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CHAPTER # 1 INTRODUCTION TO PLASMA Plasma is one of the four fundamental states of matter. OCCURRENCE OF PLASMAS IN NATURE: It has often been said that 99% of the matter in the universe is in the plasma state; that is, in the form of an electrified gas with the atoms dissociated into positive ions and negative electrons. In our everyday lives, encounters with plasmas are limited to only few examples i.e. the flash of lightning bolt, the soft glow of the Aurora Borealis, the conducting gas inside a fluorescent tube or neon sign, and the slight amount of ionization in a rocket exhaust. It would seem that we live in the 1% of the universe in which plasma do not occur naturally. The reason for this can be seen from Saha’s equation which tells us the amount of ionization to be expected in a gas with thermal equilibrium. 3

𝑛𝑖 𝑇 ⁄2 −𝑢𝑖⁄ 𝐾𝑇 ≈ 2.4 × 1021 𝑒 𝑛𝑛 𝑛𝑖 Here 𝑛𝑖 and 𝑛𝑛 are respectively the density (numbers per m3) of ionized atoms and of neutral atoms. T is the gas temperature and K is the Boltzmann’s constant, and 𝑢𝑖 is the ionization energy of the gas. As the temperature is raised, the degree of ionization remains low until 𝑢𝑖 is only a few times KT. The 𝑛𝑖 /𝑛𝑛 rises abruptly and the gas is in a plasma state. Further increase in temperature makes 𝑛𝑛 less than 𝑛𝑖 and the plasma eventually becomes fully ionized. Where KT is thermal energy.

CREATION OF PLASMA Plasma can be created by heating a gas or subjecting it to a strong electromagnetic field with laser or microwave generator. This increases or decreases the number of electrons creating positive or negative charged particles called ions and is accompanied by the dissociation of molecular bonds if present. DEFINITION OF PLASMA: Any ionized gas cannot be called plasma. Of course there is always small degree of ionization in any gas. A useful definition can be; A plasma is a quasineutral gas of charged and neutral particles which exhibit collective behavior. The word quasi-neutrality was first used by Irving Langmuir and Levi Tanks in 1929 (density of electrons = density of ions, is called quasi-neurality). Quasineutrality describes the apparent charge neutrality of plasma overall. Since the electrons are very mobile, plasma are excellent conductors of electricity, and any changes that develop are readily neutralized.

Course: Experimental Plasma Physics

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If 𝑛𝑖 >> 𝑛𝑛 , then we have a plasma state. Collective behavior means each particle in plasma interacts simultaneously with many others. These collective interactions are about ten times more important than binary collisions in ordinary gas. A macroscopic force applied to a neutral gas such as from a loud speaker generating sound waves is transmitted to the individual atoms by collisions. The situation is totally different in a plasma which has charged particles. As these charges move around, they can generate local concentration of positive or negative charge which gives rise to electric field. Motion of charge also generates current and hence magnetic field. These fields effect the motion of other charged particles far away. Also when we produce negative or positive potential then particles rush towards that it is called collective behavior. In plasma, ions are massive than electrons therefore electrons are more energetic than ions Examples of Plasma:         

Gases in discharge tube (fluorescent lamps and neon signs) Welding arcs Lightening Auroras The upper atmosphere (ionosphere) Stars and the sun Interstellar gas clouds The fireball made by a nuclear weapon Comet tails

CONCEPT OF TEMPERATURE A gas in thermal equilibrium has particles of all velocities and the most probable distribution of these velocities is known as Maxwellian distribution. For simplicity, consider a gas in which particles can move only in one dimension. The one dimensional Maxwellian distribution is given by 𝑓(𝑢) =

1 − 𝑚𝑢2⁄ 2 𝐾𝑇 𝐴𝑒 1

Where 𝑓𝑑𝑢 is the number of particles per m3 with velocity between 𝑢 & 𝑢 + 𝑑𝑢, 2 𝑚𝑣 2 is the KE and K is Boltzmann’s constant. 𝐾 = 1.38 × 10−23 𝐽/𝐾 The density ‘n’, or number of particles per m3 is given by ∞

𝑛 = ∫ 𝑓(𝑢)𝑑𝑢 −∞ 1

As 𝑓(𝑢) =

Hisham Shah

− 𝑚𝑢2⁄ 2 𝐾𝑇 , 𝐴𝑒

so

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3 ∞

𝑛 = ∫ 𝐴𝑒

−𝑚𝑢2⁄ 2𝐾𝑇

𝑑𝑢

−∞ 2

Using substitution, let 𝑥 = 𝑚𝑢 ⁄2𝐾𝑇 2𝐾𝑇

 𝑢=√

𝑚

1

𝑥2

2𝐾𝑇 1

 𝑑𝑢 = √

𝑚

1

. 2 𝑥 −2 𝑑𝑥

Putting values in integral ∞

2𝐾𝑇 1 −1 𝑛 = ∫ 𝐴𝑒 −𝑥 √ . 𝑥 2 𝑑𝑥 𝑚 2 −∞

Taking constant terms out of integral ∞

1 𝐴 2𝐾𝑇 𝑛= √ ∫ 𝑒 −𝑥 𝑥 −2 𝑑𝑥 2 𝑚 −∞

∞

1 𝐴 2𝐾𝑇 𝑛= √ . 2 ∫ 𝑒 −𝑥 . 𝑥 −2 𝑑𝑥 2 𝑚 0

∞

1 2𝐾𝑇 𝑛 = 𝐴√ ∫ 𝑒 −𝑥 𝑥 −2 𝑑𝑥 𝑚 0

Using Gamma function as ∞

∫ 𝑥 𝑛−1 . 𝑒 −𝑥 𝑑𝑥 = Γ(n) 0

So using this ∞

1 2𝐾𝑇 𝑛 = 𝐴√ ∫ 𝑥 −2 . 𝑒 −𝑥 𝑑𝑥 𝑚 0

∞

1 2𝐾𝑇 𝑛 = 𝐴√ ∫ 𝑥 2−1 . 𝑒 −𝑥 𝑑𝑥 𝑚 0

𝑛 = 𝐴√

2𝐾𝑇 1 Γ( ⁄2) 𝑚

As Γ(1⁄2) = √𝜋, so

Course: Experimental Plasma Physics

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𝑛 = 𝐴√

2𝐾𝑇 2𝜋𝐾𝑇 √𝜋 = 𝐴 √ 𝑚 𝑚 𝑚 2𝜋𝐾𝑇

=> 𝐴 = 𝑛√ Which is the relation of ‘A’ to the density ‘n’.

Now we can compute the average kinetic energy of the particles in this distribution as 𝐸𝑎𝑣 = As 𝑓(𝑢) = 𝐴𝑒

−𝑚𝑢2⁄ 2𝐾𝑇 ,

∞ 1 ∫−∞ 2 𝑚𝑢2 𝑓(𝑢)𝑑𝑢 ∞

∫−∞ 𝑓(𝑢)𝑑𝑢

so

𝐸𝑎𝑣 =

−𝑚𝑢2 ( ) ∞ 1 2 ∫−∞ 2 𝑚𝑢 𝐴𝑒 2𝐾𝑇 𝑑𝑢 −𝑚𝑢2 ( ) ∞ ∫−∞ 𝐴𝑒 2𝐾𝑇 𝑑𝑢

Taking constant terms out of integral 2

𝐸𝑎𝑣

−𝑚𝑢 ∞ 1 2 ( 2𝐾𝑇 ) 𝑚𝐴 𝑢 𝑒 𝑑𝑢 ∫ −∞ =2 −𝑚𝑢2 ∞ ( 2𝐾𝑇 ) 𝐴 ∫−∞ 𝑒 𝑑𝑢

1

2 As 𝐸 = 2 𝑚𝑉𝑡ℎ , also 𝐸 = 𝐾𝑇, so

2 2 𝑚𝑉𝑡ℎ = 2𝐾𝑇 => 𝑉𝑡ℎ =

−𝑚𝑢2 ( ) 𝑒 2𝐾𝑇

=

2𝐾𝑇 2𝐾𝑇 => 𝑉𝑡ℎ = √ 𝑚 𝑚

−𝑢2 2𝐾𝑇 𝑒 ⁄𝑚

=

𝑢2 − 2 𝑒 𝑉𝑡ℎ

Now 𝑢2

𝐸𝑎𝑣

𝑢2

− 2 ∞ 1 2 𝑉𝑡ℎ 𝑑𝑢 𝑚 𝑢 𝑒 ∫ −∞ =2 𝑢2 ∞ −𝑉 2 ∫−∞ 𝑒 𝑡ℎ 𝑑𝑢

𝑢

Let 𝑉 2 = 𝑦 2 => 𝑦 = 𝑉 , differentiation gives 𝑉𝑡ℎ 𝑑𝑦 = 𝑑𝑢 𝑡ℎ

𝑡ℎ

𝐸𝑎𝑣

2 ∞ 1 2 𝑚 ∫−∞ 𝑦 2 𝑉𝑡ℎ . 𝑒 −𝑦 . 𝑉𝑡ℎ 𝑑𝑦 2 = ∞ 2 ∫−∞ 𝑒 −𝑦 . 𝑉𝑡ℎ 𝑑𝑦

As Vth is constant with respect to integral so taking it out.

Hisham Shah

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𝐸𝑎𝑣

2 2 ∞ ∞ 1 1 2 2 𝑚𝑉𝑡ℎ . 𝑉𝑡ℎ ∫−∞ 𝑦 2 . 𝑒 −𝑦 𝑑𝑦 2 𝑚𝑉𝑡ℎ . 2 ∫0 𝑦 2 . 𝑒 −𝑦 𝑑𝑦 2 = = ∞ ∞ 2 2 𝑉𝑡ℎ ∫−∞ 𝑒 −𝑦 𝑑𝑦 2 ∫0 𝑒 −𝑦 𝑑𝑦

1

Let 𝑦 2 = 𝑥 => 𝑦 = √𝑥, differentiation gives 𝑑𝑦 = 2 𝑥



−1⁄ 2 𝑑𝑥.

So

1 2 ∞ −𝑥 1 −1⁄2 𝑚𝑉 𝑥. 𝑒 𝑑𝑥 ∫ 𝑡ℎ 0 2𝑥 𝐸𝑎𝑣 = 2 ∞ 1 −1 ∫0 𝑒 −𝑥 2 𝑥 ⁄2 𝑑𝑥 1 2 1 ∞ 1⁄2 −𝑥 𝑚𝑉 . ∫0 𝑥 .𝑒 𝑑𝑥 𝑡ℎ 𝐸𝑎𝑣 = 2 1 ∞2 −1 ∫ 𝑥 ⁄2 .𝑒 −𝑥 𝑑𝑥 2 0

 𝐸𝑎𝑣 =

1 2 ∞ 1⁄2 −𝑥 𝑚𝑉𝑡ℎ ∫0 𝑥 .𝑒 𝑑𝑥 2 ∞ −1 ∫0 𝑥 ⁄2 .𝑒 −𝑥 𝑑𝑥

Using gamma function as ∞

∫ 𝑥 𝑛−1 . 𝑒 −𝑥 𝑑𝑥 = Γ(n) 0

In numerator: 𝑛 − 1 = 1⁄2 => 𝑛 = 1⁄2 + 1 = 3⁄2 Similarly for denominator: 𝑛 − 1 = −1⁄2 => 𝑛 = −1⁄2 + 1 = 1⁄2 𝐸𝑎𝑣

1 2 ∞ 3⁄2−1 −𝑥 𝑚𝑉𝑡ℎ . 𝑒 𝑑𝑥 ∫0 𝑥 2 = ∞ 1 ∫0 𝑥 ⁄2−1 . 𝑒 −𝑥 𝑑𝑥

 𝐸𝑎𝑣 =

1 2 𝑚𝑉𝑡ℎ Γ(3/2) 2

Γ(1/2)

As Γ(𝑛 + 1) = 𝑛Γ(𝑛) 3

1

In numerator: 2 = 2 + 1 3 1 1 Γ ( ) = Γ ( + 1) = Γ(1⁄2) 2 2 2 𝐸𝑎𝑣

1 1 2 1 𝑚𝑉 . Γ( 𝑡ℎ 2 2) =2 1 Γ(2)

1

Now Γ(2) will be cancelled out, so 𝐸𝑎𝑣 = 2 As 𝑉𝑡ℎ =

2𝐾𝑇 𝑚

1 1 1 2 2 𝑚𝑉𝑡ℎ . = 𝑚𝑉𝑡ℎ 2 2 4

, so

Course: Experimental Plasma Physics

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𝐸𝑎𝑣 =

1 2𝐾𝑇 1 𝑚 = 𝐾𝑇 4 𝑚 2

For three dimensional case, 𝐸𝑎𝑣 =

3 𝐾𝑇 2

1

The general result is that 𝐸𝑎𝑣 equals 2𝐾𝑇 per degree of freedom. Since ‘T’ and ‘𝐸𝑎𝑣 ’ are so closely related, it is customary in plasma physics to give temperature in units of energy. For 𝐾𝑇 = 1 𝑒𝑉 = 1.6 × 10−19 𝐽, we have 𝐾𝑇 = 1.6 × 10−19 𝐽 𝑇=

1.6 × 10−19 𝐽 1.6 = × 104 = 11,600 𝐾 −23 1.38 × 10 𝐽/𝐾 1.38

Thus the conversion factor is 1 𝑒𝑉 = 11,600 𝐾 Now by a 2 eV plasma, we mean that KT = 2 eV , or 𝐸𝑎𝑣 = 3 𝑒𝑉 in 3D

DEBYE SHIELDING The term Debye shielding and polarization of plasma are interchangeable A fundamental characteristic of the behavior of a plasma is its ability to shield out electric potentials that are applied to it. Suppose we tried to put an electric field inside a plasma by inserting two charged balls connected to a battery. The balls would attract particles of the opposite charges, and almost immediately a cloud of ions would surround the negative ball and cloud of electrons would surround the positive ball. In case of cold plasma, there would be as many charges in the cloud as in the ball, the shielding would be perfect and no electric field will be present in the body of the plasma outside the clouds. If the temperature is finite, then particles at edge of cloud where electric field is weak have enough thermal energy to escape from the electrostatic potential well and shielding is not complete. Potentials of the order of 𝐾𝑇/𝑒 can leak into plasma and cause finite electric fields to exist there.

Hisham Shah

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When Temperature is Finite Consider a plasma of a finite temperature which contains large number of electrons and ions. Assume that the densities of both electrons and ions are initially equal but ions and electrons are not in thermal equilibrium with each other and having different temperature Ti & Te. Since ions and electrons have random thermal motion, thermally induced perturbations about the equilibrium will cause small, transient variation of electrostatic potential ϕ. From Poisson equation ∇2 𝜙 = −

𝜌 => 𝜌 = −𝜀𝑜 ∇2 𝜙 𝜀𝑜

As ions and electrons density is 𝜌 = −𝑞(𝑛𝑖 − 𝑛𝑒 ) 𝜌 = +𝑞(𝑛𝑖 − 𝑛𝑒 ) −𝜀𝑜 ∇2 𝜙 = +𝑞(𝑛𝑖 − 𝑛𝑒 ) 𝜀𝑜 ∇2 𝜙 = −𝑒(𝑛𝑖 − 𝑛𝑒 ) For 𝑛𝑒 =? Consider pressure of gas 𝑃 = 𝑛𝑒 𝐾𝑇 As total force is 𝐹 = 𝐹𝑚 − 𝐹𝑝 As 𝐹𝑚 = 𝑞𝐸 and 𝐹𝑝 = ∆𝑃/𝑛𝑒 𝐹 = 𝑞𝐸 − ∆𝑃/𝑛𝑒 If 𝐹 = 0, then 𝑞𝐸 −

∆𝑃 =0 𝑛𝑒

As 𝑃 = 𝑛𝑒 𝐾𝑇, so 𝑞𝐸 −

∆(𝑛𝑒 𝐾𝑇) =0 𝑛𝑒

𝑞𝐸 −

𝐾𝑇∆(𝑛𝑒 ) =0 𝑛𝑒

=> 𝑛𝑒 𝑞𝐸 = 𝐾𝑇∆𝑛𝑒 Or

Course: Experimental Plasma Physics

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1 𝑞𝐸 ∆𝑛𝑒 = 𝑛𝑒 𝐾𝑇 Integrating both sides 𝑛𝑒

∫ 𝑛∞

1 𝑞𝐸 ∆𝑛𝑒 = ∫ 𝑑𝑛 𝑛𝑒 𝐾𝑇

𝑞𝐸 𝑛𝑒 ln |𝑛𝑒 | |𝑛 = 𝑛 ∞ 𝐾𝑇 If 𝑛 = 1, then 𝑙𝑛 (

𝑛𝑒 𝑞𝐸 )= 𝑛∞ 𝐾𝑇

𝑞𝐸 𝑛𝑒 = 𝑒 𝐾𝑇 𝑛∞ 𝑞𝐸

𝑛𝑒 = 𝑛∞ 𝑒 𝐾𝑇 For electron 𝑞 = −𝑒, and 𝐸 = −∇𝜙 𝑒∇𝜙

𝑛𝑒 = 𝑛∞ 𝑒 𝐾𝑇 𝑎𝑛𝑑 𝑛𝑖 = 𝑛∞ As 𝜀𝑜 ∇2 𝜙 = −𝑒(𝑛𝑖 − 𝑛𝑒 ) Putting values 𝑒∇𝜙

𝜀𝑜 ∇2 𝜙 = −𝑒(𝑛∞ − 𝑛∞ 𝑒 𝐾𝑇 ) 𝑒∇𝜙

𝜀𝑜 ∇2 𝜙 = 𝑛∞ 𝑒(𝑒 𝐾𝑇 − 1) Using 𝑒 𝑥 = 1 + 𝑥 +

𝑥2 2!

+⋯ 𝜀𝑜

𝑑2𝜙 𝑒𝜙 𝑒 2 𝜙 2 = 𝑛 𝑒(1 + + + ⋯) ∞ 𝑑𝑥 2 𝐾𝑇 𝐾 2 𝑇 2

𝑒 2 𝜙2

𝑒𝜙

If 𝐾𝑇 ≪ 1, then 𝐾2 𝑇 2 ≈ 0, so neglecting higher powers. 𝑑2𝜙 𝑒𝜙 𝑛∞ 𝑒 2 𝜀𝑜 2 = 𝑛∞ 𝑒 ( ) = 𝜙 𝑑𝑥 𝐾𝑇 𝐾𝑇 𝑑 2 𝜙 𝑛∞ 𝑒 2 = 𝜙 𝑑𝑥 2 𝜀𝑜 𝐾𝑇 𝑑 2 𝜙 𝑛∞ 𝑒 2 − 𝜙=0 𝑑𝑥 2 𝜀𝑜 𝐾𝑇 As 𝜆2 =

𝜀𝑜 𝐾𝑇 𝑛∞ 𝑒 2

Hisham Shah

, so

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𝐷2 𝜙 − (𝐷2 −

1 𝜙=0 𝜆2

1 )𝜙 = 0 𝜆2

𝐷=±

1 𝜆 −|𝑥|

General solution to this equation will be 𝜙 = 𝜙𝑜 𝑒 𝜆𝐷 DEBYE LENGTH 𝜀 𝐾𝑇

𝑜 As 𝜆𝐷 = √ 𝑛𝑒 2

 𝜆𝐷 = √

𝜀𝑜 𝐾 𝑒2

.√

𝑇 𝑛

Putting values of 𝜀𝑜 , 𝐾 and 𝑒 2 , we get 1

𝑇 2 𝜆𝐷 = 69 ( ) 𝑚; T in K 𝑛 1

𝐾𝑇 2 𝜆𝐷 = 7430 ( ) 𝑚; KT in eV 𝑛 Now we can define quasineutrality. If the dimensions ‘L’ of a system are much larger than ‘𝜆𝐷 ’, then whenever local concentration of charge arise or external potential are introduced these are shielded out in a distance short compared with ‘L’. The plasma is quasineutral that is neutral enough so that one can take 𝑛𝑖 ≈ 𝑛𝑒 ≈ 𝑛, where ‘n’ is common density called plasma density, but not so neutral that all interesting electromagnetic forces vanish. A criterion for an ionized gas to be plasma is that it should be dense enough that ‘𝜆𝐷 ’ is much smaller than ‘L’. 𝜆𝐷 ≪ 𝐿 If there are only one or two particles in the sheath region, Debye shielding would not be a statistically valid concept. We can compute the number of particles in Debye sphere as 4 𝑁𝐷 = 𝑛 𝜋𝜆3𝐷 = 1.38 × 106 𝑇 3⁄2 ⁄𝑛1⁄2 3 In addition to 𝜆𝐷 ≪ 𝐿, collective behavior requires 𝑁𝐷 ≫ 1 The three conditions a plasma must satisfy are therefore 1. 𝜆𝐷 ≪ 𝐿 2. 𝑁𝐷 ≫ 1 3. 𝜔𝜏 > 1

Course: Experimental Plasma Physics

10

CHAPTER # 2 COLD PLASMA GENERATION Electric discharge is the basic method used for the generation, excitation and sustaining the cold plasmas. If the voltage is sufficiently high, electric breakdown occurs in the gas and an ionized state is formed. The methods including direct current (DC) discharges, radio frequency (RF) discharges, microwave (MW) discharges, Electron Cyclotron Resonance (ECR) discharges are most common methods for the generation, excitation and sustaining the cold plasma.

DC GLOW DISCHARGES A simple glow discharge is produced by applying direct current between two conducting electrodes inserted into a gas chamber at low pressure while a high impedance power supply provides electric field between two electrodes as shown in figure.

The DC glow discharges are easily realized in the pressure range of 10-2-10 mbar. The distance ‘d’ between electrodes in the pressure ‘P’ in the discharge tube is found to satisfy relation 𝑉𝑏 =

𝐶1 (𝑃𝑑) 𝐶2 + 𝑙𝑛(𝑝𝑑)

Which is called Paschen’s law, ‘C1’ & ‘C2’ are constants which depend upon nature of gas, ‘Vb’ is breakdown voltage, or the minimum threshold voltage required to produce the glow discharge. It is evident from Paschen’s law that no discharge is easily realized if the pressure is neither too low nor much high and similarly the distance between electrodes is neither too large nor too small. When voltage across the electrodes is small, few charge carriers are just collected and the current flowing within discharge tube is very small. With the increase of voltage the available charge carriers gain energy from the electric field between two electrodes and current starts to increase exponentially.

Hisham Shah

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Initially due to low energy of electrons, collisions is elastic one, meanwhile the electrons continue to gain energy between collisions until they attain sufficient energy to cause ionization of targets through inelastic collision. The newly formed electron ion pairs accelerated towards the anode and cathode cause electrons emission. The current in discharge tube increases exponentially.

RADIO FREQUENCY DISCHARGE A discharge generated by DC power source exhibit some serious disadvantages. For example ions and electrons gain energy and strike with the electrodes causing significant damage and sputtering of materials. The sputtered material is added into the plasma as impurity that significantly changes the characteristic of plasma. Further, discharges in some gases may result in depositing films of some insulating compounds. Therefore, the DC discharge is extinguished when surface of electrodes exposed to plasma becomes insulating. To overcome this problem, one may apply alternating voltage of high frequency across the electrodes. If an ion is moving towards momentarily cathode, the polarity of the applied electric field is reversed as it approaches to the electrode. Therefore the charged particles keep oscillating between the electrodes without colliding and sputtering of electrodes materials. The frequencies used in the high frequency discharges are in the range of radio transmission. That is why it is called RF discharges. 𝜔

The RF discharges may be operated in the frequency range of 𝑓 = 2𝜋 ≈ 1 − 100 𝑀𝐻𝑧 However the high RF power from generator may cause significant noise in the radio receivers in the vicinity of laboratory. To overcome this problem, a frequency of 13.56 MHz is fixed for cold plasma generation. If one uses RF frequency other than this particular value, one must arrange appropriate shielding of the experiment.

Course: Experimental Plasma Physics

12

If ‘L’ represents dimensions of the experimental chamber and < 𝑣 >𝑑𝑖 is average drift velocity of ions, the critical ion transition frequency may be defined as, 𝑓𝑐𝑖 =

< 𝑉 >𝑑𝑖 2𝐿

It is obvious that frequency of applied electric field between electrodes should be higher than the critical ion frequency, to avoid the collision of ions with electrodes. To avoid electrons striking the electrodes, the frequency of alternating electric field should be much higher. 𝑓𝑐𝑒 =

< 𝑉 >𝑑𝑒 2𝐿

Voltage required to initiate and maintain the RF discharge is much lower than it is required to maintain DC glow discharge. The RF plasma discharge can be created at pressure as below as 10-3 mbar because efficiency of mixing collisions is enhanced by oscillations. Consider an electron oscillating along x-axis in an AC field E of amplitude Eo and angular frequency ‘ω’. 𝐸 = 𝐸𝑜 𝑐𝑜𝑠𝜔𝑡 From equation of electron motion 𝑚𝑒 𝑥̈ = −𝑒𝐸𝑜 𝑐𝑜𝑠𝜔𝑡 𝑥̈ =

−𝑒𝐸𝑜 𝑐𝑜𝑠𝜔𝑡 𝑚𝑒

𝑥̇ =

−𝑒𝐸𝑜 𝑠𝑖𝑛𝜔𝑡 𝑚𝑒 𝜔

Integration gives

Again integrating 𝑥=

𝑒𝐸𝑜 𝑐𝑜𝑠𝜔𝑡 → (𝑎) 𝑚𝑒 𝜔 2

As electron energy “W” is given by 𝑊=

1 𝑚 𝑥̇ 2 → (𝑏) 2 𝑒

Equation (a) and (b) show that field strength and frequency are important in determining electron motion and their energies.

Hisham Shah

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MICROWAVE PLASMA Plasma generation using microwave is widely employed in many applications. A characteristic feature of microwave is the wavelength which is normally comparable to the dimensions of plasma apparatus(2.45 GHz; λ = 12.24 cm). The frequency of 2.45 GHz is commonly used for industrial or home heating applications which make suitable power supplies readily available. MW plasma & RF plasmas are similar but differ only in the large of frequencies. In collisionless plasma, the maximum amplitude “x” of electron oscillating in MW frequency range is 𝑥 < 10−3 𝑐𝑚 which can be written as 𝑥=

𝑒𝐸 𝑚𝑒 𝜔 2

Corresponding maximum energy acquired by an electron during one cycle can be written as 𝑊= 𝑥̇ =

1 𝑚 𝑥̇ 2 2 𝑒

−𝑒𝐸𝑜 𝑠𝑖𝑛𝜔𝑡 𝑚𝑒 𝜔

So 1 −𝑒𝐸𝑜 𝑠𝑖𝑛𝜔𝑡 2 𝑊 = 𝑚𝑒 ( ) 2 𝑚𝑒 𝜔 1

𝑒 2 𝐸𝑜2 sin2 𝜔𝑡

 𝑊 = 𝑚𝑒 ( 𝑊=

2 𝑚𝑒2 𝜔2 2 2 2 𝑒 𝐸𝑜 sin 𝜔𝑡

)

2𝑚𝑒 𝜔2

So 𝑊=

1 − 𝑐𝑜𝑠2𝜔𝑡 ) 𝑒 2 𝐸𝑜2 (1 − 𝑐𝑜𝑠2𝜔𝑡) 2 = → (𝐴) 2𝑚𝑒 𝜔 2 4𝑚𝑒 𝜔 2

𝑒 2 𝐸𝑜2 (

𝑐𝑜𝑠2𝜃 = 𝑐𝑜𝑠 2 𝜃 − 𝑠𝑖𝑛2 𝜃  𝑐𝑜𝑠2𝜃 = 1 − 2 𝑠𝑖𝑛2 𝜃  2 𝑠𝑖𝑛2 𝜃 = 1 − 𝑐𝑜𝑠2𝜃

Equation (A) predicts that the corresponding energy gain by an electron during one cycle is about 0.03 𝑒𝑉 that is too small to sustain plasma.

 𝑠𝑖𝑛2 𝜃 =

1−𝑐𝑜𝑠2𝜃 2

Therefore MW discharges are more difficult to sustain at lower pressures (< 1 torr) than DC or RF discharges. In a collisional discharge case, the power density at constant electric field is given by 𝑃𝑣 =

𝑛𝑒 𝑒 2 𝐸𝑜2 𝑣 ( 2 ) 2𝑚𝑒 𝑣 + 𝜔 2

Course: Experimental Plasma Physics

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This equation shows that Pv becomes maximum at 𝜔 = 𝜈. The absorption of MW power is thus a function of collision frequency “v” of electrons with heavy species and is dependent on the pressure of the gas used. A microwave plasma reactor consists in principle of a MW power supply, a circulator, the applicator, and the plasma load. In MW plasma magnitude of electric field can vary within reactor which has dimensions of the same order of magnitude as wavelength. In typical MW plasmas, the strength of the electric fields is about E = 30 v⁄cm

ELECTRON CYCLOTRON RESONANCE PLASMAS At low temperatures, MW discharges are difficult to sustain hence ECR technique is generally used to execute and sustain plasma. ECR plasmas are typical example of MW plasma in the presence of magnetic fields. 𝑒𝐵

In ECR technique, the applied magnetic field Bo of such a value the EC frequency 𝑊𝑐𝑒 = 𝑚

𝑒

resonate with the frequency of source. Where ‘B’ is magnetic field in Tesla. The radius ‘𝑟𝑙 ’ called larmour’s radius is given by 𝑟𝑙 =

𝑚𝑉⊥ 1 2𝑊⊥ √ = 𝑒𝐵 𝑒𝐵 𝑚

Where “m” is mass of charges particle & ‘𝑉⊥’ is the velocity component of particle normal to magnetic field line & 𝑊⊥ is energy component corresponding to normal component of velocity. The plasma excited in the presence of an external magnetic field which satisfies the resonance condition is called ECR plasma. In the absence of an external magnetic field, the electric field of the incident wave can penetrate the plasma if 𝜔 > 𝜔𝑝 , where 𝜔𝑝 is the electron plasma frequency. 1

𝑛𝑒 𝑒 2 2 𝜔𝑝 = ( ) 𝑚𝑜 ∈𝑜 Which sets upper limit for density 𝑛𝑒 ≤

𝜔2 ∈𝑜 𝑚𝑒 = 𝑛𝑐 𝑒2

Where ‘𝑛𝑐 ’ is called critical density

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CHAPTER # 3 PLASMA DIAGNOSTICS LANGMUIRE PROBE: AN INTRODUCTION Electric or Langmuir probes are among the most basic plasma diagnostic components. They are generally simple in construction and use, relatively inexpensive and robust enough to withstand considerable heat fluxes. Langmuir probe technique was introduced by Irving Langmuir and his colleagues in 1920s. It can provide measurement of basic cold plasma parameters such as electron temperature ‘𝑇𝑒 ’, electron number density ‘𝑛𝑒 ’, plasma potential ‘𝑉𝑝 ’, and floating potential ‘𝑉𝑓 ’. A Langmuir probe is a metallic insulated wire (electrode) except its tip which is exposed of plasma. Generally the tip of probe is several millimeters long and less than one mm in diameter. For cold plasma diagnostic it is placed near the center of the interelectrode gap and is constructed so that it would be oriented to be either parallel or perpendicular to the electrode plane. The probe material should have high melting point and usually tungsten or platinum are used. The material by which probe is insulated should be chemically inert at high temperature. It can either be quartz or a vacuum compatible ceramics. The current ‘I’ of the probe is measured as a function of applied voltage and I-V characteristic of the probe is obtained. Data obtained from I-V characteristic is used to evaluate the electron temperature and density.

A schematic diagram for a single Langmuir probe

Course: Experimental Plasma Physics

16

WORKING OF LANGMUIR PROBE The probe is inserted into the plasma reactor through an electrically insulated seal. Seal can be fixed or can permit changes in the position of the probe. As measurements are made by inserting a probe in the plasma, so this technique is called an in-situ intrusive diagnostic method that perturbs the plasma locally. Due to this local plasma perturbation, the physical conditions of the plasma are changed. The probe affects the plasma particle distribution and energy by changing the electric field. The magnitude of these disturbances depends on the dimensions of the probe and properties of plasma. Langmuir or electric probe that satisfies the condition 𝑟 > 𝜆𝑑 , where ‘r’ is probe radius and ‘𝜆𝑑 ’ is called thin sheath probe. For a single Langmuir probe, the voltage ‘V’ is referenced to a large metal surface in the plasma chamber itself. For a double Langmuir probe, the voltage is applied between the two electrodes both insulated from reactor. If plasma is placed in magnetic field measurements are interpreted with considerable difficulty. Same is true if positive ions are present. If there is no reference electrode a single probe is useless in such cases double probe circuits are used. A double probe is composed of two identically tipped probes separated by a fixed distance. The tips of the probes have to be far enough from each other so that the plasma sheaths of probes would not overlap but closed enough to sample the same region of plasma. Both probes are insulated from ground, float with plasma and are unaffected by changes in ‘V’. THE SINGLE LANGMUIR PROBE CHARACTERISTICS The typical I-V characteristics of a single Langmuir probe are presented in figure. The resulting probes I-V characteristic curve regions depends on whether the electron current or ion current dominates. The I-V characteristics of a probe can be divided into three regions, the electrons saturation region, transition region and ion saturation region. Conventionally the current flowing from probe to plasma is defined positive. So flow of electrons from plasma to probe constitute positive current. When probe is inserted, electrons being more mobile move to the probe and collected by it. The probe current due to ions is almost negligible. The magnitude of electron current depends mainly on electron number density and the average energy, the temperature. This part of curve is called electron saturation region. The continuous collection of electrons by probe makes it more and more negative. Transition region represents this behavior. When probe is charged sufficiently negative that electrons arriving at probe are repelled back and their contribution in probe current becomes equal to that of ions. This is called an ion saturation region.

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The plasma potential ‘Vpl’ is defined as the potential at which electrons arriving near probes are collected and probe current equal to electronic current. The floating potential is the potential of probe at which electrons and ion densities to the probe equals 𝐽𝑒 = 𝐽𝑖

MEASUREMENT OF PLASMA TEMPERATURE AND DENSITY WITH LANGMUIR PROBE Electrostatic probe is a convenient and reliable tool for plasma diagnostics suitable for measuring parameters of cold plasmas i.e. electron temperature and plasma density, which are important parameters in plasma processing. To use the discharges in different applications, it is essential to have information about plasma electron temperature and density. It is because the efficiency of the processes in the plasma and their reaction rates are generally dependent on the density of the charged particles and their energies.

PLASMA ELECTRON TEMPERATURE The electron current density in the transition region of the I-V characteristic can be written as −𝑒𝑉

𝐽 = 𝐽𝑜 𝑒 𝐾𝑇𝑒 → (1) Where ‘Jo’ is random current density in the plasma 𝐾𝑇𝑒 ⃗⃗⃗𝑒 = 𝑛𝑒 𝑒√ 𝐽𝑜 = 𝑛𝑒 𝑒𝑉 2𝜋𝑚𝑒 −𝑒𝑉

As 𝐽 = 𝐽𝑜 𝑒 𝐾𝑇𝑒 −𝑒𝑉

 𝐴𝐽 = 𝐴𝐽𝑜 𝑒 𝐾𝑇𝑒

−𝑒𝑉

 𝑙𝑛(𝐴𝐽) = 𝑙𝑛 (𝐴𝐽𝑜 𝑒 𝐾𝑇𝑒 ) Course: Experimental Plasma Physics

18 𝐼

since 𝐽 = 𝐴 => 𝐼 = 𝐽𝐴, similarly 𝐼𝑜 = 𝐽𝑜 𝐴, so −𝑒𝑉

𝑙𝑛 𝐼 = 𝑙𝑛 𝐼𝑜 + 𝑙𝑛 (𝑒 𝐾𝑇𝑒 )  𝑙𝑛 𝐼 = 𝑙𝑛 𝐼𝑜 −

𝑒𝑉 𝐾𝑇𝑒 𝑒𝑉

 𝑙𝑛 𝐼 − 𝑙𝑛 𝐼𝑜 = −  𝑙𝑛 𝐼𝑜 − 𝑙𝑛 𝐼 =  𝑙𝑛

𝐼𝑜 𝐼

=

𝐾𝑇𝑒 𝑒𝑉

𝐾𝑇𝑒

𝑒𝑉 𝐾𝑇𝑒

The total probe current “I” is the difference between electron and ion current. Therefore 𝐼 = 𝐼𝑒 − 𝐼𝑖

 𝑙𝑛𝐼𝑜 − ln 𝐼 =

𝑒𝑉 𝐾𝑇𝑒

 𝑙𝑛𝐼𝑜 − ln(𝐼𝑒 − 𝐼𝑖 ) =

𝑒𝑉 𝐾𝑇𝑒

For electrons 𝑙𝑛𝐼𝑒 =

𝑒𝑉 𝐾𝑇𝑒

𝑑 𝑒 𝑙𝑛𝐼𝑒 = 𝑑𝑣 𝐾𝑇𝑒 𝑑𝑣 𝐾𝑇𝑒 = 𝑑𝑙𝑛𝐼𝑒 𝑒 𝐾𝑇𝑒 ∆(𝑙𝑛𝐼𝑒 ) −1 =[ ] 𝑒 ∆𝑣

PLASMA ELECTRON DENSITY For the determination of plasma electron density we will proceed as follows. At the plasma potential the probe current will be dominantly due to electrons. Therefore 1 𝐼𝑠𝑒 = −𝑒𝐴(Γ𝑖 − Γ𝑒) ≈ 𝑒𝐴𝑝 Γ𝑒 = 𝑒𝐴𝑝 𝑛𝑒 ⃗⃗⃗ 𝑣𝑒 → (𝐴) 4 Where ‘𝐴𝑝 ’ is the probe area, ‘Γ𝑖’ is the flux of ions and ‘Γ𝑒’ is the flux of electrons arriving at the probe. The probe would thus emit a net positive current. From the elementary gas kinetic theory, the number of particles of a given species crossing unit area per unit time is 1 Γ = 𝑛𝑣̅ 4

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Where ‘𝑣̅ ’ is mean speed that may be found as follows ∞

𝑣̅ = ∫ 𝑣𝑓𝑚 (𝑣)𝑑 3 𝑣 −∞

Where ‘𝑓𝑚 ’ is the Maxwellian velocity distribution function. In 3-D, it can be written as 3

2

𝑚 2 −𝑚𝑣 𝑓𝑚 = ( ) 𝑒 2𝐾𝑇𝑒 2𝜋𝐾𝑇𝑒

The integral above can be solved in spherical coordinates more easily. Since volume of each spherical shell is 𝑑 3 𝑉 = 4𝜋𝑣 2 𝑑𝑣, so ∞

𝑣 = ∫ 𝑣𝑓𝑚 (𝑣)𝑑 3 (𝑣) −∞

As 𝑓𝑚 = (

3 2

𝑚 2𝜋𝐾𝑇𝑒

) 𝑒

−𝑚𝑣2 2𝐾𝑇𝑒 ,

∞

and 𝑑3 𝑉 = 4𝜋𝑣 2 𝑑𝑉. 3

3

2

∞

2

−𝑚𝑣 𝑚 2 −𝑚𝑣 𝑚 2 ) 𝑒 2𝐾𝑇𝑒 . 4𝜋𝑣 2 𝑑𝑣 = 4𝜋 ( ) ∫ 𝑣 2 𝑒 2𝐾𝑇𝑒 . 𝑣𝑑𝑣 𝑣̅𝑒 = ∫ 𝑣. ( 2𝜋𝐾𝑇𝑒 2𝜋𝐾𝑇𝑒 −∞

Let 𝑥

=

−∞

𝑚 2𝐾𝑇𝑒 3

∞

𝑥 2 2 𝑣̅𝑒 = 4𝜋 ( ) ∫ 𝑣 2 𝑒 −𝑥𝑣 . 𝑣𝑑𝑣 𝜋 −∞

Let 𝑥𝑣 2 = 𝑦 =>

𝑑𝑦 𝑑𝑣

= 2𝑥𝑣 =>

𝑑𝑦 2𝑥

= 𝑣𝑑𝑣 , so 3

∞

𝑥 2 𝑦 𝑑𝑦 𝑣̅𝑒 = 4𝜋 ( ) ∫ 𝑒 −𝑦 𝜋 𝑥 2𝑥 −∞

3

 𝑣̅𝑒 =

∞ 𝑥 2 1 4𝜋 ( ) 2 ∫−∞ 𝑦𝑒 −𝑦 𝑑𝑦 𝜋 2𝑥

 𝑣̅𝑒 =

∞ 𝑥 2 1 4𝜋 ( ) 2 2 ∫0 𝑦𝑒 −𝑦 𝑑𝑦 𝜋 2𝑥

 𝑣̅𝑒 =

𝑥 2 1 ∞ 4𝜋 ( ) 2 ∫0 𝑦𝑒 −𝑦 𝑑𝑦 𝜋 𝑥

3

3

∞

By using Gamma function as ∫0 𝑦𝑒 −𝑦 𝑑𝑦 = 1, so 3

1 1 𝑥 2 1 4 𝑣̅𝑒 = 4𝜋 ( ) 2 = 4𝜋 −2 𝑥 −2 = 𝜋 𝑥 √𝑥𝜋

Course: Experimental Plasma Physics

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As 𝑥 =

𝑚 2𝐾𝑇𝑒

, so

𝑣̅𝑒 =

4

2𝐾𝑇𝑒 2𝐾𝑇𝑒 = 4√ 𝑚 𝑚𝜋 √𝜋 √

Using this result in equation (A)

1 𝐼𝑠𝑒 = 𝑒𝐴𝑝 𝑛𝑒 ⃗⃗⃗ 𝑣𝑒 4 1

2𝐾𝑇𝑒

4

𝑚𝜋

 𝐼𝑠𝑒 = 𝑒𝐴𝑝 𝑛𝑒 (4√  𝑛𝑒 =

𝐼𝑠𝑒 𝑒𝐴𝑝

)

𝑚𝜋

√2𝐾𝑇

𝑒

Where ‘Ap’ is probe area, e is the magnitude of electronic charge, ‘m’ is electron mass and ‘K’ is Boltzmann’s constant. By measuring the electron saturation current and electron temperature, the electron density may be readily evaluated.

PLASMA SPECTROSCOPY It is the study of spectra of atoms and molecules. A spectrum is a chart or a graph that shows the intensity of light being emitted over a range of energies. Spectra can be produced for any energy of light from which useful information can be collected. In the field of plasma diagnostics, plasma spectroscopy has evolved as a very informative tool to study characteristics of different species. These characteristics include temperature, particle density, and history of plasma etc.

ATOMIC ABSORPTION SPECTROSCOPY When a beam of light passes through a material then intensities of certain wavelengths are found reduced and this phenomenon is attributed to absorption. Three basic objectives to study spectroscopy of the absorption of light are   

To learn which wavelengths are absorbed How much radiation is absorbed Why a radiation is absorbed

The basic instrument used for atomic absorption spectroscopy include a light source, a monochromator to isolate the specific wavelength of light, a detector, some electrons to treat signal, and a data display.

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ATOMIC EMISSION SPECTROSCOPY Atomic emission spectroscopy is used to study the spectra emitted from the excited atoms or ions especially in optical range. In plasma emission spectroscopy, the exciting medium/agent is plasma. In emission spectroscopy, outer orbital electrons in the atoms become excited from their ground state to higher energy state. After a short lifetime, the excited electrons return to the ground state, simultaneously electromagnetic radiations are emitted. The emitted radiations are analyzed by means of monochromator/spectrograph which separates various wavelengths. PES is similar to optical emission spectroscopy in principles but differs in mechanism. In OES the excitation is done by electric arc or spark mechanism in the temperature range 3000-5000 degrees kelvin. On reaching the high temperature (7000-9000 K) plasma, all type of molecular bonds break. Free atoms or ions are produced which emit their characteristics spectra. Plasma spectroscopy techniques are prepared to collect information about plasma compared to other methods because these techniques used the photons emitted from plasma and photons emissions cause negligible perturbations. Secondly, the information is more reliable quantitatively as well as qualitatively.

OPTICAL EMISSION SPECTROSCOPY OES is the spectral analysis of light emitting from plasma in optical range. It is most widely used technique to investigate glow discharges. By measuring the wavelength and intensities of emitted spectral lines one can identify the neutral particles and ions present in plasma. Although optical methods are non-intrusive, they have been of limited utility in plasmas because of selectivity and sensitivity issues. For example: OES is limited to species that emit characteristic light which is small fraction of total number of species present in the system. The most intense radiation is emitted from the species through de-excitation from first excited state E1 to ground state E0. Since every species (atoms, molecules, ion radicals) have certain energy levels, therefore, each emits a characteristic spectral line of frequency. 𝜈10 =

𝐸1 − 𝐸𝑜 ℎ

𝜆10 =

ℎ𝑐 𝐸1 − 𝐸𝑜

And wavelength

The major tools employed in optical emission spectroscopy includes monochromator and photomultiplier tube.

Course: Experimental Plasma Physics

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MONOCHROMATOR A monochromator can be used as a wavelength analyzer for unknown source of light and secondly as a source of known wavelength. A monochromator consists of and entrance slit, a dispersion device, a focusing system and an exit slit. A dispersing medium may be a prism or diffraction grating. The dispersive medium separates different wavelengths in different directions. Grating is used extensively as compared to prism. The wavelength of monochromator varies with the choice of grating. As grating is rotated, a different wavelength is selected to focus onto the exit slit.

The dispersion of monochromator is defined as the power to spread different wavelengths. Dispersion is of two types i.e. linear and angular dispersion. ∆𝜃

Mathematically, angular dispersion is ∆𝜆, where ‘∆𝜃’ is difference in emerging angles of two

rays having difference ‘∆𝜆’ in wavelengths. It is expressed in degrees or radians per angstrom. ∆𝑥

For linear dispersion ∆𝜆, where ‘∆𝑥’ is distance between two spectral lines. Linear dispersion is expressed in millimeter per angstrom. PHTOTOMULTIPLIER TUBE It is a sensitive device to detect very faint light signals and convert an extremely weak light signal into a detectable electrical signal without adding a large amount of noise. It consist of a photocathode and a series of dynodes in an evacuated enclosure. The photons that strike the photo emissive cathode emit electrons due to photoelectric effect. From there, the photoelectrons are accelerated towards the nearest electrode by means of potential difference. The photoelectrons eject secondary electrons from the first anode. This anode act as a cathode for next electrode and the process continues through several stages. The cascading effect creates 105 to 107 electrons for each photon hitting the first cathode, depending upon the number of dynodes and the accelerating voltage. Finally, this amplified signal can be recorded on a computer.

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DEDUCTIONS OF PLASMA PARAMETER WITH OES OES can be used as diagnostic tool for plasma, electron plasma temperature and plasma electron density can be found with this technique. ELECTRON PLASMA TEMPERATURE A commonly employed convenient method of temperature determines its two-line-emission ratio method, which yields electronic excitation temperature ‘𝑇𝑒𝑥 ’ that can be equated to electron temperature ‘𝑇𝑒 ’. The pair of lines which have common lower level is chosen for calculation of ‘𝑇𝑒𝑥𝑐 ’. For a system in thermodynamic equilibrium at temperature ‘T’, the relation which holds well is called Boltzmann’s equation. 𝑁𝐽 𝑔𝐽 −∆𝐸⁄𝐾𝑇 = 𝑒 → (𝐴) 𝑁𝐾 𝑔𝐾 Where 𝑁𝐽 and 𝑁𝐾 are population densities of upper and lower level J and K along with 𝑔𝐽 and 𝑔𝐾 as their statistical weights respectively. The level ‘r’ is common lower level for the upper ‘J’ and ’K’ levels.

The spectral line intensity corresponding to a transition from an upper level ‘J’ to a lower level ‘r’ is 𝐼𝐽𝑟 = 𝑁𝐽 𝐴𝐽𝑟 ℎ𝑣𝐽𝑟 = 𝑁𝐽 𝐴𝐽𝑟

ℎ𝑐 → (1) 𝜆𝐽𝑟

Similarly for K 𝐼𝐾𝑟 = 𝑁𝐾 𝐴𝐾𝑟 ℎ𝑣𝐾𝑟 = 𝑁𝐾 𝐴𝐾𝑟

ℎ𝑐 → (2) 𝜆𝐾𝑟

From equation (1) 𝑁𝐽 =

𝐼𝐽𝑟 𝜆𝐽𝑟 𝐴𝐽𝑟 ℎ𝑐

𝑁𝐾 =

𝐼𝐾𝑟 𝜆𝐾𝑟 𝐴𝐾𝑟 ℎ𝑐

Similarly from (2)

So 𝑁𝐽 𝐼𝐽𝑟 𝜆𝐽𝑟 𝐴𝐾𝑟 ℎ𝑐 𝐼𝐽𝑟 𝜆𝐽𝑟 𝐴𝐾𝑟 = . = → (𝐵) 𝑁𝐾 𝐴𝐽𝑟 ℎ𝑐 𝐼𝐾𝑟 𝜆𝐾𝑟 𝐼𝐾𝑟 𝜆𝐾𝑟 𝐴𝐽𝑟 Course: Experimental Plasma Physics

24

Comparing (A) and (B) 𝑔𝐽 −∆𝐸⁄𝐾𝑇 𝐼𝐽𝑟 𝜆𝐽𝑟 𝐴𝐾𝑟 𝑒 = 𝑔𝐾 𝐼𝐾𝑟 𝜆𝐾𝑟 𝐴𝐽𝑟 Taking natural log 𝑙𝑛 (

𝑔𝐽 −∆𝐸⁄𝐾𝑇 𝐼𝐽𝑟 𝜆𝐽𝑟 𝐴𝐾𝑟 𝑒 ) = 𝑙𝑛 ( ) 𝑔𝐾 𝐼𝐾𝑟 𝜆𝐾𝑟 𝐴𝐽𝑟

𝑔𝐽 (𝐸𝐽 − 𝐸𝐾 ) 𝐼𝐽𝑟 𝜆𝐽𝑟 𝐴𝐾𝑟 𝑙𝑛 ( ) − = 𝑙𝑛 ( ) 𝑔𝐾 𝐾𝑇 𝐼𝐾𝑟 𝜆𝐾𝑟 𝐴𝐽𝑟 (𝐸𝐾 − 𝐸𝐽 ) 𝐼𝐽𝑟 𝜆𝐽𝑟 𝐴𝐾𝑟 𝑔𝐾 = 𝑙𝑛 ( ) 𝐾𝑇 𝐼𝐾𝑟 𝜆𝐾𝑟 𝐴𝐽𝑟 𝑔𝐽 𝐾𝑇 =

(𝐸𝐾 − 𝐸𝐽 ) 𝐼𝐽𝑟 𝜆𝐽𝑟 𝐴𝐾𝑟 𝑔𝐾 𝑙𝑛 ( ) 𝐼𝐾𝑟 𝜆𝐾𝑟 𝐴𝐽𝑟 𝑔𝐽

Where 𝐼𝐾𝑟 , 𝜆𝐾𝑟 , 𝐴𝐾𝑟 and 𝑔𝐾 are total intensity (integrated over profile), the wavelength, the transition probability and the statistical weight, respectively of like ‘K’ with ′𝐸𝐾 ′ its excitation energy. The corresponding quantities for other line ‘J’ are 𝐼𝐽𝑟 , 𝜆𝐽𝑟, 𝐴𝐽𝑟 and 𝑔𝐽 . If we take K=1 and J=2, the above equation takes the form 𝐾𝑇 =

(𝐸1 − 𝐸2 ) 𝐴 𝑔𝐼𝜆 𝑙𝑛 ( 1 1 2 2 ) 𝐴2 𝑔2 𝐼1 𝜆1

ELECTRON PLASMA DENSITY The electron number density can be determined from the measurement of the relative intensities of atomic and ionic lines by using Boltzmann and Saha’s equations.

(2𝜋𝑚𝐾𝑇)3⁄2 2𝐴+ 𝑔+ 𝜆0 𝐼 0 −(𝐸 +−𝐸0 +𝐸0 −∆𝐸0 )⁄𝐾𝑇 𝑒𝑥 𝑖 𝑖 𝑛𝑒 = ( )( 0 0 + + )𝑒 3 ℎ 𝐴 𝑔 𝜆 𝐼 Where (0,+) represent the neutral and singly ionized atom. Tex = excitation temperature E = energy of emitting level Ei0 = ionization energy of neutral atom ∆Ei0 = lowering of ionization energy

The great probabilist Mark Kac (1914-1984) once gave a lecture at Caltech, with Feynman in the audience. When Kac finished, Feynman stood up and loudly proclaimed, "If all mathematics disappeared, it would set physics back precisely one week." To that outrageous comment, Kac shot back with that yes, he knew of that week; it was "Precisely the week in which God created the world."

Hisham Shah

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CHAPTER # 4 PINCH EFFECT The constriction of a plasma through which a large electric current is flowing, caused by the attractive force of the current’s own magnetic field. Consider an infinite cylindrical column of conducting fluid with an axial current density J and a resulting azimuthal magnetic induction ‘B’. The 𝐽 × 𝐵 force acting on a plasma forces the column to contract radially. This radial constriction of the plasma column is known as the pinch effect. There are two types of pinch effect 1. EQUILIBRIUM PINCH When plasma is compressed radially, the plasma number density and the temperature increase. The plasma kinetic pressure counteracts to hinder the constriction of the plasma column. Whereas the magnetic force acts to confine plasma. When these counteracting forces are balanced, a steady state condition results in which the plasma is mainly confined within a certain radius ‘R’ which remains constant in this. This situation is called equilibrium pinch. 2. DYNAMIC PINCH When the self-magnetic pressure exceeds the plasma kinetic pressure the column radius changes with time resulting in a situation known as dynamic pinch. As charged particles are present in plasma having own magnetic field known as selfmagnetic pressure. First we investigate the equilibrium pinch and afterwards the dynamic pinch.

EQUILIBRIUM PINCH For simplicity the current density, the magnetic field and plasma magnetic pressure are assumed only on the distance from cylindrical axis. For steady state conditions, none of the variable changes with time. Since the system is cylindrically symmetric, only radial component must be considered. For equilibrium pinch kinetic and magnetic pressures are equal, then rate of change of total pressure equals to the product of current density and magnetic field. 𝑑𝑃(𝑟) = −𝐽𝑧 (𝑟)𝐵𝜃 (𝑟) → (1) 𝑑𝑟 Course: Experimental Plasma Physics

26

Now total enclosed current 𝐼𝑧 (𝑟) is 𝑟

𝐼𝑧 (𝑟) = ∫ 𝐽𝑧 (𝑟)2𝜋𝑟𝑑𝑟 0

We can write 𝑑𝐼𝑧 (𝑟) = 2𝜋𝑟𝐽𝑧 (𝑟) 𝑑𝑟 𝐽𝑧 (𝑟) =

1 𝑑𝐼𝑧 (𝑟) . → (2) 2𝜋𝑟 𝑑𝑟

Ampere’s law in integral form relates 𝐵𝜃 (𝑟) to the total enclosed current. As 𝐵𝜃 (𝑟) =

𝜇𝑜 𝐼 (𝑟) → (3) 2𝜋𝑟 𝑧

Using value of 𝐼𝑧 (𝑟) we get 𝑟

𝑟

𝜇𝑜 𝜇𝑜 𝐵𝜃 (𝑟) = . ∫ 𝐽𝑧 (𝑟). 2𝜋𝑟𝑑𝑟 = . ∫ 𝐽𝑧 (𝑟). 𝑟𝑑𝑟 2𝜋𝑟 𝑟 0

0

Using (2) and (3) in (1), we get 𝑑𝑃(𝑟) = −𝐽𝑧 (𝑟)𝐵𝜃 (𝑟) 𝑑𝑟

   

𝑑𝑃(𝑟) 𝑑𝑟 𝑑𝑃(𝑟)

=( =

−1 𝑑𝐼𝑧 (𝑟)

2𝜋𝑟 −𝜇𝑜

4𝜋2 𝑟 2

𝑑𝑟 𝑑𝑃(𝑟) 4𝜋 2 𝑟 2 𝑑𝑟 𝑑𝑃(𝑟) 4𝜋 2 𝑟 2 𝑑𝑟

.

𝑑𝑟

𝐼𝑧 (𝑟).

𝜇

𝑜 𝐼𝑧 (𝑟)) ) (2𝜋𝑟

𝑑𝐼𝑧 (𝑟) 𝑑𝑟

= −𝜇𝑜 𝐼𝑧 (𝑟). =

𝑑𝐼𝑧 (𝑟)

𝑑𝑟 − ( 𝜇𝑜 𝐼𝑧2 (𝑟)) 𝑑𝑟 2 𝑑

1

Integrating this equation from 𝑟 = 0 to 𝑟 = 𝑅 𝑅

𝑅

𝑟 2 𝑑𝑃(𝑟) 𝑑 1 2 4𝜋 ∫ 𝑑𝑟 = − ∫ ( 𝜇𝑜 𝐼𝑧2 (𝑟)) 𝑑𝑟 𝑑𝑟 𝑑𝑟 2 0

0

𝑅

𝑅

𝑑𝑃(𝑟) 𝑑 𝑑𝑃(𝑟) 1 4𝜋 2 [𝑟 2 ∫ 𝑑𝑟 − ∫ ( 𝑟 2 ) (∫ 𝑑𝑟) 𝑑𝑟] = − 𝜇𝑜 𝐼02 𝑑𝑟 𝑑𝑟 𝑑𝑟 2 0

4𝜋 2 [𝑟 2 . 𝑃(𝑟) |

0

𝑅 0

Hisham Shah

𝑅

1 − ∫ 2𝑟. 𝑃(𝑟)𝑑𝑟] = − 𝜇𝑜 𝐼02 2 0

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4𝜋 2 𝑟 2 𝑃(𝑟) |

𝑅 0

𝑅

1 − 4𝜋 ∫ 2𝜋𝑟. 𝑃(𝑟)𝑑𝑟 = − 𝜇𝑜 𝐼02 2 0

Where 𝐼0 = 𝐼𝑧 (𝑅) is total current flowing through the entire cross section of plasma column. Considering plasma column to be confined to the range 0 ≤ 𝑟 < 𝑅, it follows that 𝑃(𝑟) is zero for 𝑟 ≥ 𝑅 and finite for 0 ≤ 𝑟 < 𝑅. So as 𝑃(𝑟) = 0, 𝑅

1 −4𝜋 ∫ 2𝜋𝑟𝑃(𝑟)𝑑𝑟 = − 𝜇𝑜 𝐼02 2 0

Or 𝑅

8𝜋 𝐼𝑜2 = ∫ 2𝜋𝑟𝑃(𝑟)𝑑𝑟 → (4) 𝜇𝑜 0

Now if the partial pressures of electrons and ions are governed by ideal gas law, then 𝑃𝑒 (𝑟) = 𝑛(𝑟)𝐾𝑇𝑒 𝑃𝑖 (𝑟) = 𝑛(𝑟)𝐾𝑇𝑖 The total pressure will be 𝑃(𝑟) = 𝑃𝑒 (𝑟) + 𝑃𝑖 (𝑟) 𝑃(𝑟) = 𝑛(𝑟)𝐾𝑇𝑒 + 𝑛(𝑟)𝐾𝑇𝑖 𝑃(𝑟) = 𝑛(𝑟)𝐾(𝑇𝑒 + 𝑇𝑖 ) Using this in equation (4) we get 𝑅

8𝜋 ∫ 2𝜋𝑟𝑛(𝑟)𝐾(𝑇𝑒 + 𝑇𝑖 )𝑑𝑟 = 𝐼𝑜2 → (4) 𝜇𝑜 0

𝑅

𝐼𝑜2

8𝜋 = 𝐾(𝑇𝑒 + 𝑇𝑖 ) ∫ 2𝜋𝑟𝑛(𝑟)𝑑𝑟 𝜇𝑜 0

𝑅

As 𝑁𝑙 = ∫0 2𝜋𝑟𝑛(𝑟)𝑑𝑟 is the number of particles per unit length of plasma column, so 𝐼𝑜2 =

8𝜋 𝐾(𝑇𝑒 + 𝑇𝑖 )𝑁𝑙 𝜇𝑜

This relation is called Bennet relation. It gives the total current that must be discharged through the plasma column in order to confine a plasma at a specified temperature and a given number of particles ‘𝑁𝑙 ’ per unit length. The current required for the confinement of hot plasma is usually very large. Course: Experimental Plasma Physics

28

 𝐓𝐨 𝐨𝐛𝐭𝐚𝐢𝐧 𝐭𝐡𝐞 𝐫𝐚𝐝𝐢𝐚𝐥 𝐝𝐢𝐬𝐭𝐫𝐢𝐛𝐮𝐭𝐢𝐨𝐧 𝐨𝐟 𝐏(𝐫)𝐢𝐧 𝐭𝐞𝐫𝐦𝐬 𝐨𝐟 𝐁𝛉 (𝐫) From Maxwell’s equation ∇ × B = μo J In cylindrical coordinates with radial dependence, we have 1𝑑 [𝑟𝐵𝜃 (𝑟)] = 𝜇𝑜 𝐽𝑧 (𝑟) 𝑟 𝑑𝑟 1 𝑑 𝑑 [𝑟 𝐵𝜃 (𝑟) + 𝐵𝜃 (𝑟) 𝑟] = 𝜇𝑜 𝐽𝑧 (𝑟) 𝑟 𝑑𝑟 𝑑𝑟 𝑑 𝐵𝜃 (𝑟) 𝑑 𝐵𝜃 (𝑟) + 𝑟 = 𝜇𝑜 𝐽𝑧 (𝑟) 𝑑𝑟 𝑟 𝑑𝑟 𝐽𝑧 (𝑟) =

1 𝑑 1 𝐵𝜃 (𝑟) 𝐵𝜃 (𝑟) + 𝜇𝑜 𝑑𝑟 𝜇𝑜 𝑟

As 𝑑𝑃(𝑟) = −𝐽𝑧 (𝑟)𝐵𝜃 (𝑟) 𝑑𝑟 Using value of 𝐽𝑧 (𝑟)

 

𝑑𝑃(𝑟) 𝑑𝑟 𝑑𝑃(𝑟) 𝑑𝑟

= −[ = −[

1 𝑑

𝐵𝜃 (𝑟) +

𝜇𝑜 𝑑𝑟 𝐵𝜃 (𝑟) 𝑑 𝜇𝑜

𝑑𝑟

1 𝐵𝜃 (𝑟) 𝜇𝑜

] 𝐵𝜃 (𝑟)

𝑟 𝐵𝜃2 (𝑟)

𝐵𝜃 (𝑟) +

𝜇𝑜 𝑟

]

Multiply and divide by 2𝑟 2 and taking 𝜇𝑜 common

 

𝑑𝑃(𝑟) 𝑑𝑟 𝑑𝑃(𝑟) 𝑑𝑟

=− =−

2𝑟 2 1 2𝑟 2 𝜇𝑜 1 2𝑟 2 𝜇𝑜

[𝐵𝜃 (𝑟)

𝑑 𝑑𝑟

𝐵𝜃 (𝑟) +

[2𝑟 2 𝐵𝜃 (𝑟)

The term in bracket can be written as

𝑑 𝑑𝑟

𝑑 𝑑𝑟

𝐵𝜃2 (𝑟) 𝑟

]

𝐵𝜃 (𝑟) + 2𝑟𝐵𝜃2 (𝑟)]

𝑟 2 𝐵𝜃2 (𝑟).

As 𝑑 2 2 𝑑 𝑟 𝐵𝜃 (𝑟) = 𝑟 2 . 2𝐵𝜃 (𝑟) 𝐵𝜃 (𝑟) + 𝐵𝜃2 (𝑟). 2𝑟 𝑑𝑟 𝑑𝑟 So 𝑑𝑃(𝑟) 1 𝑑 2 2 =− 2 𝑟 𝐵𝜃 (𝑟) 𝑑𝑟 2𝑟 𝜇𝑜 𝑑𝑟 Integrating this equation from 𝑟 = 0 to a general radius 𝑟 𝑟

𝑟

𝑑𝑃(𝑟) 1 𝑑 ∫ = − 2 ∫ 𝑟 2 𝐵𝜃2 (𝑟)𝑑𝑟 𝑑𝑟 2𝑟 𝜇𝑜 𝑑𝑟 0

Hisham Shah

0

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29 𝑟

𝑟

1 1 𝑑 𝑃(𝑟) | = − ∫ 2 𝑟 2 𝐵𝜃2 (𝑟)𝑑𝑟 2𝜇𝑜 𝑟 𝑑𝑟 0 0 𝑟

1 1 𝑑 𝑃(𝑟) − 𝑃(0) = − ∫ 2 𝑟 2 𝐵𝜃2 (𝑟)𝑑𝑟 → (𝐴) 2𝜇𝑜 𝑟 𝑑𝑟 0

Since for 𝑟 = 𝑅 we have 𝑃(𝑅) = 0, so 𝑅

1 1 𝑑 𝑃(0) = ∫ 2 𝑟 2 𝐵𝜃2 (𝑟)𝑑𝑟 2𝜇𝑜 𝑟 𝑑𝑟 0

Using this in equation A 𝑅

𝑟

1 1 𝑑 1 1 𝑑 𝑃(𝑟) = ∫ 2 𝑟 2 𝐵𝜃2 (𝑟)𝑑𝑟 − ∫ 2 𝑟 2 𝐵𝜃2 (𝑟)𝑑𝑟 → (𝐴) 2𝜇𝑜 𝑟 𝑑𝑟 2𝜇𝑜 𝑟 𝑑𝑟 0

0

0

𝑅

1 1 𝑑 1 1 𝑑 𝑃(𝑟) = [∫ 2 𝑟 2 𝐵𝜃2 (𝑟)𝑑𝑟 + ∫ 2 𝑟 2 𝐵𝜃2 (𝑟)𝑑𝑟] 2𝜇𝑜 𝑟 𝑑𝑟 2𝜇𝑜 𝑟 𝑑𝑟 0

0

𝑅

𝑟

𝑅

We can write ∫𝑟 + ∫0 = ∫𝑟 , so 𝑅

1 1 𝑑 𝑃(𝑟) = ∫ 2 𝑟 2 𝐵𝜃2 (𝑟)𝑑𝑟 2𝜇𝑜 𝑟 𝑑𝑟 𝑟

The average pressure 𝑃̅ inside the cylinder can be related to the total current 𝐼𝑜 and column radius R as 𝑅

1 𝑃̅ = ∫(2𝜋𝑟)𝑃(𝑟)𝑑𝑟 𝜋𝑅 2 0

𝑅

1 𝑃̅ = 2𝜋 ∫ 𝑟𝑃(𝑟)𝑑𝑟 𝜋𝑅 2 0

𝑅

2 𝑟2 𝑅 𝑑𝑃(𝑟) 𝑃̅ = 2 [𝑃(𝑟). | − ∫ ( ∫ 𝑟𝑑𝑟) 𝑑𝑟] 𝑅 2 𝑑𝑟 0 0 As 𝑃(𝑅) = 0 and 𝑃(0) = 0 𝑅

2 1 𝑑𝑃(𝑟) 𝑃̅ = 2 [− ∫ 𝑟 2 𝑑𝑟] 𝑅 2 𝑑𝑟 0

Course: Experimental Plasma Physics

30 𝑅

1 𝑑𝑃(𝑟) 𝑃̅ = − 2 [∫ 𝑟 2 𝑑𝑟] 𝑅 𝑑𝑟 0

As

𝑑𝑃(𝑟) 𝑑𝑟

=−

1

𝑑

2𝑟 2 𝜇𝑜 𝑑𝑟

𝑟 2 𝐵𝜃2 (𝑟) 𝑅

1 1 𝑑 2 2 𝑃̅ = 2 ∫ 𝑟 2 . 2 (𝑟 𝐵𝜃 (𝑟)) 𝑑𝑟 𝑅 2𝑟 𝜇𝑜 𝑑𝑟 0

𝑅

1 𝑑 𝑃̅ = ∫ (𝑟 2 𝐵𝜃2 (𝑟)) 𝑑𝑟 2 2𝜇𝑜 𝑅 𝑑𝑟 0

𝑃̅ = 𝑃̅ =

𝑅 1 2 2 (𝑟) 𝑟 𝐵 | 𝜃 2𝜇𝑜 𝑅2 0

1 (𝑅 2 𝐵𝜃2 (𝑅) − 0) 2𝜇𝑜 𝑅 2 𝑃̅ =

As 𝐵𝜃 (𝑅) =

𝜇𝑜 𝐼𝑜 2𝜋𝑅

=> 𝐵𝜃2 (𝑅) = 𝑃̅ =

𝐵𝜃2 (𝑅) 2𝜇𝑜

𝜇𝑜2 𝐼𝑜2 4𝜋2 𝑅 2

1 𝜇𝑜2 𝐼𝑜2 𝜇𝑜 𝐼𝑜2 . 2 2 = 2 2 → (𝐵) 2𝜇𝑜 4𝜋 𝑅 8𝜋 𝑅

This result shows that the average kinetic pressure in this equilibrium plasma column is balanced by the magnetic pressure at the boundary. As an example, consider the case in which the current density 𝐽𝑧 (𝑟) is constant for 𝑟 < 𝑅. 𝐼 Taking 𝐽𝑧 = 𝑜⁄𝜋𝑅 2 As 𝑟

𝜇𝑜 𝐵𝜃 (𝑟) = ∫ 𝐽𝑧 (𝑟)𝑟𝑑𝑟 𝑅 0

𝑟

𝜇𝑜 𝐼𝑜 𝐵𝜃 (𝑟) = ∫ 𝑟𝑑𝑟 𝑟 𝜋𝑅 2 0

𝐵𝜃 (𝑟) =

𝜇𝑜 𝐼𝑜 𝑟 2 𝜇𝑜 𝐼𝑜 𝑟 𝜇𝑜 𝐼𝑜 𝑟 = = (𝑟 < 𝑅) 𝑟 𝜋𝑅 2 2 𝜋𝑅 2 2 2𝜋𝑅 2

As

Hisham Shah

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31 𝑅

1 1 𝑑 𝑃(𝑟) = ∫ 2 𝑟 2 𝐵𝜃2 (𝑟)𝑑𝑟 2𝜇𝑜 𝑟 𝑑𝑟 𝑟

So 𝑅

𝑅

1 1 𝑑 𝜇2𝐼2𝑟 2 𝜇𝑜2 𝐼𝑜2 1 1 𝑑 2 𝑜 𝑜 => 𝑃(𝑟) = ∫ 2 (𝑟 . 2 4 ) 𝑑𝑟 = . 2 4 ∫ 2 𝑟 4 𝑑𝑟 2𝜇𝑜 𝑟 𝑑𝑟 4𝜋 𝑅 2𝜇𝑜 4𝜋 𝑅 𝑟 𝑑𝑟 𝑟

𝑟

𝑅

𝑅

𝜇𝑜 𝐼𝑜2 1 𝜇𝑜 𝐼𝑜2 𝜇𝑜 𝐼𝑜2 𝑟 2 𝑅 𝜇𝑜 𝐼𝑜2 => 𝑃(𝑟) = 2 4 ∫ 2 4𝑟 3 𝑑𝑟 = 2 4 ∫ 𝑟𝑑𝑟 = 2 4 | = 2 4 (𝑅 2 − 𝑟 2 ) 8𝜋 𝑅 𝑟 2𝜋 𝑅 2𝜋 𝑅 2 4𝜋 𝑅 𝑟 𝑟 𝑟 𝜇𝑜 𝐼𝑜2 𝑅 2 𝑟 2 𝜇𝑜 𝐼𝑜2 𝑟2 => 𝑃(𝑟) = 2 2 ( 2 − 2 ) = 2 2 (1 − 2 ) 4𝜋 𝑅 𝑅 𝑅 4𝜋 𝑅 𝑅 In this case the axial pressure 𝑃(0) is twice the average pressure given in (B). 𝑃̅ = 𝑃(𝑟) =

𝜇𝑜 𝐼𝑜2 8𝜋 2 𝑅 2

𝜇𝑜 𝐼𝑜2 𝑟2 (1 − ) 4𝜋 2 𝑅 2 𝑅2

2𝜇𝑜 𝐼𝑜2 𝑟2 𝑃(𝑟) = 2 2 (1 − 2 ) 8𝜋 𝑅 𝑅 𝑃(𝑟) = 2𝑃̅ (1 −

𝑟2 ) 𝑅2

This equation shows that axial pressure 𝑃(𝑟) is twice the average pressure.

THE BENNET PINCH W. H. Bennet, the discoverer of pinch effect, investigated a special model of the equilibrium longitudinal pinch in which the radial distribution of various quantities are such that the drift velocity of plasma particle is constant throughout the column cross section. Ion mass is much larger than the electron mass, so drift velocity of ion is much smaller than that of electrons and therefore neglected in a first approximation. We consider the current density to be given by, 𝐽(𝑟) = −𝑒𝑛(𝑟)𝑢𝑒 Since the applied electric field is in 2 directions, we have 𝐽(𝑟) = 𝐽𝑧 (𝑟)𝑧̅, and 𝑢𝑒 − 𝑢𝑒𝑧 𝑧̂ , where 𝑢𝑒𝑧 is positive and constant, independent of r, therefore, 𝐽𝑧 (𝑟) = 𝑒𝑛(𝑟)𝑢𝑒𝑧 As 𝑝𝑒 (𝑟) + 𝑝𝑖 (𝑟) = 𝑝(𝑟) = 𝑛(𝑟)𝐾(𝑇𝑒 + 𝑇𝑖 ) And 𝑑𝑃(𝑟) = −𝐽𝑧 (𝑟)𝐵𝜃 (𝑟) 𝑑𝑟 Course: Experimental Plasma Physics

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Using values or 𝑃(𝑟) and 𝐽𝑧 (𝑟), we get 𝐾(𝑇𝑒 + 𝑇𝑖 )

𝑑 𝑛(𝑟) = −𝑒𝑛(𝑟)𝑢𝑒𝑧 𝐵𝜃 (𝑟) 𝑑𝑟

Multiplying both sides by 𝑟⁄𝑛(𝑟)𝐾(𝑇 + 𝑇 ) 𝑒 𝑖 𝑟 𝑑 𝑟 . 𝐾(𝑇𝑒 + 𝑇𝑖 ) 𝑛(𝑟) = . (−𝑒𝑛(𝑟)𝑢𝑒𝑧 𝐵𝜃 (𝑟)) 𝑛(𝑟)𝐾(𝑇𝑒 + 𝑇𝑖 ) 𝑑𝑟 𝑛(𝑟)𝐾(𝑇𝑒 + 𝑇𝑖 ) 𝑟 𝑑 −𝑒𝑢𝑒𝑧 . 𝑛(𝑟) = . 𝑟(𝐵𝜃 (𝑟)) 𝑛(𝑟) 𝑑𝑟 𝐾(𝑇𝑒 + 𝑇𝑖 ) Differentiating both sides w.r.t “r” 𝑑 𝑟 𝑑 −𝑒𝑢𝑒𝑧 𝑑 [𝑟𝐵𝜃 (𝑟)] [ . 𝑛(𝑟)] = 𝑑𝑟 𝑛(𝑟) 𝑑𝑟 𝐾(𝑇𝑒 + 𝑇𝑖 ) 𝑑𝑟 From equation … 1𝑑 [𝑟𝐵𝜃 (𝑟)] = 𝜇𝑜 𝐽𝑧 (𝑟) 𝑟 𝑑𝑟 And as 𝐽𝑧 (𝑟) = 𝑒𝑛(𝑟)𝑢𝑒𝑧 𝑑 [𝑟𝐵𝜃 (𝑟)] = 𝑟𝜇𝑜 𝑒𝑛(𝑟)𝑢𝑒𝑧 𝑑𝑟 Using this result, we get 2 𝑑 𝑟 𝑑 𝑒 2 𝑢𝑒𝑧 𝜇𝑜 [ . 𝑛(𝑟)] − [ ] 𝑟𝑛(𝑟) = 0 → (𝐶) 𝑑𝑟 𝑛(𝑟) 𝑑𝑟 𝐾(𝑇𝑒 + 𝑇𝑖 )

The solution of this nonlinear differential eqn gives radial dependence of number density ‘𝑛(𝑟)’. Bennet obtained the solution of this nonlinear equation subjected to the boundary condition that 𝑛(𝑟) is symmetric about z-axis, where 𝑟 = 0 [

𝑑𝑛(𝑟) ] =0 𝑑𝑟 𝑟=0

Solution of eq C subjected to boundary condition is knows as Bennet distribution and is given by 𝑛(𝑟) =

𝑛0 (1 + 𝑛𝑜 𝑏𝑟 2 )2

Where 𝑛𝑜 = 𝑛(0) which is number density on the axis, and 𝑏=

Hisham Shah

2 𝜇𝑜 𝑒 2 𝑢𝑒𝑧 8𝐾(𝑇𝑒 + 𝑇𝑖 )

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33

The Bennet distribution shows that the particles are present upto density but since ‘𝑛(𝑟)’ falls off very rapidly with increasing values of ‘r’. We obtain number of particles 𝑁𝑙 (𝑅) per unit length contained a cylindrical column of radius ‘R’. 𝑅

𝑅

𝑁𝑙 (𝑅) = ∫ 𝑛(𝑟)2𝜋𝑟𝑑𝑟 = 2𝜋𝑛𝑜 ∫ 0

0

𝑟 𝑑𝑟 (1 + 𝑛𝑜 𝑏𝑟 2 )2

Let 𝑡 = 1 + 𝑛𝑜 𝑏𝑟 2 𝑑𝑡 𝑑𝑡 = 2𝑛𝑜 𝑏𝑟 => 2𝑟𝑑𝑟 = 𝑑𝑟 𝑛𝑜 𝑏 𝑅

𝑁𝑙 (𝑅) = 𝜋𝑛𝑜 ∫ 0

2𝑟 𝑑𝑟 (1 + 𝑛𝑜 𝑏𝑟 2 )2

1+𝑛𝑜 𝑏𝑟 2

𝑁𝑙 (𝑅) = 𝜋𝑛𝑜

∫ 1

𝑁𝑙 (𝑅) =

𝜋 𝑏

1 1 𝑑𝑡 2 𝑡 𝑛𝑜 𝑏

1+𝑛𝑜 𝑏𝑟 2

∫

𝑡 −2 𝑑𝑡

1

2 𝜋 𝑡 −1 1 + 𝑛𝑜 𝑏𝑟 𝑁𝑙 (𝑅) = | 𝑏 −1 1 𝜋 𝑁𝑙 (𝑅) = − [(1 + 𝑛𝑜 𝑏𝑅 2 )−1 − (1)−1 ] 𝑏

𝜋 1 𝜋 −𝑛𝑜 𝑏𝑅 𝑁𝑙 (𝑅) = − [ − 1] = − [ ] 2 𝑏 (1 + 𝑛𝑜 𝑏𝑅 ) 𝑏 (1 + 𝑛𝑜 𝑏𝑅 2 ) Particles are present upto infinity, total number of particles per unit length can be obtained by taking 𝑅 → ∞

Course: Experimental Plasma Physics

34

𝜋 1 𝜋 𝑁𝑙 (𝑅) = − [ − 1] = 𝑏 ∞ 𝑏 If we let and denote the fraction of the number of particles per unit length that is contained in a cylinder of radius R, that is 𝛼=

𝑁𝑙 (𝑅) 𝑏 = 𝑁 (𝑅) 𝑁𝑙 (∞) 𝜋 𝑙

𝛼=

𝑏 𝑛𝑜 𝜋𝑅 2 𝜋 (1 + 𝑛𝑜 𝑏𝑅 2 )

𝛼(1 + 𝑛𝑜 𝑏𝑅 2 ) = 𝑏𝑛𝑜 𝑅 2 𝛼 = 𝑏𝑛𝑜 𝑅 2 − 𝛼𝑛𝑜 𝑏𝑅 2 𝛼 = 𝑏𝑛𝑜 𝑅 2 (1 − 𝛼) 𝛼 = 𝑏𝑛𝑜 𝑅 2 1−𝛼 Taking under root 𝛼 1−𝛼

√𝑏𝑛𝑜 𝑅 = √

Therefore 90% of plasma particles are confined within the cylindrical plasma column of radius ‘R’ that is 𝛼 = 0.9 0.9 0.9 =√ = √9 = 3 √𝑏𝑛𝑜 𝑅 = √ 1 − 0.9 0.1 If we assume arbitrary that a plasma is confined within a cylindrical surface of radius ‘R’. If 90% of particles are within this cylindrical column, then this radius must satisfy the above equation.

INSTABILITIES IN A PINCHED PLASMA COLUMN Although it is possible to achieve an equilibrium state for plasma confinement with the pinch effect, this equilibrium style is not stable. The growth of instabilities is the reason why it is difficult to sustain reasonably long lived pinched plasma in the laboratory. In the following discussion of instabilities we shall consider a perfectly diamagnetic plasma column confined by a static magnetic field. Since the plasma is perfectly diamagnetic, there is no magnetic field and consequently no magnetic pressure inside the plasma column. The plasma kinetic pressure is assumed to be uniform inside the plasma and vanishes outside it.

Hisham Shah

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In the equilibrium state, the magnetic pressure of the plasma surface ‘𝑃𝑚𝑜 ’ must be equal to the kinetic pressure ‘P’ of plasma. 𝑃 = 𝑃𝑚𝑜 =

𝐵𝜃2 2𝜇𝑜

Where ‘𝐵𝜃 ’ is magnitude of magnetic flux density at the plasma surface. This situation of a sharp plasma is an idealized one and is difficult to create in laboratory.

THE SAUSAGE INSTABILITY Consider equilibrium state of the pinched plasma column is distributed by a wave perturbation with the crests and troughs on the surface of plasma column. We shall consider that the plasma is constricted in some locations and expanded at others in such a way that its volume doesn’t change consequently the uniform kinetic pressure of plasma is left unchanged. At the locations where radius has decreased in relation to equilibrium value, the magnetic pressure at the constricted plasma surface radially inward, thus enhancing the constrictor. At the locations where radius has become larger than the equilibrium value, the plasma kinetic pressure will be larger than magnetic pressure at the expanded plasma surface and will force the surface radially outwards. Therefore the troughs will become deeper and crests higher. So the equilibrium state is unstable. When the constrictor reach the axis, the column appears like a surface of sausages, for this reason this type of instability is known as sausage instability.

Course: Experimental Plasma Physics

36

The sausage instability can be inhibited by a longitudinal magnetic field applied inside the plasma column. This longitudinal magnetic field can be produced by passing a current through a solenoidal coil wound around the column. When sausage distortion starts to grow the longitudinal magnetic field lines are compressed at the constrictions causing increase in total pressure inside plasma at the location where radius has increased thus decreasing the total internal pressure. If the radius ‘r’ of the column at the column at the constriction is decreased by an amount ‘dr’, considering magnetic flux 𝜙𝑚 = 𝐵𝑧 𝜋𝑟 2 𝑑𝜙𝑚 𝑑𝐵𝑧 = 𝜋𝑟 2 + 𝐵𝑧 . 2𝜋𝑟 = 0 𝑑𝑟 𝑑𝑟 𝑑𝜙𝑚 = 𝜋𝑟 2 𝑑𝐵𝑧 + 𝐵𝑧 . 2𝜋𝑟𝑑𝑟 = 0 𝜋𝑟 2 𝑑𝐵𝑧 = −𝐵𝑧 . 2𝜋𝑟𝑑𝑟 𝑑𝐵𝑧 = −2𝐵𝑧

𝑑𝑟 → (1) 𝑟

Consequently the corresponding internal magnetic pressure increases as 𝑃𝑧 =

𝐵𝑧2 1 => 𝑑𝑃𝑧 = 𝐵 𝑑𝐵 2𝜇𝑜 𝜇𝑜 𝑧 𝑧

Using 𝑑𝐵𝑧 from (1) 𝑑𝑃𝑧 =

1 𝑑𝑟 1 𝑑𝑟 𝐵𝑧 (−2𝐵𝑧 ) = − 2𝐵𝑧2 𝜇𝑜 𝑟 𝜇𝑜 𝑟

Considering now the azimuthal flux density ‘𝐵𝜃 ’. From Ampere’s law, 𝑟𝐵𝜃 (𝑟) = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑟𝑑𝐵𝜃 (𝑟) + 𝐵𝜃 (𝑟)𝑑𝑟 = 0 𝑑𝐵𝜃 (𝑟) = −𝐵𝜃 (𝑟)

𝑑𝑟 → (3) 𝑟

As 𝑃𝜃 =

𝐵𝜃2 𝑑𝑃𝜃 2𝐵𝜃 𝐵𝜃 => = => 𝑑𝑃𝜃 = 𝑑𝐵𝜃 2𝜇𝑜 𝑑𝐵𝜃 2𝜇𝑜 𝜇𝑜

Using 𝑑𝐵𝜃 from eq (3) 𝑑𝑃𝜃 =

𝐵𝜃 𝑑𝑟 (−𝐵𝜃 (𝑟) ) 𝜇𝑜 𝑟

𝐵𝜃2 (𝑟) 𝑑𝑟 𝑑𝑃𝜃 = 𝜇𝑜 𝑟 From (2) and (4) we can say that 𝑑𝑃𝑧 > 𝑑𝑃𝜃 . Condition in order to plasma column be stable 1 against sausage distortion or 𝐵𝑧2 > 2 𝐵𝜃2 Hisham Shah

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37

THE KINK INSTABILITY The kink distortion consists of a perturbation in the form of a bend or kink in the column, but with the disturbed column maintaining its uniform circular cross-section.

Usually there may be several kinks along the column length. In the neighborhood of the column, where kink has developed the magnetic field lines are brought closer together on convex side. Therefore the changes in the external magnetic pressure are in such a way as to accentuate the distortion still. Thus the type of distortion is therefore unstable.

FAMOUS QUOTES ON FAILURES We should not get afraid of failing; we should embrace failure with open arms. It’s only through failure that we learn. Here are some of my favorite quotes on failure. “I’ve missed more than 9,000 shots in my career. I’ve lost almost 300 games. 26 times I’ve been trusted to take the game's winning shot and missed. I’ve failed over and over and over again in my life and that's why I succeed.”

– Michael Jordan, NBA Hall of Famer “She turned him down because he had no prospects, he grew up to become an oil baron and the don of Wall Street.”

– John D. Rockefeller, Standard Oil – Soichiro Honda, Honda Motors

“He was turned down for an engineering job by Toyota”

"When everything seems to be going against you, remember that the airplane takes off against the wind, not with it." – Henry Ford, Ford Motors “It’s important to question your sanity because at the point which you stop questioning your sanity, you’re probably insane.” – Elon Musk: Tesla, SpaceX, SolarCity, PayPal, OpenAI, Hyperloop “Don’t be embarrassed by your failures, learn from them and start again.” – Richard Branson, Virgin "I have not failed. I’ve just found 10,000 ways that won’t work."

– Thomas Edison

“If you’re not embarrassed by first version of your product, you’ve launched too late.” – Reid Hoffman “You have to be willing to be misunderstood if you’re going to innovate.”

– Jeff Bezos, Amazon

Course: Experimental Plasma Physics

38

CHAPTER #5 PHYSICAL CONDITIONS FOR THERMONUCLEAR REACTIONS RATES OF THERMONUCLEAR REACTIONS (2.15-2.18) Consider a binary reaction in a system containing n1 nuclei per cm3 of one reacting species and n2 nuclei per cm3 of the other. To determine the rate at which the two nuclear species interact, it may be supposed that the nuclei of the first kind form a stationary lattice within which the nuclei of second kind move at random with a constant velocity 𝑣 cm/sec, equal to the relative velocity of the nuclei. The rate of thermonuclear energy production is readily obtained from the reaction rate “R 12” for a system of two species of nuclei of number densities n1 and n2 and is given by 𝑅12 = 𝑛1 𝑛2 < σ𝑣 > interactions⁄cm3 . 𝑠 → (1) If the reaction occurs between the two nuclei of the same kind, e.g. two deuterons, so that n1 and n2 are equal, the expression for nuclear reaction rate, represented by R12 becomes 1 𝑅11 = 𝑛2 < σ𝑣 > interactions⁄cm3 . 𝑠 → (2) 2 Where ‘n’ is number of reactant nuclei per cm3. In order that each interaction between identical nuclei should not be counted twice, a factor of ½ is introduced in above equation. If the velocity distribution is maxwellian, the equation for the distribution in terms of relative velocity is obtained upon substituting the reduced mass ‘M’ of the interacting nuclei for the individual masses. Thus 3

2

𝑀 2 (−𝑀𝑣 )𝑣 2 𝑑𝑣 𝑑𝑛 = 𝑛 ( ) 𝑒 2𝑘𝑇 → (3) 2𝜋𝑘𝑇 Where dn is the number of particles whose velocities relative to that of a given particle lie in the range from 𝑣 𝑡𝑜 𝑣 + 𝑑𝑣. Hence it follows that ̅̅̅̅ = σ𝑉

∞ ∫0 σ𝑣𝑑𝑛 ∞ ∫0 𝑑𝑛

=

𝑀𝑣 2 (− ) ∞ 2𝑘𝑇 𝑣 2 𝑑𝑣 σ𝑣𝑒 ∫0 𝑀𝑣 2 ∞ (− 2𝑘𝑇 ) 2 𝑒 𝑣 𝑑𝑣 ∫0

→ (4)

The denominator of the above equation is ∞

∫𝑒 0

Hisham Shah

(−

𝑚𝑣 2 ) 2𝑘𝑇 𝑣 2 𝑑𝑣

3

2𝑘𝑇 2 √𝜋 =( ) 𝑀 4

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39 3 ∞

2

𝑀𝑣 𝑀 2 (− ) ̅σ𝑣 ̅̅̅ = ( ) ∫ σ𝑒 2𝑘𝑇 𝑣 3 𝑑𝑣 → (5) √𝜋 2𝑘𝑇

4

0

The integral in equation (5) can be evaluated by changing the variable. Since nuclear cross sections are always determined and expressed as a function of the energy of the bombarding particle, the bombarded particle being essentially at rest in the target, the actual velocity of bombarding nucleus is also its relative velocity. Hence if ‘W’ is the actual energy, in the laboratory system, of the bombarding nucleus of mass ‘m’, then 1

1 2𝑊 2 𝑊 = 𝑚𝑣 2 ; 𝑉 = ( ) 2 𝑚 1



𝑑𝑣 𝑑𝑊

1 2𝑊 −2 2

= ( 2

 𝑑𝑣 = (

)

𝑚

𝑚 2𝑊

)

=(

𝑚

1 2 1

𝑚

 𝑣 𝑑𝑣 = (  𝑣 3 𝑑𝑣 =

2𝑊 2

𝑀 2𝑊 𝑚2

2𝑊

)

1 2 1

𝑚

𝑑𝑊

3

3

𝑚

) (

𝑀 2𝑊

)

1 2 1

𝑚

𝑑𝑊

𝑑𝑊

Now eq (5) implies that 3 ∞

2

𝑀𝑣 2𝑊 𝑀 2 (− ) ̅̅̅̅ σ𝑉 = ( ) ∫ σ𝑒 2𝑘𝑇 𝑑𝑊 𝑚2 √𝜋 2𝑘𝑇

4

0

 ̅̅̅̅ σ𝑉 =  ̅̅̅̅ σ𝑉 =

4

(

3 2

𝑀

√𝜋 2𝐾𝑇 8

(

𝑀

√𝜋 2𝐾𝑇

) . 3 2

) .

2

∞

2𝑊

(−𝑀 ⁄2𝐾𝑇 ) 𝑚 𝑊 𝑑𝑊 ∫ σ𝑒 𝑚2 0 1

∞

𝑀𝑊

(− ) ∫ σ𝑒 𝑚𝐾𝑇 𝑊 𝑑𝑊 𝑚2 0

Where σ in the integrand is the cross section for a bombarding nucleus of mass ‘m’ and energy ‘W’.

NUCLEAR FUSION REACTIONS (2.21) Thermonuclear fusion reaction offers and inexhaustible source of energy for the future. In this process two light nuclei combine to form a heavier one, the total final mass being slightly less than the total initial mass. The mass difference ∆𝑚 appears as energy ‘E’ according to Einstein’s famous equation 𝐸 = ∆𝑚𝑐 2 . It is believed that such a source of energy will provide easy, cheap and relatively radiative free energy for our future needs. There are two possible ways to obtain fusion. 1. An energetic light nuclei beam may be directed to stationary nuclei in solid or gaseous form. The beam and target nuclei undergo fusion reactions and this process is called

Course: Experimental Plasma Physics

40

beam target mechanism. However, this technique does not work, because most of deuterons lose energy by scattering before undergoing a fusion reaction. 2. The light nucleic gas may be heated to sufficiently high temperature and confined for sufficiently long time. The gas obtains the Maxwellian distribution. The nuclei undergo fusion reactions and this process is known as thermonuclear fusion. To achieve thermonuclear fusion energy, one must solve two problems  

Produce and heat a plasma fuel to thermonuclear fusion temperatures Confine it long enough to produce more fusion energy than expanded in heating and containing the fuel.

These twin requirements are quantified by a relation known as Lawson’s criterion. The reactions of highest interest are controlled thermonuclear fusion are as follows 2 1D 1D

+ 1D2 →

2

2 1D

+ 1D2 → + 1T 3 →

2He 1T

3

+ 0n1 + 3.27 MeV

3

+ 1H1 + 4.03 MeV

4

+ 0n1 + 17.6 MeV

2He

Here the first two reactions are respectively the neutron branch and the proton branch of the DD-reaction. The tritium produced in the proton branch or obtained in other way as explained above, can then react, at a considerably faster rate, with deuteron nuclei in the D-T-reaction. Among all these nuclear fusion reactions, the deuterium-tritium 1D2 + 1T 3 → 2He4 + 1 4 0n has large cross sections even at relatively low (~10 𝐾𝑒𝑉) temperature, and has large Qvalue as compared to 1D2 + 1D2 → reactions. For those reasons. CHARGED PARTICLE ENERGY (2.24-2.25) To access how much energy will be carried by the end products, one can simply use the conservation laws of energy and momentum. If m and m’ are the masses of particles produced in a given fusion reaction and V and V’ are their respective velocities, then the energy carried by the end products can be written by using conservative laws. 𝑚𝑣 = 𝑚′ 𝑣 ′ 2

2

 𝑚2 𝑣 2 = 𝑚′ 𝑣 ′ 2  𝑚(𝑚𝑣 2 ) = 𝑚′ (𝑚′ 𝑣 ′ )  𝑚𝑞 = 𝑚′𝑞′ Where q and q’ are the energies carried by the particles of masses m and m′. The total energy release of the nuclear reaction is Q. then As 𝑚𝑞 = 𝑚′ 𝑞 ′

 𝑞′ =

𝑚𝑞 𝑚′

 𝑄=𝑞+

Hisham Shah

𝑚𝑞 𝑚′

=

𝑚′ 𝑞+ 𝑚𝑞 𝑚′

=

𝑞(𝑚′ + 𝑚) 𝑚′

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41

 𝑞=

𝑚′ 𝑄 𝑚′ +𝑚

Similarly 𝑞 =

𝑚′ 𝑞 ′ +𝑞′𝑚 𝑚

=

𝑞′(𝑚+𝑚′) 𝑚

𝑚𝑄

 𝑞 ′ = 𝑚+𝑚′; where Q is the total energy released in a typical fusion reaction. Using this result, one can re-write the reactions as 2 1D

+ 1D2 →

2He

3

3

2 1D

+ 1D2 →

1T

2 1D

+ 1T 3 →

2He

2

+ 2He3 →

1D

(0.82 MeV) + 0n1 (2.45 MeV)

(1.01 MeV) + 1H1 (3.02 MeV) 4

2He

(3.5 MeV) + 0n1 (14.1 MeV)

4

(3.6 MeV) + 1H1 (14.7 MeV)

Notice that in thermonuclear fusion reaction high energy neutrons are emitted which would practically escape from reacting system and would deposit their energy elsewhere. Only energy of charged particles will be retained within reaction region. THE LAWSON CRITERION The usual Lawson criterion can be obtained by balancing the fusion energy release against the energy by investment in bringing the fuel to the required thermonuclear fusion temperatures and the energy loss through Bremsstrahlung and Cyclotron radiations. 𝐸𝑓𝑢𝑠𝑖𝑜𝑛 = 𝐸𝑡ℎ𝑒𝑟𝑚𝑎𝑙 + 𝐸𝑟𝑎𝑑 The fusion energy release can be expressed in terms of fusion reactions rate and confinement time ‘𝜏’as 𝐸𝑓𝑢𝑠𝑖𝑜𝑛

𝑛2 = < σ𝑉 > 𝑄𝜏 4

Where < σ𝑉 > is the Maxwellian averaged reaction rate parameter, Q is the fusion energy released per fusion reaction and 𝜏 is the plasma confinement time. We have assumed equimolar density of the fusion fuel. The thermal energy assuming ideal gas behavior of the plasma is given by 3 3 𝐸𝑡ℎ𝑒𝑟𝑚𝑎𝑙 = 𝑛𝐾𝑇𝑖 + 𝑛𝐾𝑇𝑒 = 3𝑛𝐾𝑇 2 2 Where we assume 𝑇𝑒 = 𝑇𝑖 = 𝑇 If we balance the fusion energy released against thermal energy and neglect radiation energy loss for simplicity, then we have 𝑛2 < σ𝑉 > 𝑄𝜏 > 3𝑛𝐾𝑇 4 𝑛𝜏 >

12𝑛𝐾𝑇 < σ𝑉 > 𝑄

Course: Experimental Plasma Physics

42

It is evident from the above expression that 𝑛𝜏 product is a function of temperature alone. At suitable temperature, the 𝑛𝜏 criterion becomes nτ > 1014 sec⁄cm3 for DT reactions nτ > 1016 sec⁄cm3 for DD reactions The criterion provides confinement time ‘𝜏’ for a given number density ‘m’ of the plasma. In Magnetic Confinement Fusion, 𝑛 𝑖𝑠 ~(1015 − 1016 )𝑐𝑚−3 and therefore the confinement time is (0.1 − 10)𝑠𝑒𝑐

2.47: Radiation losses from Plasma (Bremsstrahlung) 2.77: Cyclotron radiation

ASSIGNMENT QUESTIONS Q No: 1 (2.3-2.4) (a) Lets we have two nuclei of charge 𝑍1 𝑒 𝑎𝑛𝑑 𝑍2 𝑒. The separation between the nuclei ≈ 5 Fermi. So the energy required to overview the Coulomb repulsion so that the fusion can occur is given by 𝐸=

𝑍! 𝑍2 𝑒 2 𝑅

As 𝑅 = 5 𝐹𝑒𝑟𝑚𝑖 = 5 ∗ 1013 𝑐𝑚 and 𝑒 = 4.8 × 10−10 esu (statcoulomb) 𝐸=

𝑍! 𝑍2 (4.8 × 10−10 )2 = 𝑍1 𝑍2 . 4.608 ∗ 10−7 𝑒𝑟𝑔𝑠 5 × 1013

1 𝑒𝑟𝑔 =

1 𝑒𝑉 1.602 × 10−12

4.608 × 10−7 𝐸 = 𝑍1 𝑍2 𝑒𝑉 1.602 × 10−12  𝐸 = 𝑍1 𝑍2 × 2.876 ∗ 105 𝑒𝑉  𝐸 = 𝑍1 𝑍2 × 0.2876 ∗ 106 𝑒𝑉  𝐸 = 𝑍1 𝑍2 × 0.2876𝑀𝑒𝑉 For hydrogen isotopes 𝑍1 = 𝑍2 = 1  𝐸 = 0.2876𝑀𝑒𝑉 This is the required amount of energy to surround the Coulomb’s barrier (b) “One atomic mass unit (1 amu) is equal to the 1/16 of the mass of 8𝑂16 ”. that is 1 1 1 𝑎𝑡𝑜𝑚𝑖𝑐 𝑚𝑜𝑙𝑒 𝑜𝑓 8𝑂16 1 16 𝑔 𝑜𝑓 8𝑂16 16 1 𝑎𝑚𝑢 = mass of 8𝑂 = = 16 16 𝐴𝑣𝑜𝑔𝑎𝑑𝑟𝑜′ 𝑠 𝑛𝑢𝑚𝑏𝑒𝑟 16 6.02 × 1023 Hisham Shah

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43 1𝑔

 1 𝑎𝑚𝑢 = 6.02∗1023 = 0.166 × 10−23 𝑔 Now 𝐸 = 𝑚𝑐 2  𝐸 = (0.166 × 10−23 )(3 × 1010 )2 𝑒𝑟𝑔𝑠  𝐸 = 0.166 × 10−23 ∗ 9 × 1020 𝑒𝑟𝑔𝑠  𝐸 = 1.494 × 10−3 𝑒𝑟𝑔𝑠 1

As 1 𝑒𝑟𝑔 = 1.602∗10−12 𝑒𝑉  𝐸=

1.494×10−3 1.602×10−12

𝑒𝑉 = 0.9325 × 109 𝑒𝑉 = 932.5 𝑀𝑒𝑉 1 𝑎𝑚𝑢 = 932.5 𝑀𝑒𝑉

Q No: 2 “It is the amount of energy released in a nuclear reaction” Generally a reaction is written as 𝑋+𝑎 =𝑌+𝑏+𝑄 Where Q is the energy released in the reaction and is called Q-value of the reaction. (1) For D-D Reaction To calculate the ‘Q’ value of this reaction, we proceed as follows 2 1D

+ P2 → →

2He 1 1H

3

+ 0n1 + 1T 3

mass of 1D2 = 2.014743 amu mass of 2He3 = 3.016986 amu mass of 0n1 = 1.008987 amu mass of 1H1 = 1.008145 amu mass of 1T 3 = 3.017005 amu (I) Now ∆𝑚 = (𝑚𝐷 + 𝑚𝐷 ) − (𝑚𝐻𝑒 3 + 𝑚𝑛 )  ∆𝑚 = 4.029486 − 4.025973 = 0.003513 𝑎𝑚𝑢 So 𝑄 = ∆𝑚𝑐 2  𝑄 = 0.003513 𝑎𝑚𝑢. 𝑐 2 = 0.003513 ∗ 931.5𝑀𝑒𝑉 = 3.27 𝑀𝑒𝑉 (II) ∆𝑚 = (𝑚𝐷 + 𝑚𝐷 ) − (𝑚𝐻 1 + 𝑚 𝑇 3 )  ∆𝑚 = 4.029486 − 4.02515 = 0.004336 𝑎𝑚𝑢 So 𝑄 = ∆𝑚𝑐 2

Course: Experimental Plasma Physics

44

 𝑄 = 0.004336 × 931.5𝑀𝑒𝑉 = 4.03 𝑀𝑒𝑉 To calculate the energy level shared among the reaction products 𝑞𝐻𝑒 3 =

𝑚𝑛 𝑄 1.008987 × 3.27 3.299 = = = 0.82 𝑀𝑒𝑉 𝑚𝑛 + 𝑚𝐻𝑒 3 1.008797 × 3.016986 4.025973

Now 𝑞𝑛 = 𝑞𝐻 1 =

𝑚𝐻𝑒 3 𝑄 3.016986 × 3.27 = = 2.45 𝑀𝑒𝑉 𝑚𝑛 + 𝑚𝐻𝑒 3 4.025973

𝑚𝑇3 𝑄 3.017005 × 4.03 12.15853015 = = = 3.02 𝑀𝑒𝑉 𝑚𝐻 1 + 𝑚 𝑇 3 1.008145 + 3.017005 4.02515

And 𝑚𝐻 1 𝑄 1.008145 × 4.03 = = 1.01 𝑀𝑒𝑉 𝑚𝐻 1 + 𝑚 𝑇 3 4.02515

𝑞𝑇3 = (2) D-T reaction

1D

2

+ T3 →

2He

4

+ 0n1 (17.6 𝑀𝑒𝑉)

Now ∆𝑚 = (𝑚𝐷2 + 𝑚 𝑇 ) − (𝑚𝐻𝑒 4 + 𝑚𝑛 )  ∆𝑚 = (2.014743 + 3.017005) − (4.003874 + 1.008987) = 0.018887 𝑎𝑚𝑢 Now 𝑄 = ∆𝑚𝑐 2  𝑄 = 0.018887 ∗ 931.5 𝑀𝑒𝑉 = 17.58 𝑀𝑒𝑉 And 𝑞𝐻𝑒 4 = 𝑞

1 0𝑛

=

𝑚𝑛 𝑄 1.008987 × 17.6 = = 3.54 𝑀𝑒𝑉 𝑚𝑛 + 𝑚𝐻𝑒 4 5.012861 𝑚𝐻𝑒 4 𝑄 4.003874 × 17.6 = = 14.05 𝑀𝑒𝑉 𝑚𝑛 + 𝑚𝐻𝑒 4 5.012861

(3) D-He3 reaction 1𝐷

2

+ 2𝐻𝑒 3 →

2𝐻𝑒

4

→ 1𝐻1 (18.3 𝑀𝑒𝑉)

∆𝑚 = (𝑚𝐷 + 𝑚𝐻𝑒 3 ) − (𝑚𝐻𝑒 4 + 𝑚𝐻 1 ) ∆𝑚 = (2.014743 + 3.016986) − (4.003874 + 1.008145) = 0.01971 𝑎𝑚𝑢 Now 𝑄 = ∆𝑚𝑐 2  𝑄 = 0.01971 ∗ 931.5 𝑀𝑒𝑉 = 18.35 𝑀𝑒𝑉 Now

Hisham Shah

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45

𝑞𝐻𝑒 4 =

𝑚𝐻 1 𝑄 1.008145 × 18.3 = = 3.69 𝑀𝑒𝑉 𝑚𝐻 1 + 𝑚𝐻𝑒 4 5.012019

𝑞𝐻𝑒 1 =

𝑚𝐻𝑒 4 𝑄 4.003874 × 18.3 = = 14.65 𝑀𝑒𝑉 𝑚𝐻 1 + 𝑚𝐻𝑒 4 5.012019

(4) P-B11 reaction 1𝑃

2

+ 5𝐵11 → 2𝐻𝑒 4 → 4𝐵𝑒 8 (8.590 𝑀𝑒𝑉) (8.682 𝑀𝑒𝑉) → 3𝐻𝑒 4

(I) ∆𝑚 = (𝑚𝑃 + 𝑚𝐵11 ) − (𝑚𝐻𝑒 4 + 𝑚𝐵𝑒 8 )  ∆𝑚 = (1.007593 + 11.009305) − (1.002603 + 8.012053) = 0.002242 𝑎𝑚𝑢

Now 𝑄 = ∆𝑚𝑐 2  𝑄 = 0.002242 ∗ 931.5 𝑀𝑒𝑉 = 2.08 𝑀𝑒𝑉 (II) ∆𝑚 = (𝑚𝑃 + 𝑚𝐵11 ) − (𝑚3𝐻𝑒 4 )  ∆𝑚 = (1.007593 + 11.009305) − 3(4.003874) = 0.008782 𝑎𝑚𝑢 Now 𝑄 = ∆𝑚𝑐 2  𝑄 = 0.008782 ∗ 931.5 𝑀𝑒𝑉 = 8.18 𝑀𝑒𝑉 𝑚𝐵𝑒 8 𝑄 8.012053 × 2.08 𝑞𝐻𝑒 4 = = = 1.38 𝑀𝑒𝑉 𝑚𝐵𝑒 8 + 𝑚𝐻𝑒 4 4.003374 + 8.012053 𝑚𝐻𝑒 4 𝑄 4.003374 × 2.08 𝑞𝐵𝑒 8 = = = 0.69 𝑀𝑒𝑉 𝑚𝐻𝑒 + 𝑚𝐵𝑒 12.015927 Q No: 3 (I)

To calculate 𝑛𝜏 fir D-D reactions at 100 KeV we will proceed as follows.

We know that 𝑛𝜏 =

12𝐾𝑇 < 𝜎𝜈 > 𝑊

Where W is the Q-value for the reaction. < σν >DD for 100 KeV temperature = 4.5 ∗ 10−23 m3 /sec 12 × 100𝐾𝑒𝑉 12 × 100 × 103 𝑒𝑉 𝑛𝜏 = = 3 m3 −23 −23 m × 106 𝑒𝑉 4.5 × 10 × 3.65 𝑀𝑒𝑉 16.42 × 10 sec sec

 𝑛𝜏 =

12×105 𝑒𝑉 𝑐m3 16.42×10−17 ×106 𝑒𝑉 sec

= 0.73 × 1016 sec⁄cm3

 𝑛𝜏 ≈ 1016 sec⁄cm3

Course: Experimental Plasma Physics

46

(II)

𝑛𝜏 for D-T reactions: 𝑛𝜏 =

12𝐾𝑇 < 𝜎𝜈 >𝐷𝑇 𝑊

Here < 𝜎𝜈 >𝐷𝑇 = 7.0 ∗ 10−22 𝑚3 ⁄𝑠 and 𝑊 = 17.6 𝑀𝑒𝑉, so 12 × 100𝐾𝑒𝑉 12 × 105 𝑒𝑉 = m3 𝑐m3 7 × 10−22 sec × 17.6 𝑀𝑒𝑉 7 × 10−16 sec × 17.6 × 106 𝑒𝑉 𝑛𝜏 = 0.09 ∗ 1015 𝑠𝑒𝑐 ⁄𝑐𝑚3 𝑛𝜏 =

(III)

D-He3 𝑛𝜏 =

12𝐾𝑇 < 𝜎𝜈 > 𝑊

Here < 𝜎𝜈 >= 1.0 × 10−22 𝑚3 ⁄𝑠 and 𝑊 = 18.3 𝑀𝑒𝑉, so 𝑛𝜏 = (IV)

12 × 105 𝑒𝑉 = 0.65 × 1015 𝑠𝑒𝑐 ⁄𝑐𝑚3 3 𝑐m 1 × 10−16 sec × 18.3 × 106 𝑒𝑉

P-Li6 𝑛𝜏 =

12𝐾𝑇 < 𝜎𝜈 > 𝑊

Here < 𝜎𝜈 >= 0.8 ∗ 10−23 𝑚3 ⁄𝑠 and 𝑊 = 4.023 𝑀𝑒𝑉, so 𝑛𝜏 = (V)

12 × 105 𝑒𝑉 = 3.71 × 1016 𝑠𝑒𝑐 ⁄𝑐𝑚3 3 𝑐m 0.8 × 10−17 sec × 4.023 × 106 𝑒𝑉

P-B reaction 𝑛𝜏 =

12𝐾𝑇 < 𝜎𝜈 > 𝑊

Here < 𝜎𝜈 >= 5.0 × 10−23 𝑚3 ⁄𝑠 and 𝑊 = 8.682 𝑀𝑒𝑉, so 12 × 105 𝑒𝑉 𝑛𝜏 = = 0.27 × 1016 𝑠𝑒𝑐 ⁄𝑐𝑚3 3 𝑐m 5.0 × 10−17 sec × 8.682 × 106 𝑒𝑉

Hisham Shah

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