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.