Transcript One of the most practical applications of occurs with the resulting from

We’ve noted a collision reaction that produces free neutrons:
Cross Section
Total
,3n
,4n
,2n
,n
Alpha particle energy
One of the most practical applications of nuclear reactions occurs with
the compound nucleus resulting from A>230 nuclei absorbing neutrons.
Often split into two medium mass nuclear fragments
plus additional neutrons.
NUCLEAR FISSION
1930 Bothe & Becker
Studying -rays bombarding beryllium
produced a very penetrating non-ionizing form of radiation
-rays?
Irène and Frédéric Joliot-Curie
knocked protons free from paraffin targets
the proton energy range revealed
the uncharged radiation from Be
to carry 5.3 MeV
1932 James Chadwick
in discussions with Rutherford
became convinced could not be s
since assuming Compton Scattering
to be the mechanism, E>52 MeV!
Neutron
chamber
ionization
(cloud)
chamber
Replacing the paraffin with other light substances,
even beryllium, the protons were still produced.
Nature, February 27, 1932
Chadwick developed the theory explaining the phenomena
as due to a 5.3 MeV neutral particle
with mass identical to the proton
undergoing head-on collisions with nucleons in the target.
1935 Nobel Prize in Physics
9Be
has a loosely bound neutron
(1.7 MeV binding energy)
above a closed shell:
4
He Be C  n
9
5-6 MeV 
from some
other decay
12
Q=5.7 MeV
Neutrons produced by many nuclear reactions
(but can’t be steered, focused or accelerated!)
Natural sources of neutrons
Mixtures of 226Ra ( source) and 9Be 
~constant rate of neutron production
also strong  source
so often replaced by 210Po, 230Pu or 241Am
Spontaneous fission, e.g. 252Cf ( ½ = 2.65 yr)
only 3% of its decays are through fission
97% -decays
Yield is still 2.31012 neutrons/gramsec !
A possible (and observed) spontaneous fission reaction
U 2 46 Pd
238
119
92
8.5 MeV/A
7.5 MeV/A
Gains ~1 MeV per nucleon!
2119 MeV = 238 MeV
released by splitting
119Pd
238U
Atomic (chemical) processes ~few eV
Fission involves 108 as much energy as chemical reactions!
Yet
U 2 46 Pd
is a rare decay:  ½ = 1016 yr
238
119
92
not as probable as the much more common -decay
 ½ = 4.510 9 yr
From the curve of binding energy per nucleon the most
stable form of nuclear matter is as medium mass nuclei.
Consider:
M ( A, Z )  M ( A1 , Z1 )  M ( A2 , Z 2 )
The Q value (energy release) of this process is
Q  B( A1 , Z1 )  B( A2 , Z2 )  B( A, Z )
The mass differences cancel since the total number of constituents remains unchanged.
For simplicity, if we assume the protons and neutrons
divide in the same ratio as the total nucleons:
A1 / A  Z1 / Z  y1
A2 / A  Z 2 / Z  y2
y1  y2  1
The difference in binding energy comes from the
surface and coulomb terms
1 / 3
Es  as A
Ec  ac Z ( Z  1) A
2/3
so the energy released can then be expressed in terms of the
surface energy Es and the coulomb energy Ec
of the original nucleus (A,Z).
2/3
as A
 as [ y1 A]
2/3
1 / 3
ac Z ( Z  1) A
 as A
 as [ y2 A]
2/3
2/3
1 / 3
 ac y1Z ( y1Z  1)[ y1 A]
1 / 3
 ac Z ( Z  1) A
[1  y1
2/3
 y2
2/3
]
1 / 3
 ac y2 Z ( y2 Z  1)[ y2 A]
1 / 3
[1  y1 y1 y1
 Ec [1  y1
5/ 3
 y2
 y2 y2 y2
5/3
]
1 / 3
]
Expressing the energy released in terms of the
surface energy Es and the coulomb energy Ec
of the original nucleus (A,Z).
Q  ES (1  ( y1 )  ( y2 ) )  EC (1  ( y1 )  ( y2 ) )
2/3
2/3
5/3
maximum Q is found by setting dQ/dy1
5/3
=0
dQ
1 / 3
1 / 3
2 / 3
2 / 3
  23 ES y1  23 ES y2 ( 1)  53 EC y1
 53 EC y2 ( 1)  0
dy1
Note:
1 / 3
 ES [ y1
2
3
y1  y2  1
 (1  y1 )
1 / 3
dy2 / dy1  1
2 / 3
]  EC [ y1
5
3
maximum occurs when y1
 (1  y1 )
= y2 = 1/2.
Q  0.37 EC  0.26ES
2 / 3
]
Fission into two equal nuclei (symmetric fission)
produces the largest energy output or Q value
The process is exothermic (Q > 0) if Ec/Es > 0.7.
in terms of the fission parameter, x
2
( Z / A)
>0.35
x  EC /(2 ES )  ( Z / A)(2aS / aC ) 
50
2
Suggesting all nuclei with (Z2/A) > 18 (i.e. heavier than 90Zr)
should spontaneously release energy
by undergoing symmetric fission.
However
Half-life of spontaneous fission as a function of x
where
Z2 / A
x 2
( Z / A)critical
and
( Z 2 / A)critical  49
R.Vandenbosch
and
J.R.Huizenga.
Nuclear Fusion,
Academic Press,
New York, 1973.
There is
a competition between
the nuclear force
binding the nucleus together
and the
coulomb repulsion
trying to tear it apart
Induced fission as nuclear reaction
n 92 U  92 U  57 La  35 Br  2n
235
236
139
95
n 92 U  92 U 37 Rb 55 Cs  2n
235
236
93
141
suggests the absorption of the neutron (and its energy)
may induce such distortions/vibrations in the nucleus.
The surface if any arbitrary figure can be expanded as

l
R  R0 [1     lmY (, )]
l 0 m   l
m
l
If lm time-independent: permanent deformation of the nucleus
If lm time-dependent: an oscillation of the nucleus
The Spherical Harmonics Yℓ,m(,)
ℓ=0
ℓ=1
1
Y00 
4
3
Y11  
8
3
Y10 
4
sin 
15
4
2
Y21  
15
8
i
e
Y32 
2
sin 
e
2i
sin  cos 
3

Y20 
cos2 

4  2
15
Y33  
cos 
1
Y22 
ℓ=2
ℓ=3
i
e
1
 
2
1
35
4
4
1
105
4
2
Y31  
3
sin 
3i
2
sin  cos 
1
21
4
4
e
2i
 i

sin  5 cos2   1 e
5

Y30 
cos3 

4  2
7
e
3

 cos  
2

ℓ=0
z

Nuclear Charge Density
R ~ 1  sin 
ℓ=1
R ~ 1  cos 
Lowest order to be considered:
ℓ=2
quadrupole deformation
For which we write the nuclear radius
2
R  R0 [1    2 mY2 (, )]
m
m  2
The l=2, m=0 mode:
1/ 2
5 

R (t )  R0 [1   20  
 16  
2
( 3 cos   1)]
Z

1/ 2
5 

R (t )  R0 [1   20  
 16  
2
( 3 cos   1)]
Nuclei do show spectra for such vibrational modes
Example of a vibrational spectrum (levels denoted by the number of phonons, N)
O.Nathan and S.G.Nilsson, Alpha- Beta- and Gamma-Ray Spectroscopy,
Vol.1, (K. Siegbahn, ed.) North Holland, Amsterdam, 1965.
We can approximate any small elongation from a spherical shape by
semi-major axis
a  R0 (1  )
semi-minor axis b
The semi-empirical mass formula
1
 R0 (1  2 )
B  aV A  aS A  asym (Z  N ) A  aC Z A
2
B
(Z  N )
1 / 3
2
4 / 3
 aV  aS A  asym
 aC Z A
2
A
A
2/3
From which:
2
1
ES  aS A (1    )
1 / 3
2
5
2
EC  aC Z A (1    )
2
1 / 3
1
5
2
1 / 3
2
surface of spheroid
2

b
1  e 

2
 2a  ln

e  1  e 


e  1  (b / a )
2
2
With the surface energy (strong nuclear binding force) proportional to area
E  EC  ES
 aC Z A
2
1 / 3
1
(5
 )  aS A
2
2/3
Coulomb force
deforming nucleus
where
  [aC Z A
> 0
Notice
(so the Coulomb
force wins out) for:
1
5
2
2
1/ 3
 )
2
surface tension
holding spherical
shape
which we can write in the form
E  
2
(5
2
 2aS A
2/3
]
Same fission parameter
Z
2 aS
when

 49. introduced
estimating available Q
A aC
in symmetric fission
E  
2
comes from considering small perturbations from a sphere.

V(r)
r
for small r
As long as these disturbances are slight, the
Separation, r, of distinct fragments linearly follows
r  2
  r 
V (r)  Q 
2
4  R0 
At zero separation the potential
just equals the release energy Q
separation r
For Z2/A<49,
 is negative.
While for large r, after the fragments have been scissioned

V(r)
r
for small r
r
r
Z1 Z 2 e
V (r ) 
r
for large r
separation r
2
For such quadrupole
distortions the figure
shows the energy of
deformation (as a factor
of the original sphere’s
surface energy Es)
plotted against 
for different values of
the fission parameter x.
When x > 1
(Z2/A>49)
the nuclei are
completely unstable
to such distortions.
Z2/A=49
such unstable states
decay in characteristic
nuclear times ~10-22 sec
Z2/A=36
Tunneling does allow spontaneous
fission, but it must compete with
other decay mechanisms (-decay)
The potential energy V(r) = constant-B
as a function of the separation, r, between fragments.
No stable states
with Z2/A>49!
Tunneling
probability
drops as
Z2/A drops
(half-life
increases).
At smaller values of x, fission by barrier penetration can occur,
However recall that the transmission factor (e.g., for -decay) is

X e
where
2
 
h
2m[V (r )  E]dr
while for  particles (m~4u)
this gave reasonable, observable
probabilities for tunneling/decay
for the masses of the nuclear fragments we’re talking about,
 can become huge and X negligible.
Neutron absorption by heavy nuclei can create
a compound nucleus in an excited state
above the activation energy barrier.
As we have seen, compound nuclei have many final states into which they can decay:
n 92 U  92 U * 55 Cs 37 Rb  2n
235
236
141
93
n 92 U  92 U * 57 La 35 Br  2n
235
236
139
95
n 92 U  92 U *.  54 Xe38 Sr  2n
235
236
139
95
..
in general:
n 92 U  92 U *Z 1 X  Z 2Y  n
235
236
A1
A2
where Z1+Z2=92, A1+A2+=236
PROMPT
NEUTRONS
Experimentally find the average A1/A2 peaks at 3/2
Thermal neutrons
E< 1 eV
Slow neutrons
E ~ 1 keV
Fast neutrons
E ~ 100 keV – 10 MeV
“Thermal neutrons”
(slowed by interactions
with any material they
pass through) have been
demonstrated to be
particularly effective.
Cross section 
The incident neutron itself need not be of high energy.
Typical
of decay
Products
& nuclear
reactions
incident particle velocity, v
This merely reflects the general ~1/v behavior
we have noted for all cross sections!
At such low excitation there may be barely enough available
energy to drive the two fragments of the nucleus apart.
Division can only proceed
if as much binding energy as possible
is transformed into the kinetic energy separating them out.
(so MOST of the available Q goes into the kinetic energy of the fragments!)
Thus the individual nucleons
settle into the lowest possible energy configurations
involving the most tightly bound final states.
There is a strong tendency to produce a heavy fragment of
A ~ 140 (with double magic numbers N = 82 and Z = 50).
A possible (and observed) spontaneous fission reaction
U 2 46 Pd
238
119
92
8.5 MeV/A
7.5 MeV/A
Gains ~1 MeV per nucleon!
2119 MeV = 238 MeV
released by splitting
119Pd
238U
238 MeV represented an estimate of the maximum available energy
for symmetric fission.
For the observed
distribution
of final states
the typical average is
~200 MeV per fission.
This 200 MeV is distributed approximately as:
Fragment kinetic energy
Prompt neutrons
Prompt gamma rays
Radioactive decay fragments
165 MeV
5 MeV
7 MeV
25 MeV
235U
Isobars off the valley of stability
(dark squares on preceding slide)
b-decay to a more stable state.
 and b decays can leave a daughter in an excited nuclear state
1/2
187W
2
b
b
198Au
0.68610
0.61890
b
1.088 MeV
b
0.20625
0.412 MeV
0.13425
5/2
187Re
0
198Hg
n 92 U  92 U * 56 Ba36 Kr  3n
235
236
143
90
With the fission fragments radioactive, a decay sequence to stable nuclei must follow
Ba

Kr

143
56
90
36
La  e  
143

Ce  e  
58
143

Pr  e  
59
143

Nd  e  
59
90

Rb  e  
37
90

Sr  e  
38
90

Y e 
39
90

Zr  e  
40
143

57
n 92 U  92 U * 60 Nd  40 Zr  8e  8  3n
235
236
143
90
n 92 U  92 U * 55 Cs 37 Rb  2n
235
236
141
93
With the fission fragments radioactive, a decay sequence to stable nuclei must follow
141
55
Cs

b,
25 sec
0.03%
141
56
Ba
Cs  n 
b,
65 sec
140
55
93
37
Rb

b,
18 min

b,
6 sec
1.40%
92
37
93
38
Sr

7 min
b,
Rb  n 
b,
5 sec

b,
140
La
Ba
13 d
57
b,
140
56

141
4 hr
141
93
Y
39

b,
92
38
Sr
10 hr

3 hr
b,
58
57
Ce
La
93
40
Zr
92
Y
39

b,
141

b,
140
33 d
40 h

b,
106 yr

b,
4h
59
58
Pr
Ce
93
Nb
41
92
40
Zr
n 92 U  92 U * 58 Ce 59 Pr  8e  8  8  2n
sometimes  3n or  4n
235
236
140
141
For 235U fission, average number of prompt neutrons ~ 2.5
n 92 U  92 U * 56 Cs 36 Kr  3n
235
236
143
90
n 92 U  92 U * 55 Cs 37 Rb  2n
235
236
141
93
n 92 U  92 U * 57 La 35 Br  2n
235
236
139
95
n 92 U  92 U * 54 Xe38 Sr  2n
235
236
139
95
with a small number of additional delayed neutrons.
with every neutron freed comes the possibility of additional fission events
This avalanche is the chain reaction.
235U
will fission (n,f)
at all energies of the absorbed neutron.
It is a FISSILE material.
However such a reaction cannot occur in
natural uranium (0.7% 235U, 99.3% 238U)
Total (t) and fission (f) cross sections of 235U.
1 b = 10-24 cm2