Fission and Fusion
3224
Nuclear and Particle Physics
Ruben Saakyan
UCL
Induced fission
• Recall that for a nucleus with A240, the Coulomb
barrier is 5-6 MeV
• If a neutron with Ek  0 MeV enters 235U, it will form
236U with excitation energy of 6.5 MeV which as above
fission barrier
• To induce fission in 238U one needs a fast neutron with
Ek  1.2 MeV since the binding energy of last neutron
in 239U is only 4.8 MeV
• The differences in BE(last neutron) in even-A and
odd-A are given by pairing term in SEMF.
Fissile materials
233
92
U,
235
92
U,
239
94
Pu,
241
94
Pu
“Fissile” nuclei
232
90
Th,
238
92
U,
240
94
Pu,
242
94
Pu
“Non-Fissile” nuclei
(require an energetic neutron to induce fission)
238U
and 235U
Natural uranium: 99.3% 238U + 0.7% 235U
235U
238U
 f ~ 1eV   f  /  f  1014 s
prompt neutrons: n  2.5. In addition decay products will decay
by b-decay (t  13s) + delayed component.
235U
Fission chain reaction
• In each fission reaction large amount of energy and
secondary neutrons produced (n(235U)2.5)
• Sustained chain reaction is possible
Neutrons(n  1)
k
Neutrons(n)
• If k = 1, the process is critical (reactor)
• If k < 1, the process is subcritical (reaction dies out)
• If k > 1, the process is supercritical (nuclear bomb)
Fission chain reactions
• Neutron mean free path
 tot  c
235
tot
 (1  c)
238
tot
l
1
 nucl tot
for 2 MeV neutron l  3cm
• which neutron travels in 1.5 ns
• Consider 100% enriched 235U. For a 2 MeV neutron there
is a 18% probability to induce fission. Otherwise it will
scatter, lose energy and Pinteraction . On average it will
make ~ 6 collisions before inducing fission and will move
a net distance of 6 ×3cm 7cm in a time tp=10 ns
• After that it will be replaced with ~2.5 neutrons
Fission chain reactions
• From above one can conclude that the critical mass of
235U corresponds to a sphere of radius ~ 7cm
• However not all neutrons induce fission. Some escape
and some undergo radiative capture
• If the probability that a new neutron induces fission is
q, than each neutron leads to (nq-1) additional neutrons
in time tp
N (t   t )  N (t ) 1  ( nq  1)( t / t p ) 
In the limit  t  0,
N (t )  N (0)e
( nq 1) t / t p
dN (nq  1)

N (t )
dt
tp
Fission chain reactions
• N(t)  if nq > 1; N(t)  if nq < 1
• For 235U, N(t)  if q > 1/n  0.4 In this case since tp =
10ns explosion will occur in a ~1 ms
• For a simple sphere of 235U the critical radius (nq=1) is 
8.7 cm, critical mass  52 kg
Nuclear Reactors
Core
To increase fission probability:
1. 235U enrichment (~3%)
2. Moderator (D2O, graphite)
Delayed neutron may be a problem
To control neutron density, k = 1
retractable rods are used (Cd)
Single fission of 235U ~ 200 MeV ~ 3.210-11 j
1g of 235U could give 1 MW-day. In practice efficiency much lower
due to conventional engineering
Fast Breeder Reactor
• 20% 239Pu(n3) + 80%238U used in the core
• Fast neutrons are used to induce fission
• Pu obtained by chemical separation from spent
fuel rods
• Produces more 239Pu than consumes. Much
more efficient.
• The main problem of nuclear power industry is
radioactive waste.
– It is possible to convert long-lived isotopes into shortlived or even stable using resonance capture of
neutrons but at the moment it is too expensive
Nuclear Fusion
Two light nuclei can fuse to produce
a heavier more tightly bound nucleus
Although the energy release is smaller
than in fission, there are far greater
abundance of stable light nuclei
The practical problem:
ZZ ' e2
VC 
4 0 R  R '
1
E=kBT  T~3×1010 K
Fortunately, in practice you do not need
that much
For A  8, VC  4 MeV
The solar pp chain
pp
p+p  2H + e+ + ne + 0.42 MeV
(99.77%)
2H+p

3He
p+p+e-  2H + ne
pep
(0.23%)
+ g+ 5.49 MeV
(84.92%)
(~10-5%)
(15.08%)
3He+3He
a+2p + 12.86 MeV
3He+p
3He+a
 7Be + g
(15.07%)
7Be+e-
7Li
Overall:
hep
(0.01%)
 7Li + ne
+p  a+a
a+ e+ + ne
7Be+p  8B
7Be
+g
8B
8B
2a e+ + ne
4  1H   4 He  2e   2n e  2g  24.68 MeV
Solar neutrino spectra
Fusion Reactors
Main reactions:
Or even better:
2
1
H  12 H  23 He  n  3.27 MeV
2
1
H  12 H  13 H  p  4.03 MeV
H  13 H  24 He  n  17.62 MeV
More heat
Cross-section much larger
Drawback: there is no much tritium around
2
1
A reasonable cross-section at ~20 keV  3×108 K
The main problem is how to contain plasma at such temperatures
• Magnetic confinement
• Inertial confinement (pulsed laser beams)
Fusion reactors
Tokamak
Input of energy  4 d (3kBT / 2)
Reaction rate   d 2 
Lawson criterion
energy output  d2 tc (17.6 MeV )
L

 (1019 m3 s 1 )  d tc
energy input
6  d k BT
 d - number density of 12 H ions,  - prop. to  , tc - plasma confinement time
 d tc  1019 m 3 s
ITER
Construction to start in 2008
First plasma in 2016
20 yr of exploitation after that