e., not all mj-values for the 2nd transition are equally probable!
Ji
Jf
J1
2
As a simple example, consider the special casewhere Ji = Jf = 0
with an intermediate state of J.
For the initial transition to the intermediate nuclear state mj = J, J-1, … (J-1), J
are equally likely.
Suppose 1 carries off angular momentum 1.
It must leave the intermediate nuclear state with mj = 1.??
Since mf can only = 0, it follows 2 = mj = 1
Nature’s randomly selected 1st step, fixes the nature of the 2nd step.
Ji
Jf
J1
2
For example suppose
J = 1mj = 1, 0, +1
The 1st transition emits 1
leaving the nucleus in one state of mj.
source
Ei-counter
Positioning a detector effectively fixes the z-axis, by selecting a direction!
Ef -counter
When the Ei-counter registers a hitit is selecting p1 as the z-axis.^
Any it detects can’t be froma ~sin2 distribution, only the
~ 1 + cos2.
The Ei-counter preferentially selects out the m=1 decays!
With 1 = 0, the Yjm=Yj
0 transition is not even detected. The remaining (equally likely) cases m = 1 have been selected out.
The distribution resulting from these two contributions of coincident radiation
1,1,)(
jjSSW
22
21
20
)1(
)1)(2(
)1(
2jjj
Yjj
jjY
jjY
So if Ji=0, J=1, Jf=0 (the dipole transition Krane uses as an example)
221
1
20 cos1)( YYWj
or if J=2
1cos3cos43
2
3
1)( 2422
2
21
2
20
2 YYYW