4. Coulomb barrier

An obstacle called the Coulomb barrier caused by the strongly repulsive electrostatic forces between the positively charged nuclei prevents them from fusing

under normal circumstances. However, fusion can occur under conditions of extreme pressure and temperature. That is why fusion reaction is often termed as thermonuclear reaction.

where kB <sup>¼</sup> <sup>8</sup>:<sup>62</sup> � <sup>10</sup>�<sup>11</sup> MeV/K is the Boltzmann constant. By equating the average thermal energy to the Coulomb barrier height and solving for T gives a value for the

The above "back of the envelop" calculation, using classical physics, does not take into consideration the quantum effect of tunneling, which predicts there will be a small probability that the Coulomb barrier will be overcome by nuclei tunnel-

<sup>P</sup><sup>∝</sup> exp � <sup>Z</sup>1Z2α<sup>c</sup>

where Z<sup>1</sup> and Z<sup>2</sup> are the atomic numbers of the interacting particles, α ¼ 1=137 is the fine structure constant, and v is the relative velocity of the colliding nuclei. Using the classical turning point r<sup>0</sup> and de Broglie wavelength λ ¼ h=p ¼ h=mv, the

<sup>P</sup><sup>∝</sup> exp � <sup>r</sup><sup>0</sup>

Thus, large values of v (high energies), or small λ, favor a fusion reaction. Taking into account the tunneling probability, we can now estimate the temperature for fusion to occur. In terms of de Broglie wavelength, the kinetic

<sup>K</sup> <sup>¼</sup> <sup>p</sup><sup>2</sup>

If we use this wavelength as the distance of closest approach to calculate the

<sup>2</sup> kBT <sup>¼</sup> <sup>e</sup><sup>2</sup>

<sup>T</sup> <sup>¼</sup> <sup>4</sup>me<sup>4</sup> 3h<sup>2</sup> kB

For two hydrogen nuclei (mc<sup>2</sup> <sup>¼</sup> 940 MeV), this gives a temperature of about 20

Since the 1950s, scientists have been working tirelessly to develop a reactor in order to harness the nearly inexhaustible energy produced during fusion [5]. The

e2 <sup>λ</sup> <sup>¼</sup> <sup>h</sup><sup>2</sup>

3

goals of fusion research at present include the following:

<sup>2</sup><sup>m</sup> <sup>¼</sup> <sup>h</sup><sup>2</sup>

If we require that the nuclei must be closer than the de Broglie wavelength for tunneling to take over and the nuclei to fuse, then the Coulomb barrier is given by

<sup>2</sup>mλ<sup>2</sup> ! <sup>λ</sup> <sup>¼</sup> <sup>h</sup><sup>2</sup>

<sup>λ</sup> <sup>¼</sup> <sup>2</sup>me<sup>4</sup>

<sup>¼</sup> mc<sup>2</sup>α<sup>2</sup> 3π<sup>2</sup>kB

v

λ

, (6)

: (7)

<sup>2</sup>mλ<sup>2</sup> : (8)

<sup>2</sup>me<sup>2</sup> : (9)

<sup>h</sup><sup>2</sup> : (10)

: (11)

temperature of around 10 billion Kelvin (K).

DOI: http://dx.doi.org/10.5772/intechopen.82335

Nuclear Fusion: Holy Grail of Energy

above expression can be written as

energy is

temperature, we obtain

million Kelvin.

7

6. Fusion reactor

Solving for the temperature, we get

ing through it. The probability P of such an event happening is

Nuclei, which have positive charges, must collide at extremely high speeds to overcome the Coulomb barrier. The speed of particles in a gas is governed by the temperature. At the very center of the Sun and other stars, it is extremely hot and density is very high. For the Sun, the temperature is around 15 million degrees Celsius, and the central density is about 150 times that of water. Under such extreme conditions, electrons in an atom become completely detached from the atomic nucleus, thereby forming an ionized fluid called plasma—a "soup" of hot gas, with bare, positively charged atomic nuclei and negatively charged electrons whizzing about at extremely high speeds. The plasma as a mixture of positive ions (nuclei) and negative electrons is overall electrically neutral.

Without the high pressure of the overlying layers, the hot plasma at the solar core would simply explode into space, shutting off the nuclear reactions. The pressure, which is about 250 billion atmospheres at the Sun's core, squeezes the nuclei so that they are within 1 fm (10�<sup>15</sup> m) of each other. At this distance, the attractive strong nuclear force that binds protons and neutrons together in the nucleus becomes dominant and pulls the incoming particles together, causing them to fuse.

Additionally, massive gravitational force causes nuclei to be crowded together very densely. This means collisions occur very frequently, another requirement if a high fusion rate is to occur. A quick and crude calculation suggests that we need about 10<sup>38</sup> collisions per second to keep the Sun going, while within the core we get about 10<sup>64</sup> collisions per interactions per second, implying only one in 10<sup>26</sup> collisions needs to be a successful fusion event.

### 5. Nuclear fusion on Earth

One of the major challenges in initiating a fusion reaction in a laboratory environment on Earth is to create conditions similar to that in the Sun—extremely high temperatures, perhaps more than 100 million degrees Celsius (equivalent to mean particle kinetic energies of �10 keV) while simultaneously maintaining a high enough density for a long enough time so that the rate of fusion reactions will be large enough to generate the desired power.

#### 5.1 "Ignition" temperature

We can estimate the minimum temperature required to initiate fusion by calculating the Coulomb barrier which opposes two protons approaching each other to fuse. With <sup>e</sup><sup>2</sup> <sup>¼</sup> <sup>1</sup>:44 MeV-fm, where <sup>e</sup> is the charge of a proton, and <sup>r</sup> <sup>¼</sup> <sup>1</sup>:0 fm (separation between two protons), the height of the Coulomb barrier is

$$U = \frac{e^2}{r} = 1.44 \text{ MeV}.\tag{4}$$

The kinetic energy of the nuclei moving with a speed v is related to the temperature T by

$$\frac{1}{2}mv^2 = \frac{3}{2}\ \,\,k\_BT\,,\tag{5}$$

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under normal circumstances. However, fusion can occur under conditions of extreme pressure and temperature. That is why fusion reaction is often termed as

Nuclear Fusion - One Noble Goal and a Variety of Scientific and Technological Challenges

(nuclei) and negative electrons is overall electrically neutral.

sions needs to be a successful fusion event.

large enough to generate the desired power.

5. Nuclear fusion on Earth

5.1 "Ignition" temperature

temperature T by

6

Nuclei, which have positive charges, must collide at extremely high speeds to overcome the Coulomb barrier. The speed of particles in a gas is governed by the temperature. At the very center of the Sun and other stars, it is extremely hot and density is very high. For the Sun, the temperature is around 15 million degrees Celsius, and the central density is about 150 times that of water. Under such extreme conditions, electrons in an atom become completely detached from the atomic nucleus, thereby forming an ionized fluid called plasma—a "soup" of hot gas, with bare, positively charged atomic nuclei and negatively charged electrons whizzing about at extremely high speeds. The plasma as a mixture of positive ions

Without the high pressure of the overlying layers, the hot plasma at the solar core would simply explode into space, shutting off the nuclear reactions. The pressure, which is about 250 billion atmospheres at the Sun's core, squeezes the nuclei so that they are within 1 fm (10�<sup>15</sup> m) of each other. At this distance, the attractive strong nuclear force that binds protons and neutrons together in the nucleus becomes dominant and pulls the incoming particles together, causing them

Additionally, massive gravitational force causes nuclei to be crowded together very densely. This means collisions occur very frequently, another requirement if a high fusion rate is to occur. A quick and crude calculation suggests that we need about 10<sup>38</sup> collisions per second to keep the Sun going, while within the core we get about 10<sup>64</sup> collisions per interactions per second, implying only one in 10<sup>26</sup> colli-

One of the major challenges in initiating a fusion reaction in a laboratory environment on Earth is to create conditions similar to that in the Sun—extremely high temperatures, perhaps more than 100 million degrees Celsius (equivalent to mean particle kinetic energies of �10 keV) while simultaneously maintaining a high enough density for a long enough time so that the rate of fusion reactions will be

We can estimate the minimum temperature required to initiate fusion by calculating the Coulomb barrier which opposes two protons approaching each other to fuse. With <sup>e</sup><sup>2</sup> <sup>¼</sup> <sup>1</sup>:44 MeV-fm, where <sup>e</sup> is the charge of a proton, and <sup>r</sup> <sup>¼</sup> <sup>1</sup>:0 fm

<sup>r</sup> <sup>¼</sup> <sup>1</sup>:44 MeV: (4)

<sup>2</sup> kBT, (5)

(separation between two protons), the height of the Coulomb barrier is

<sup>U</sup> <sup>¼</sup> <sup>e</sup><sup>2</sup>

1 2

The kinetic energy of the nuclei moving with a speed v is related to the

mv<sup>2</sup> <sup>¼</sup> <sup>3</sup>

thermonuclear reaction.

to fuse.

where kB <sup>¼</sup> <sup>8</sup>:<sup>62</sup> � <sup>10</sup>�<sup>11</sup> MeV/K is the Boltzmann constant. By equating the average thermal energy to the Coulomb barrier height and solving for T gives a value for the temperature of around 10 billion Kelvin (K).

The above "back of the envelop" calculation, using classical physics, does not take into consideration the quantum effect of tunneling, which predicts there will be a small probability that the Coulomb barrier will be overcome by nuclei tunneling through it. The probability P of such an event happening is

$$P \propto \exp\left(-\frac{Z\_1 Z\_2 a v}{v}\right),\tag{6}$$

where Z<sup>1</sup> and Z<sup>2</sup> are the atomic numbers of the interacting particles, α ¼ 1=137 is the fine structure constant, and v is the relative velocity of the colliding nuclei. Using the classical turning point r<sup>0</sup> and de Broglie wavelength λ ¼ h=p ¼ h=mv, the above expression can be written as

$$P \propto \exp\left(-\frac{r\_0}{\lambda}\right). \tag{7}$$

Thus, large values of v (high energies), or small λ, favor a fusion reaction.

Taking into account the tunneling probability, we can now estimate the temperature for fusion to occur. In terms of de Broglie wavelength, the kinetic energy is

$$K = \frac{p^2}{2m} = \frac{h^2}{2m\lambda^2}.\tag{8}$$

If we require that the nuclei must be closer than the de Broglie wavelength for tunneling to take over and the nuclei to fuse, then the Coulomb barrier is given by

$$\frac{e^2}{\lambda} = \frac{h^2}{2m\lambda^2} \to \lambda = \frac{h^2}{2m\varepsilon^2}.\tag{9}$$

If we use this wavelength as the distance of closest approach to calculate the temperature, we obtain

$$\frac{3}{2}\ \,\,k\_BT = \frac{e^2}{\lambda} = \frac{2me^4}{h^2}.\tag{10}$$

Solving for the temperature, we get

$$T = \frac{4me^4}{3h^2k\_B} = \frac{mc^2a^2}{3\pi^2k\_B}.\tag{11}$$

For two hydrogen nuclei (mc<sup>2</sup> <sup>¼</sup> 940 MeV), this gives a temperature of about 20 million Kelvin.

### 6. Fusion reactor

Since the 1950s, scientists have been working tirelessly to develop a reactor in order to harness the nearly inexhaustible energy produced during fusion [5]. The goals of fusion research at present include the following:


To date, much headway has been made toward achieving these goals.

#### 6.1 Fuel

Just like the Sun, the fuel for a fusion reactor is hydrogen, the most abundant element in the Universe. But without the benefit of gravitational force that is at work in the Sun, achieving fusion on Earth requires a different approach. The simplest reaction in which enormous amount of energy will be released is the fusion of the hydrogen isotopes deuterium (2H) and tritium (3H) producing 4He and a neutron. For the sake of brevity, we will use the notation d and t for deuterium and tritium, respectively.

Deuterium is found aplenty in ocean water, enough to last for billions of years. This makes it an attractive source of alternative energy relative to other sources of energy. Naturally occurring tritium, on the other hand, is extremely rare. It is radioactive with a half-life of around 12 years. Trace quantities of tritium can be found in cosmic rays. Nevertheless, it can be produced inside a reactor by neutron (n) activation of lithium (Li), the other raw material for fusion found in brines, minerals, and clays. Because of the abundance of fusion fuel, the amount of energy that can be released in controlled fusion reactions is virtually unlimited.

For d-t reaction, we must first create the tritium from either flavor of lithium:

$$^6\text{Li} + \text{n} \rightarrow ^4\text{He} + \text{t},\tag{12}$$

reaction that releases its energy entirely in the form of charged particles, rather than neutrons, thereby offering the possibility, at least in principle, of direct conversion of fusion energy into electrical energy. However, the cross sections and reaction rates for both the reactions are as much as a factor of 10 lower than the d-t reaction. Moreover, because of the higher Coulomb barrier (�2.88 MeV), the ignition temperatures required for 2H <sup>þ</sup> 3He reaction are much higher than those of d-t fusion. An interesting fusion reaction is a proton colliding with boron (B). The proton fuses with 11B to form 12C which immediately decays into three alpha (4H nucleus) particles. A total energy of 8.7 MeV is released in the form of kinetic energy of the alpha particles. Since it is relatively easy to control the energy of the proton with today's accelerator technology, this fusion reaction can be easily initiated without

In order to attain the temperature for fusion to occur, the plasma has to meet

In addition to providing a sufficiently high temperature to enable the particles to overcome the Coulomb barrier, a critical density of the ions in the plasma must be maintained to make the probability of fusion high enough to achieve a net yield of energy from the reaction. The condition which must be met for a yield of more energy than is required for the heating of the plasma is stated in terms of the product of the plasma density (nd) and confinement time (τ). The product has to

<sup>n</sup>d<sup>τ</sup> <sup>≥</sup><sup>3</sup> � <sup>10</sup><sup>20</sup> <sup>s</sup>=m<sup>3</sup>

This relation is called the Lawson criterion [6]. Researchers sometimes use the triple product of nd, τ, and the plasma temperature T. Called the fusion product, the

<sup>n</sup>dτ<sup>T</sup> <sup>≥</sup><sup>5</sup> � <sup>10</sup><sup>21</sup> <sup>s</sup>: keV=m<sup>3</sup>

To summarize, three main conditions are necessary for nuclear fusion:

1. The temperature must be hot enough to allow the ions to overcome the

Coulomb barrier and fuse together. This requires a temperature of at least 100

2. The ions have to be confined together in close proximity to allow them to fuse.

3. The ions must be held together in close proximity at high temperature long

At higher densities, charged particles in the plasma moving at high speeds may give rise to bremsstrahlung—radiation given off by a charged particle (most often an electron) due to its acceleration caused by an electric field of another charged particle (most often a proton or an atomic nucleus). Bremsstrahlung could become so dominant that all the energy in the plasma may radiate away. Other radiation

: (15)

: (16)

some conditions. They are Lawson criterion and Debye length.

involving other interaction channels.

Nuclear Fusion: Holy Grail of Energy

DOI: http://dx.doi.org/10.5772/intechopen.82335

6.2 Conditions for fusion reaction

6.2.1 Lawson criterion

satisfy the inequality:

condition for fusion to take place is

million degrees Celsius.

9

enough to avoid plasma cooling.

A suitable ion density is 2 � <sup>3</sup> � <sup>10</sup><sup>20</sup> ions=m3.

or

$$\text{\textquotedblleft Li} + \text{n} \rightarrow \, ^4\text{He} + \, ^4\text{H} \rightarrow \, ^4\text{He} + \text{t} + \text{n}. \tag{13}$$

The next step in the reaction is

$$\mathbf{d} + \mathbf{t} \to \,^5\mathbf{He} \to \,^4\mathbf{He} + \mathbf{n}.\tag{14}$$

The neutrons generated from the d-t fusion can be used to bombard lithium to produce helium and tritium, thereby starting a controlled, sustainable chain reaction.

The mass of the resulting helium atom and neutron is not the exact sum of the masses of deuterium and tritium. Once again, because of mass defect, each lithium nucleus converted to tritium will end up yielding about 18 MeV of thermal energy. Compared to fission, where each split of uranium releases about 200 MeV of energy, it might appear that the energy released during fusion is rather small. The discrepancy in the energies lies in the number of nucleons involved in the reactions —more than 200 for fission and 5 for fusion. On a per nucleon basis, fusion releases 18/5 = 3.6 MeV, while fission releases 200/236 = 0.85 MeV. So, fusion wins hands down, by greater than a factor of 4.

The other fusion scheme for which the required fuel (4He) will be produced is <sup>d</sup> <sup>þ</sup> <sup>d</sup> ! 4He. Another reaction, 2Hþ3He!4He <sup>þ</sup> <sup>p</sup>, is an example of a fusion

Nuclear Fusion: Holy Grail of Energy DOI: http://dx.doi.org/10.5772/intechopen.82335

reaction that releases its energy entirely in the form of charged particles, rather than neutrons, thereby offering the possibility, at least in principle, of direct conversion of fusion energy into electrical energy. However, the cross sections and reaction rates for both the reactions are as much as a factor of 10 lower than the d-t reaction. Moreover, because of the higher Coulomb barrier (�2.88 MeV), the ignition temperatures required for 2H <sup>þ</sup> 3He reaction are much higher than those of d-t fusion.

An interesting fusion reaction is a proton colliding with boron (B). The proton fuses with 11B to form 12C which immediately decays into three alpha (4H nucleus) particles. A total energy of 8.7 MeV is released in the form of kinetic energy of the alpha particles. Since it is relatively easy to control the energy of the proton with today's accelerator technology, this fusion reaction can be easily initiated without involving other interaction channels.

#### 6.2 Conditions for fusion reaction

In order to attain the temperature for fusion to occur, the plasma has to meet some conditions. They are Lawson criterion and Debye length.

#### 6.2.1 Lawson criterion

1. To achieve the required temperature to ignite the fusion reaction.

Nuclear Fusion - One Noble Goal and a Variety of Scientific and Technological Challenges

amounts of energy out of the thermonuclear fusion reactions.

To date, much headway has been made toward achieving these goals.

the plasma to the ignition temperature.

6.1 Fuel

or

reaction.

8

The next step in the reaction is

down, by greater than a factor of 4.

tritium, respectively.

2. To keep the plasma together at this temperature long enough to get useful

3. To obtain more energy from the thermonuclear reactions than is used to heat

Just like the Sun, the fuel for a fusion reactor is hydrogen, the most abundant element in the Universe. But without the benefit of gravitational force that is at work in the Sun, achieving fusion on Earth requires a different approach. The simplest reaction in which enormous amount of energy will be released is the fusion of the hydrogen isotopes deuterium (2H) and tritium (3H) producing 4He and a neutron. For the sake of brevity, we will use the notation d and t for deuterium and

Deuterium is found aplenty in ocean water, enough to last for billions of years. This makes it an attractive source of alternative energy relative to other sources of energy. Naturally occurring tritium, on the other hand, is extremely rare. It is radioactive with a half-life of around 12 years. Trace quantities of tritium can be found in cosmic rays. Nevertheless, it can be produced inside a reactor by neutron (n) activation of lithium (Li), the other raw material for fusion found in brines, minerals, and clays. Because of the abundance of fusion fuel, the amount of energy

For d-t reaction, we must first create the tritium from either flavor of lithium:

The neutrons generated from the d-t fusion can be used to bombard lithium to produce helium and tritium, thereby starting a controlled, sustainable chain

The mass of the resulting helium atom and neutron is not the exact sum of the masses of deuterium and tritium. Once again, because of mass defect, each lithium nucleus converted to tritium will end up yielding about 18 MeV of thermal energy. Compared to fission, where each split of uranium releases about 200 MeV of energy, it might appear that the energy released during fusion is rather small. The discrepancy in the energies lies in the number of nucleons involved in the reactions —more than 200 for fission and 5 for fusion. On a per nucleon basis, fusion releases 18/5 = 3.6 MeV, while fission releases 200/236 = 0.85 MeV. So, fusion wins hands

The other fusion scheme for which the required fuel (4He) will be produced is

<sup>d</sup> <sup>þ</sup> <sup>d</sup> ! 4He. Another reaction, 2Hþ3He!4He <sup>þ</sup> <sup>p</sup>, is an example of a fusion

<sup>6</sup> Li <sup>þ</sup> <sup>n</sup> !4He <sup>þ</sup> <sup>t</sup>, (12)

<sup>7</sup> Li <sup>þ</sup> <sup>n</sup> ! 4He <sup>þ</sup> 4H ! 4He <sup>þ</sup> <sup>t</sup> <sup>þ</sup> <sup>n</sup>: (13)

<sup>d</sup> <sup>þ</sup> <sup>t</sup> ! 5He ! 4He <sup>þ</sup> <sup>n</sup>: (14)

that can be released in controlled fusion reactions is virtually unlimited.

In addition to providing a sufficiently high temperature to enable the particles to overcome the Coulomb barrier, a critical density of the ions in the plasma must be maintained to make the probability of fusion high enough to achieve a net yield of energy from the reaction. The condition which must be met for a yield of more energy than is required for the heating of the plasma is stated in terms of the product of the plasma density (nd) and confinement time (τ). The product has to satisfy the inequality:

$$\mathbf{n}\_d \tau \ge 3 \times 10^{20} \text{ s/m}^3. \tag{15}$$

This relation is called the Lawson criterion [6]. Researchers sometimes use the triple product of nd, τ, and the plasma temperature T. Called the fusion product, the condition for fusion to take place is

$$\mathbf{n}\_d \tau T \ge \mathbf{5} \times \mathbf{10}^{21} \text{ s. keV/m}^3. \tag{16}$$

To summarize, three main conditions are necessary for nuclear fusion:


At higher densities, charged particles in the plasma moving at high speeds may give rise to bremsstrahlung—radiation given off by a charged particle (most often an electron) due to its acceleration caused by an electric field of another charged particle (most often a proton or an atomic nucleus). Bremsstrahlung could become so dominant that all the energy in the plasma may radiate away. Other radiation

losses, including synchrotron radiation from charged particles orbiting about magnetic fields would be negligible. A fusion reactor, therefore, has to be operated at a temperature where the power gain from fusion would exceed the bremsstrahlung losses.

### 6.2.2 Debye length

A parameter that determines the electrostatic properties of a plasma is called the Debye length LD [7]:

$$L\_D \propto \sqrt{\frac{k\_B T}{n\_d}}.\tag{17}$$

Many techniques have been developed, but the two main experimental approaches that seem capable of doing this task are magnetic confinement and

from coming into contact with the reactor walls. The magnetic fields keep the plasma in perpetually looping paths because the electrical charges on the separated ions and electrons mean that they follow the magnetic field lines. As a consequence,

the plasma does not touch the wall of the container.

This method uses strong magnetic fields to contain the hot plasma and prevent it

There are several types of magnetic confinement system, but the approaches that have been developed to the point of being used in a reactor are tokamak and stellarator devices. Because of its versatility, tokamak is considered to be the most developed magnetic confinement system. Hence, it is the workhorse of fusion.

The tokamak, acronym for the Russian phrase toroidál'naja kámera s magnitnymi katúškami meaning toroidal chamber with magnetic coils, was designed in 1951 by Soviet physicists Andrei Sakharov and Igor Tamm [8]. It is a doughnut-shaped device in which the combination of two sets of magnetic coils, known as toroidal and poloidal field coils, creates a field in both vertical and horizontal directions. The magnetic fields hold and shape the charged particles of the plasma by forcing them to follow the magnetic field lines. They essentially create a "cage," a magnetic bottle, inside which the plasma is confined. A strong electric current is induced in the plasma using a central solenoid, and this induced current also contributes to the

Unlike tokamaks, stellarators [9, 10] do not require a toroidal current to be induced in the plasma. Instead, the plasma is confined and heated by means of helical magnetic field lines. They are produced by a series of coils which may themselves be helical in shape. As a result, plasma stability is increased compared with tokamaks. Since heating the plasma can be more easily controlled and monitored with stellarators, they have an intrinsic potential for steady-state, continuous operation. The disadvantage is that, due to their more complex shape, stellarators

After the invention of laser in 1960 at Hughes Research Laboratory in California, researchers sought to heat the fusion fuels with a laser so suddenly that the plasma would not have time to escape before it was burned in the fusion reaction. It would be trapped by its own inertia, hence the name "inertial confinement," because it relies on the inertia of the implosion to bring nuclei close together. This approach to confinement was developed at Lawrence Livermore National Laboratory in

Within the context of inertial confinement, laser beams with an intensity of the

order of 10<sup>14</sup> <sup>10</sup>15W=cm2 are fired on a solid pellet filled with a low-density mixture of deuterium and tritium. The energy of the laser vaporizes the pellet

are much more complicated than tokamaks to design and build.

inertial confinement.

7.1.1 Tokamak

poloidal field.

7.1.2 Stellarators

7.2 Inertial confinement

California [11].

11

7.1 Magnetic confinement

Nuclear Fusion: Holy Grail of Energy

DOI: http://dx.doi.org/10.5772/intechopen.82335

It is a length scale over which electrons screen out electric fields in the plasma. In other words, it is the distance over which significant charge separation can occur and how far its electrostatic effect persists. For distances greater than the Debye length, the energy of the particles in the plasma balances the electrostatic potential energy.

Using nd <sup>¼</sup> <sup>10</sup><sup>28</sup> particles/m3, the Debye length for a 10 keV plasma is of the order of 10 nm, and the number of particles in a volume of the plasma of one Debye length is about 104. For a more rarefied plasma, say nd <sup>¼</sup> <sup>10</sup><sup>22</sup> particles/m3, LD ¼ 10 μm, and the number of particles in a volume of dimension of one Debye length is 10<sup>7</sup> : In either of these two extreme cases, there are two basic properties: the physical size of the plasma is far larger than the Debye length, and there are many particles in a spherical volume of radius equal to one Debye length. They are these two properties that describe the hot thermonuclear fuel.
