what temperature would provide a mean kinetic energy of 0.5 mev? by comparison, the temperature of...
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What temperature would provide a mean kinetic energy of 0.5 MeV?
MeVTeV/K.kT 5.0)106178( 5
2
3
2
3
KT 9109.3
By comparison, the temperature of the surface
of the sun 6000 K.
Interior Zones of the Sun
6,000 KConvective Zone < 0.01 g/cm3
500,000 K
Radiative Zone < 0.01-10 g/cm3
8,000,000 K
Core < 10-160 g/cm3
15,000,000 K
xkxk DeCex 22)(II
E
I II III
V
)(2
22 EVm
k
where
In simple 1-dimensional case an exponential decay connects the wave functions either side of the barrier
The Nuclear pp cycle
4 protons 4He + 6+ 2e + 2p 26.7 MeV
The effective kinetic energy spectrum of nuclei in a gas (at thermal equilibrium) is given by the high energy part
of the Maxwell-Boltzmann distribution
kTEeEN /~)(
kTmv
xxevN kT
m 2/22/1
2)(
vx
vy
vzAnd for each value of energyE = ½ mv2 = ½ m(vx
2 + vy2 + vz
2 )
Notice for any fixed E, m this definesa sphere of velocity points all which give the same kinetic energy.
The number of “states” accessible by that energy are within the infinitesimal volume (a shell a thickness dv on that sphere).
dV = 4v2dv
(1.7)
Maxwell Boltzmann distribution
kTmvevvf kT
m 2/222/3
24)(
mkTdvvfv
8
0
)(
The probability distribution
With a root mean square speed of
While the cross-section will have an energy dependence dominated by
the barrier penetration probability
eE)(
0
2
21
4
21
eZZ
E
m
where
the details follow…
222222II |)(| rkeDrx
E
I II III
V
probability of tunneling to here
x = r1 x = r2
In simple 1-dimensional case an exponential decay connects the wave functions either side of the barrier
R
E
Where this time we’re tunneling in with an energy from r2 where:
2
2
21
04
1
r
eZZE
r2
r
eZZrV
2
21
04
1)(
Er
rrV 2)( which we can just write as
minimum energy to reach the barrier
hence
drr
rmT
drErVm
12
2
)(2
2
22
2
22 cos/ rr
drdr cossin2 2
1tan 2 r
r
then with the substitutions:
with E=T becomes
drmE
drmE
2
22
22
sin22
2
)cossin2)((tan2
2
222
1
0
2
212
/1//cos4
22 rRrRrR
E
eZZmE
and for R « r2 the term in the square brackets reduces to
2
Performing the integral yields:
24
22
0
2
212
eZZ
E
m
v
ZZeeZZ
v21
0
2
0
2
212 24
4
0
2
21
4
21
eZZ
E
m
with E = ½ mv2 can also write as
eE ~)(
0
2
21
4
21
eZZ
E
m
Cross sections for a number of fusion reactions
which I willabbreviate as
2/1
)( EeE
The probability of a fusion event at kinetic energy E is proportional to the product of these two functions.
kinetic energy EEm
N(E)
P (E)
(E)
)/( 2/1
~)( kTEEeE P
)/( 2/1
~)( kTEEeE Pwhich has a maximum where
0)(/ )/(2/3 2/1121 kTEEeEdEd kT
P
kTE 121 2/3
3/2)( 2kTE
]/)2/()2/([ 3/23/1
)( kTkTkTeE
Pand at that maximum:
3/13/23/13/1 )(])4/1()2[( kTe
3/13/2 )(]25992.162996.0[ kTe
3/13/2 )(88988.1 kTe
]/)2/()2/([ 3/23/1
)( kTkTkTeE
P])4/()/2[( 3/1332223/12 TkTkkTe
3/12 )4/( kT
3/13/13 )/2()( kT
0
2
21
4
21
eZZ
E
m
E Since
we can evaluate , for example, for the case of 2 protons:
)439976.1(1058217.6
/280.9382
4
222
2
0
2
fmMeVsMeV
cMeVemp
keVsfmcMeV 19.22//10977263.2 223
3/1)(14)( kTeEP with kT in keV
from which we can see:
If T ~ 106 K, kT ~ 0.1 keV
for T ~ 108 K, kT ~ 10 keV
3/1)(14)( kTeEP with kT in keV
]101.0[14)1.0(14)10(14 3/13/13/13/1
/
eee
This factor of 100 change in the temperature leads to a change in the fusion rate > 1010 !!!
10]101.0[14 1089.13/13/1
e
There are actually two different sequences of nuclear reactions which lead to the conversion of protons into helium nuclei.
1
3
20
1
11
2
1 HeHH Q=5.49 MeV
eHHH1
2
10
1
10
1
1 Q=0.42 MeV
The sun 1st makes deuterium through the weak (slow) process:
)(2 0
1
12
4
21
3
21
3
2 HHeHH Q=12.86 MeV
then
2 passes through both of the above steps then can allow
This last step won’t happen until the first two steps have built up sufficient quantities of tritium that the last step even becomes possible.
I. The proton-proton cycle
2(Q1+Q2)+Q3=24.68 MeVplus two positrons whose
annihilation brings an extra
4mec 2 = 40.511 MeV
eCN 7
13
66
13
7Q=1.20 MeV
6
13
70
1
16
12
6 NHC Q=1.95 MeV
7
14
70
1
17
13
6 NHC Q=7.55 MeV
II. The CNO cycle
7
15
80
1
17
14
7 OHN Q=7.34 MeV
eNO 8
15
77
15
8Q=1.68 MeV
2
4
26
12
60
1
18
15
7 HeCHN Q=4.96 MeV
carbon, nitrogen and oxygen are only catalysts
Interior Zones of the Sun
6,000 KConvective Zone < 0.01 g/cm3
500,000 K
Radiative Zone < 0.01-10 g/cm3
8,000,000 K
Core < 10-160 g/cm3
15,000,000 K
The 1st generation of stars (following the big bang) have no C or N.The only route for hydrogen burning was through the p-p chain.
Shown are curvesfor solar densities
105 kg m-3 for protons and 103 kg m-3 for 12C.
Rat
e of
en
ergy
pro
du
ctio
n
In later generationsthe relative importance of the two processes depends upon temperature.
T4
T17
CNO cycle
p-p chain
sun
The heat generated by these fusion reactions raises the temperature of the core of the star.
The pressure of this "black body" radiation is sufficient to counteract gravitational collapse.
However once the hydrogen in the central region is exhausted gravitational collapse resumes.
The temperature will rise as gravitational potential energy converts to kinetic energy of the nuclei.
At 108 K helium burning starts fusing:
6
12
66
12
64
8
42
4
2* CCBeHe Q=190 keV
4
8
42
4
22
4
2BeHeHe Q=-91.9 keV
At T= 108 K the fraction of helium nuclei meeting this thresholdis given by the Boltzmann factor e91.9/kT ~ 2.2 105
(with kT= 8.6 keV, the mean thermal energy).
Note: these reactions are reversible.
8Be exists as a resonance decaying with 10-16 sec. Its formation requires 91.9 keV kinetic energy shared between the initial states.
the small branching ratio for the -decay
6
12
66
12
6* CC
*6
12
64
8
42
4
2CBeHe
makes it only 4 x 10-4 as likely as a return to the initial state:
Once stable 12C has been produced, further absorption can occur through
10
20
102
4
28
16
8NeHeO Q=4.73 MeV
8
16
82
4
26
12
6OHeC Q=7.16 MeV
12
24
122
4
210
20
10MgHeNe Q=9.31 MeV
As the helium supply in the core is exhausted further collapse leads to even higher temperatures.
At ~5 x 108 109 K carbon and oxygen fusion can take place.
12
24
126
12
66
12
6MgCC Q=13.9 MeV
16
32
168
16
88
16
8SOO Q=16.5 MeV
and others, which yield protons, neutrons or helium nuclei
During the silicon burning phase (2109 K)elements up to iron are finally produced.
28
56
2814
28
1414
28
14NiSiSi
Even at these temperaturesthe Coulomb barrier remains
too high to allow direct formation:
Instead it is done in an equilibrium process of successive alpha particle absorptions balanced against photo-disintegration:
2
4
212
24
1214
28
14HeMgSi
16
32
162
4
214
28
14SHeSi
At these temperatures the thermal photons have an average energy of 170 keV and their absorption can easily lead to the break up of nuclei.
Then absorption of the 4He by other 28Si nuclei eventually leads to the build up of 56Ni etc.
Beyond iron, nickel and cobalt there are no more exothermic fusion reactions possible.
Heavier elements cannot be built by this process.