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Solar energy production Solar

33 Solar radiation spectrum M =1.989·10 g 10 R=6.96·10 cm 33 L=3.847·10 erg/sec Tsurface  6000K

Math and Units

kg m2 T :Temperature in units K 1 J  1 2 s E :Energy in units J 1 erg  107 J S :Power in units W 1 eV  1.6022 1019 J 23 W A g  6.022 10 particles F :Flux in units 2 c  2.998108 m / s m MeV J 1 amu  1.661027kg  931.49 L :Luminosity in units c2 s

http://en.wikipedia.org/wiki/Electronvolt Example

33 26 Solar luminosity: Lsun  3.8510 erg / s  3.8510 J / s kg m2 1J  1 s2

E According to Einstein: E  m  c2 m  c2

 kg m2  3.851026   / s  2  1  s  9 kg tons ms  2  4.2810  4,280,000 2  m  s s 3108      2   s  That amount of energy can only be generated by processes! Nuclear masses and energy

Isotopes: nuclei with Z=constant, N varies!

The mass M of the nucleus is smaller than the mass of its and neutron constituents!

2 2 M c  Z mp  N mn c

The mass difference is the binding energy B

2 2 B  M c  Z mp  N mn c http://ie.lbl.gov/toimass.html

http://nucleardata.nuclear.lu.se/database/masses/

Nuclear Binding Energy

Yield from nuclear fission

56Fe, 56Ni, maximum binding energy

Yield from nuclear fusion Example: What is the binding energy of the oxygen isotope 18O?

-24 mp= 1.007825 · 1.66 · 10 g -24 mn= 1.008665 · 1.66 · 10 g

M(18O) = 17.99916 · 1.66 · 10-24 g

Z=8, N=10

B = (Z · m + N · m - M) · c2 p n B(18O) = 1.398 · 105 keV = 2.24·10-11 J

1g 18O contains 7.5·1011 J (W·s)

Atomic Mass Unit: 1 amu=1/12(M12C)=1.66 · 10-24 g Nuclear Reactions and Q-values

2 A(a,b)B (mA+ma-mB-mb)·c =Q • Mass difference between initial particles and final particles is called: Q-value (energy) • If a reaction needs energy to take place it is called: Endothermic (Q<0) • If a reaction releases energy when taking place it is called: Exothermic (Q>0) Product b http://ie.lbl.gov/toi2003/MassSearch.asp(p, n, ) Projectile a (p, n, ) http://nucleardata.nuclear.lu.se/database/masses/

Reaction occurs with a certain probability (cross section), which depends on the energy dependent Coulomb interaction and the probability for forming a Target A (17N, 17O, 14C) new quantum system ! Recoil B & burning Calculate the energy release of solar hydrogen fusion:

4 1H1 4He

-24 mp = 1.007825 · 1.66 · 10 g -24 m4He = 4.002603 · 1.66 · 10 g

27 29 M 4H 1He  41.007825  4.0026031.6610 kg  4.7610 kg 2 12 7 E4H 1He  Q4H 1He  M 4H 1He c  4.2810 J  2.67 10 eV

E4H 1He  Q4H 1He  41.007825  4.002603931.48MeV  26.7MeV This is the energy release per fusion reaction MeV 2.41039 J MeV reactions L  3.851026  2.41039  R  s  91037 s s 26.7MeV s reactions tons M  41.0078251.661027kg91037  6.02 108 H s s Stellar helium burning

Helium burning process in

3 4He 1 12C

Betelgeuse In Orion

-24 m4He = 4.002603 · 1.66 · 10 g -24 m12C = 12.000000 · 1.66 · 10 g

27 29 M 4 12  34.00260312.00 1.6610 kg 1.3010 kg 3 He1 C   2 12 6 E 4 12  Q 12  M c 1.1710 J  7.2710 eV 3 He1 C 3He1 C 4H 1He

E 4 12  Q 12  34.00260312.00 931.48MeV  7.27MeV 3 He1 C 3He1 C   Solar Energy Source 1H 1H 1H 1H 4 1H 1 4He Q Q  26.7MeV

4 2 2.410 X   1/ 3   H e3.380 T erg s1 g 1 1H 1H pp T 2/3 2H=D 2H=D The energy generation rate in the pp chain,

where XH is hydrogen mass fraction. pp-I 3He 3He A  N 1 N 4 N X  i i X  H X  He i   N H   N He   N A A A 1H 1H Particle density: Ni 23 Avogadro’s number: NA=6.022·10 part/A g Hydrogen mass fraction: X  0. 5, 4 H He Solar core density (g/cm3):   160 g/cm3 Burning temperature (GK): T  0.015 GK 4 2 2.4 10 X   1/ 3   H  e3.380 T erg s1 g 1 pp T 2 / 3 4 2 3 2.4 10 0.5 160g / cm 1/ 3    e3.380 0.015 erg s1 g 1 pp 0.0152 / 3 1 1  pp  17.6 erg s g 33 1 Lsun  3.8510 erg s 3.851033erg s1 M   2.19 1032g core 17.6 erg s1 g 1

33 M sun  1.989 10 g  M core  0.11 M sun

About 10% of the solar material is undergoing hydrogen burning reactions in the solar core ! Additional pp-chain sources

3He 4He

18% e-

1H 7Be pp-III pp-II

8B 1 7Li H e-

8Be 4He 4He

4He 4He Additional energy sources

25 4.410 X  Z   1/ 3   H e15.228 T erg s1 g 1 CNO T 2/3

The energy generation rate in the pp chain,

where XH is hydrogen mass fraction and Z the average CNO mass fraction ().

 Xi X H  X He  Z  1 Z  0.02 i

This requires CNO seed abundance as catalyst! More massive stars in CNO mode

Vega in Lyra Sirius in Canis Majoris

Distance 25 Ly Distance 8.6 Ly

Mvega=2.5Msun Mvega=2.0Msun Lvega=50Lsun Lvega=25Lsun Homework 1

The solar hydrogen fuel will eventually get exhausted. The material contracts to increase the internal pressure to balance the gravitational forces. Assuming the stellar core material follows the ideal gas law, at what temperature will the CNO energy production be equal to the energy production by the pp-chains?.

 pp  CNO

T  0.017GK