# The Sun as a Fusion Reactor

The sun is the central point of our solar system; it has probably been in existence for 5 billion years and is expected to survive for a further 5 billion

Table 2.1 Important Radiant Physical Quantities and Daylight Quantities

Table 2.1 Important Radiant Physical Quantities and Daylight Quantities

 Name Symbol Unit Name Symbol Unit Radiant energy Qe Ws = J Quantity of light v lm s Radiant flux/radiant power W Luminous flux ®v lm Specific emission Me W/m2 Luminous exitance Mv lm/m2 Radiant intensity Ie W/sr Luminous intensity Iv cd = lm/sr Radiance Le W/(m2 sr) Luminance Lv cd/m2 Irradiance Ee G W/m2 Illuminance Ev lx = lm/m2 Irradiation He Ws/m2 Light exposure Hv lx s
 Table 2.2 Data for the Sun and the Earth Sun Earth Ratio Diameter (km) 1,392,520 12,756 1 109 Circumference (km) 4,373,097 40,075 1 109 Surface (km2) 6.0874^1012 5.10M08 1 11,934 Volume (km3) 1.4123^1018 1.0833^1012 1 1,303,670 Mass (kg) 1.9891 •1030 5.9742^1024 1 332,946 Average density (g/cm3) 1.409 5.516 1 0.26 Gravity (surface) (m/s2) 274.0 9.81 1 28 Surface temperature (K) 5777 288 1 367 Centre temperature (K) 15,000,000 6700 1 2200

years. The sun consists of about 80 per cent hydrogen, 20 per cent helium and only 0.1 per cent other elements. Table 2.2 contains data about the sun in comparison to the Earth.

Nuclear fusion processes create the radiant power of the sun. During these processes, four hydrogen nuclei (protons 1p) fuse to form one helium nucleus (alpha particle 4a). The alpha particle consists of two neutrons 1n and two positively charged protons 1p. Furthermore, this reaction produces two positrons e+ and two neutrinos ve and generates energy. The equation of the gross reaction illustrated in Figure 2.1 is:

Comparing the masses of the atomic particles before and after the reaction shows that the total mass is reduced by the reaction. Table 2.3 shows the necessary particle masses for the calculation of the mass difference. The mass of the neutrinos ve can be ignored in this calculation and the mass of a positron e+ is the same as that of an electron e-.

The mass difference Am will be calculated by: Figure 2.1 Fusion of Four Hydrogen Nuclei to Form One Helium Nucleus

(Alpha Particle)

Figure 2.1 Fusion of Four Hydrogen Nuclei to Form One Helium Nucleus

(Alpha Particle)

 Particle or nuclide Mass Particle or nuclide Mass Electron (e-) 0.00054858 u Hydrogen (1H) 1.007825032 u Proton (1p) 1.00727647 u Helium (4He) 4.002603250 u Neutron (1n) 1.008664923 u Alpha particle (4a) 4.0015060883 u

For this reaction the result is:

Am = 4 • 1.00727647 u - 4.0015060883 u - 2 • 0.00054858 u = 0.02650263 u

Thus, the total mass of all particles after the fusion is less than that before. The mass difference is converted into energy AE, with the relationship:

With the speed of light c = 2.99792458-108 m/s, this equation determines the energy released by the fusion as AE = 3.955-10"12 J = 24.687 MeV. The binding energy Eb of a nucleus explains the different masses after the fusion as well as the energy difference. An atomic nucleus consists of N neutrons 1n and Z protons 1p. To maintain equilibrium, this binding energy has to be released during the assembly of a nucleus with protons and neutrons. The mass difference of the alpha particle and the two neutrons together with the two protons determines the binding energy of a helium nucleus.

So far, only the atomic nuclei have been considered; the electrons in the atomic shell have not been taken into account. There is one electron in the atomic shell of a hydrogen atom 1H, while there are two electrons in the helium atom 4He. During the nuclear fusion process, two of the four electrons of the hydrogen atoms become the atomic shell electrons of the helium atom. The two other electrons and the positrons convert directly into energy. This radiative energy is four times the equivalent mass of an electron of 2.044 MeV. The total energy released during the reaction is thus 26.731 MeV. This very small amount of the energy does not appear to be significant at first glance; however, the enormous number of fusing nuclei results in the release of vast quantities of energy.

The sun loses 4.3 million metric tonnes of mass per second (Am 4.3-109 kg/s). This results in the solar radiant power 0eS of:

This value divided by the sun's surface area, AS, provides the specific emission of the sun:

Every square metre of the sun's surface emits a radiant power of 63.11 MW. One fifth of a square kilometre of the sun's surface emits radiant energy of 400 EJ per year. This amount of energy is equal to the total primary energy demand on Earth at present.

The sun's irradiance can be approximated to that of a black body. The Stefan-Boltzmann law:

can be used to estimate the surface temperature of the sun, Tsun. With the Stefan-Boltzmann constant ct = 5.67051-10"8 W/(m2 K4), it becomes:

Me,S

The surface, ASE, of a sphere with the sun as its centre and with a radius equal to the average distance from the Earth to the centre of the sun (rSE = 1.5 • 108 km) receives the same total radiant power as the surface of the sun AS (Figure 2.2). However, the specific emission, MeS, or the energy density measured over one square metre, is much higher at the sun's surface than at the sphere surrounding the sun.

With Me s • AS = Ee • Ase and substituting ASE = 4 • n • rSE, the irradiance at the Earth, Ee, finally becomes: 