Nuclear Engineering

Basic Fusion Process, Stellar Evolution and Fusion Reaction Rate

Let us now discuss another important nuclear reaction, i.e. nuclear fusion to generate energy and electrical power. From binding energy curve (Please see module 1), it is clear that the B.E./A value of light nuclei on the steep portion of the curve is less than that for nuclei of intermediate mass numbers. So, if two lighter nuclei combine or fuse together to produce a relatively heavier nucleus, there would be a greater binding energy and a consequent decrease in nuclear mass. This would thus result in a positive Q-value and release of energy. Such type of nuclear reaction is known as nuclear fusion, a process opposite to that of fission.


Typical examples of fusion reaction are


with energy associated as shown.

The reactions appear to be very simple, still an useful fusion reactor could not be built, why? The reason behind this is that to drive a fusion a reaction, nuclei needs high kinetic energy to overcome the Coulomb barrier. This kinetic energy can be provided by raising the temperature of the reactant nuclei. Let me clarify this point. Fusion reaction is possible only when two interacting particles can come near each other within a distance of the order of 10^-14 m. This means that they will have to overcome the mutual repulsive Coulomb barrier acting between them due to positive charges and come within a distance of the above order. Therefore, their relative velocity should be very large, i.e., the temperature of the medium must be very high. Balancing the kinetic energy and mutual electrostatic energy, one can write

or
where R ^~10^-14 m.

Therefore, the necessary condition for the occurrence of fusion process is that the medium must be very hot. This is why it is called thermonuclear reaction. These thermonuclear reactions have a great impact in energy generation in stars and also in evolution of stars. Considering sun as a medium size star, it radiates energy at the rate of about 10²⁶ Js^-1. This can be considered as typical hot star whose interiors have a temperature of about 20 x 10⁶ K. The age of the sun is at least 5 x 10⁹ year, so that the total loss of energy during this period is exceedingly high. Therefore, the natural question is how did the sun maintain this energy output for so long – what is the source of all stellar energy? Helmholtz and Kelvin suggested that a slow gravitational contraction of the stars might be the source; Jeans proposed annihilation of high-energy protons and electrons in the stars as the source. In 1938, Hans Bethe suggested that the stellar energy is produced by thermonuclear reactions in which protons are combined and transformed into helium nuclei. This is known as the proton-proton cycle, and is applicable for relatively low stellar temperatures. The cycle is:

by addition, we have


For the main sequence – stars having higher temperatures, Bethe suggested an alternative to the proton-proton cycle – the carbon nitrogen cycle. The cycle is

Addin,

The conversion of hydrogen to helium will continue until whole of star’s supply of protons is used up. In the case of the sun, both the above cycles take place with roughly equal probability, and it is estimated that it will be about 3 x 10¹⁰ years before the protons have all been converted to helium. Fusion of four protons into helium and overcoming the coulomb repulsion is only possible because of the extremely high initial temperature. It is now believed that the carbon-nitrogen cycle starts at a later stage in the life of a star. According to the theory of stellar evolution, a star is formed by way of an enormous amount of matter at certain point in space. As the mass contracts, its temperature increases, and when it reaches the value of about 2,00,000 ⁰C, the proton-proton cycle starts operating. When the deuterons, formed as a product in the cycle, are completely exhausted, the star shrinks and the temperature again increases. At a temperature of about million degrees, protons interact with heavier elements such as beryllium, lithium, and boron, forming a helium nucleus. At this stage the star looks very bright and is called a red giant. When the above type is also exhausted, further contraction takes place and the temperature rises to about 20 x 10⁶ ⁰C. The carbon-nitrogen cycle operates at this stage and supplies the energy for the major portion of the radiating life of a star. Let us now discuss the probability of various fusion reactions with temperature.

Fusion reaction rate


Let us consider a binary reaction in which there are n₁ nuclei per cc of one species interacting with n₂ nuclei per cc of another species. To determine the rate at which two nuclear systems interact, it may be supposed that nuclei of the first system forms a stationary system within which the nuclei of the second system moves with a velocity of v cm/sec, which is the relative velocity of the nuclei. Let σ be the cross section for the given reaction. Then n₁σ gives the number of nuclei of the first system, which will react, with each nucleus of the second system while traversing a distance of 1 cm. The total distance traveled by all the nuclei of the second system in one second is n₂v. Hence the reaction rate is⁵


If the reaction takes place between two nuclei of the same kind the reaction rate becomes


The interaction between two nuclei should not be counted twice. Hence

Considering M.B. statistics, the distribution of the nuclei can be written as

Since σ is a function of velocity

where <σv> is the average value of σv and

Putting w = 1/2 Mv ² , we get


The value of <σv> assuming Maxwellian distribution of velocities for the D-D, D-T and D-He3 reactions have been calculated for w varying from 1 keV to 1000 keV and are represented in Fig. m3.6.^1,5

FIGURE m3.6 <σv> versus w (1 to 1000 keV)

It is clear from this distribution that <σv> falls at lower temperature although it is always finite.

Nuclear Engineering

Nuclear Engineering

Curriculum

Basic Fusion Process, Stellar Evolution and Fusion Reaction Rate

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