IPhO 2014 · Theory problem 3

Simplest model of gas discharge

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An electric current flowing through a gas is called a gas discharge. There are many types of gas discharges including glow discharge in lighting lamps, arc discharge in welding and the well known spark discharge that occurs between the clouds and the earth in the form of lightning.

Part A. Non-self-sustained gas discharge (4.8points)

In this part of the problem the so-called non-self-sustained gas discharge is studied. To maintain it a permanent operationan external ionizer is needed, which creates \(Z_{\mathrm{ext}}\) pairs of singly ionized ions and free electrons per unit volume and per unit time uniformly in the volume.

When an external ionizer is switched on, the number of electrons and ions starts to grow. Unlimited increase in the number densities of electrons and ions in the gas is prevented by the recombination process in which a free electron recombineswith an ion to form a neutral atom. The number of recombining events \(Z_{\mathrm{rec}}\) that occurs in the gas per unit volume and per unit time is given by

\[ Z_{\mathrm{rec}}=rn_en_i, \]

where \(r\) is a constant called the recombination coefficient, and \(n_e,n_i\) denotethe electron and ion number densities, respectively.

Suppose that at time \(t=0\) the external ionizer is switched on and the initial number densities of electrons and ions in the gas are both equal to zero. Then, the electron number density \(n_e(t)\) depends on time \(t\) as follows:

\[ n_e(t)=n_0+a\tanh bt, \]

where \(n_0,a\) and \(b\) are some constants, and \(\tanh x\) stands for the hyperbolic tangent.

A1 Find \(n_0,a,b\) and express them in terms of \(Z_{\mathrm{ext}}\) and \(r\). (1.8points)

Assume that there are two external ionizers available. When the first one is switched on, the electron number density in the gas reaches its equilibrium value of be \(n_{e1}=12\cdot10^{10}\,\mathrm{cm}^{-3}\). When the second external ionizer is switched on, the electron number density reaches its equilibrium value of \(n_{e2}=16\cdot10^{10}\,\mathrm{cm}^{-3}\).

A2 Find the electron number density \(n_e\) at equilibriumwhen both external ionizers are switched on simultaneously. (0.6 points)

Attention! In what follows it is assumed that the external ionizer is switched on for quite long period of time such that all processes have become stationary and do not depend on time. Completely neglect the electric field due the charge carriers.

Assume that the gas fills in the tube between the two parallel conductive plates of area \(S\) separated by the distance \(L\ll\sqrt S\) from each other. The voltage \(U\) is applied across the plates to create an electric field between them. Assume that the number densities of both kinds of charge carriers remain almost constant along the tube.

Assumethat both the electrons (denoted by thesubscript \(e\)) and the ions (denoted by the subscript \(i\)) acquire the same ordered speed \(v\) due to the electric field strength \(E\) found as

\[ v=\beta E, \]

where \(b\) isa constant called charge mobility.

A3 Express the electric current \(I\) in the tube in terms of \(U,\beta,L,S,Z_{\mathrm{ext}},r\) and \(e\) whichis the elementary charge. (1.7 points)
A4 Find the resistivity \(\rho_{\mathrm{gas}}\) of the gas at sufficiently small values of the voltage applied and express it in terms of \(\beta,L,Z_{\mathrm{ext}},r\) and \(e\). (0.7 points)

PartB. Self-sustained gas discharge (5.2 points)

In this part of the problem the ignition of the self-sustained gas discharge is consideredto show how the electric current in the tube becomes self-maintaining.

Attention! In the sequel assume that the external ionizer continues to operatewith the same \(Z_{\mathrm{ext}}\) rate, neglect the electric field due to the charge carriers such that the electric field is uniform along the tube, and the recombination can be completely ignored.

For the self-sustained gas discharge there are two important processes not considered above. The first process isasecondary electron emission, and the second one isa formation of electron avalanche. The secondary electron emission occurs when ions hit on the negative electrode, called a cathode, and the electrons are knocked out of it to move towards the positive electrode, called an anode. The ratio of the number of the knocked electrons \(\dot N_e\) per unit timeto the number of ions \(\dot N_i\) hitting the cathode per unit time is called the coefficient of the secondary electron emission, \(\gamma=\dot N_e/\dot N_i\). The formation of the electron avalanche isexplained as follows. The electric field accelerates free electrons which acquire enough kinetic energy to ionize the atoms in the gas by hitting them. As a result the number of free electrons moving towards the anodesignificantly increases. This process is described by the Townsend coefficient \(\alpha\), which characterizes an increase in the number of electrons \(dN_e\) due to moving \(N_e\) electrons that have passed the distance \(dl\), i.e.

\[ \frac{dN_e}{dl}=\alpha N_e. \]

Figure 1

The total current \(I\) in any cross section of the gas tube consists of the ion \(I_i(x)\) and the electron \(I_e(x)\) currents which, in the steady state, depend on the coordinate \(x\), shown in the figureabove. The electron current \(I_e(x)\) varies along the \(x\)-axis according to the formula

\[ I_e(x)=C_1e^{A_1x}+A_2, \]

where \(A_1,A_2,C_1\) are some constants.

B1 Find \(A_1,A_2\) and express them in terms of \(Z_{\mathrm{ext}},\alpha,e,L,S\). (2 points)

The ion current \(I_i(x)\) varies along the \(x\)-axis according to the formula

\[ I_i(x)=C_2+B_1e^{B_2x}, \]

where \(B_1,B_2,C_2\) are some constants.

B2 Find \(B_1,B_2\) and express them in terms of \(Z_{\mathrm{ext}},\alpha,e,L,S,C_1\). (0.6 points)
B3 Write down the condition for \(I_i(x)\) at \(x=L\). (0.3 points)
B4 Write down the condition for \(I_i(x)\) and \(I_e(x)\) at \(x=0\). (0.6 points)
B5 Find the total current \(I\) and express it in terms of \(Z_{\mathrm{ext}},\alpha,\gamma,e,L,S\). Assume that it remains finite. (1.2 points)

Let the Townsend coefficient \(\alpha\) be constant. When the length of the tube turns out greater than some critical value, i.e. \(L>L_{\mathrm{cr}}\), the external ionizer can be turned off and the discharge becomes self-sustained.

B6 Find \(L_{\mathrm{cr}}\) and express it in terms of \(Z_{\mathrm{ext}},\alpha,\gamma,e,L,S\). (0.5 points)

Official English text of the problem.

Answers

Currents below denote magnitudes, and \(U\geq0\).

  • A1 \[ \boxed{n_0=0,\qquad a=\sqrt{\frac{Z_{\mathrm{ext}}}{r}}, \qquad b=\sqrt{rZ_{\mathrm{ext}}}.} \]

  • A2 \[ \boxed{n_e=2.0\times10^{11}\,\mathrm{cm}^{-3} =2.0\times10^{17}\,\mathrm{m}^{-3}.} \]

  • A3, using the stated uniform-density collection approximation: \[ \boxed{ I=\frac{e\beta SU}{rL} \left[ \sqrt{\frac{\beta^2U^2}{L^4}+4rZ_{\mathrm{ext}}} -\frac{\beta U}{L^2} \right].} \]

  • A4 \[ \boxed{\rho_{\mathrm{gas}} =\frac{1}{2e\beta}\sqrt{\frac{r}{Z_{\mathrm{ext}}}}.} \]

  • B1 \[ \boxed{A_1=\alpha,\qquad A_2=-\frac{eSZ_{\mathrm{ext}}}{\alpha}.} \]

  • B2 \[ \boxed{B_1=-C_1,\qquad B_2=\alpha.} \]

  • B3 \[ \boxed{I_i(L)=0.} \]

  • B4 \[ \boxed{I_e(0)=\gamma I_i(0).} \]

  • B5 \[ \boxed{ I=\frac{eSZ_{\mathrm{ext}}}{\alpha} \frac{(1+\gamma)(e^{\alpha L}-1)} {1-\gamma(e^{\alpha L}-1)}.} \] A physical finite stationary solution with \(Z_{\mathrm{ext}}>0\) requires \[ \boxed{\gamma(e^{\alpha L}-1)<1.} \]

  • B6, for \(\alpha>0,\gamma>0\): \[ \boxed{L_{\mathrm{cr}}=\frac{1}{\alpha} \ln\!\left(1+\frac{1}{\gamma}\right).} \]

Solution

A1. Growth of the carrier density

Ionization creates one electron and one ion per event, while recombination removes one of each. Since their initial densities are equal, they remain equal:

\[ n_e(t)=n_i(t)=n(t). \]

The particle balance per unit volume is therefore

\[ \frac{dn}{dt}=Z_{\mathrm{ext}}-rn^2, \qquad n(0)=0. \]

The prescribed expression gives \(n(0)=n_0\), so

\[ n_0=0. \]

Substituting \(n=a\tanh(bt)\) into the balance equation,

\[ ab\left[1-\tanh^2(bt)\right] =Z_{\mathrm{ext}}-ra^2\tanh^2(bt). \]

Matching the constant terms and the coefficients of \(\tanh^2(bt)\) yields

\[ ab=Z_{\mathrm{ext}}, \qquad ab=ra^2. \]

Taking \(a,b\) positive,

\[ \boxed{a=\sqrt{\frac{Z_{\mathrm{ext}}}{r}}, \qquad b=\sqrt{rZ_{\mathrm{ext}}}.} \]

Thus

\[ n_e(t)=\sqrt{\frac{Z_{\mathrm{ext}}}{r}}\, \tanh\!\left(\sqrt{rZ_{\mathrm{ext}}}\,t\right). \]

Its initial slope is \(Z_{\mathrm{ext}}\), and its limiting density is \(\sqrt{Z_{\mathrm{ext}}/r}\), as required by generation–recombination balance. Simultaneously reversing the signs of \(a\) and \(b\) describes the same function.

A2. Two ionizers

Let the individual production rates be \(Z_1\) and \(Z_2\). At equilibrium,

\[ Z_1=rn_{e1}^2,\qquad Z_2=rn_{e2}^2. \]

The independent production rates add when both ionizers operate:

\[ rn_e^2=Z_1+Z_2=r(n_{e1}^2+n_{e2}^2). \]

Consequently,

\[ n_e=\sqrt{n_{e1}^2+n_{e2}^2} =\sqrt{12^2+16^2}\times10^{10}\,\mathrm{cm}^{-3}. \]

Therefore,

\[ \boxed{n_e=2.0\times10^{11}\,\mathrm{cm}^{-3} =2.0\times10^{17}\,\mathrm{m}^{-3}.} \]

The densities themselves do not add because the recombination rate is quadratic in density.

A3. Stationary current with recombination and collection

Neglecting edge effects and the field of the carriers,

\[ E=\frac UL,\qquad v=\frac{\beta U}{L}. \]

Apply the stipulated uniform-density approximation to the volume \(SL\). In a time \(dt\), a collecting electrode removes \(n_eSv\,dt\) electrons; the opposite electrode removes \(n_iSv\,dt\) ions. The stationary particle balances are

\[ Z_{\mathrm{ext}}SL-rn_en_iSL-n_eSv=0, \]

\[ Z_{\mathrm{ext}}SL-rn_en_iSL-n_iSv=0. \]

For \(v>0\), subtraction gives \(n_e=n_i=n\). Hence

\[ rn^2+\frac vL n-Z_{\mathrm{ext}}=0. \]

Only the nonnegative root is physical:

\[ n=\frac{1}{2r} \left[ \sqrt{\left(\frac vL\right)^2+4rZ_{\mathrm{ext}}} -\frac vL \right]. \]

Although the electrons and ions move in opposite directions, their charges also have opposite signs. Their conventional currents therefore add:

\[ I=eSn_ev+eSn_iv=2eSnv. \]

Substitution gives

\[ I=\frac{eSv}{r} \left[ \sqrt{\left(\frac vL\right)^2+4rZ_{\mathrm{ext}}} -\frac vL \right]. \]

Using \(v=\beta U/L\),

\[ \boxed{ I=\frac{e\beta SU}{rL} \left[ \sqrt{\frac{\beta^2U^2}{L^4}+4rZ_{\mathrm{ext}}} -\frac{\beta U}{L^2} \right]. } \]

This is the result of applying the spatially uniform collection model literally. Strictly, the densities cannot stay uniform up to the plates: the electron density must vanish at the cathode and the ion density at the anode. The formula is therefore reliable while these depleted layers are thin compared with \(L\), as in the small-voltage regime of A4; at high voltage it tends to \(2eZ_{\mathrm{ext}}SL\), twice the true saturation current \(eZ_{\mathrm{ext}}SL\).

A4. Small-voltage resistivity

For sufficiently small voltage,

\[ \frac{\beta U}{L^2}\ll\sqrt{rZ_{\mathrm{ext}}}, \]

collection perturbs the generation–recombination equilibrium only slightly. To leading order,

\[ n\simeq\sqrt{\frac{Z_{\mathrm{ext}}}{r}}. \]

Thus the current becomes proportional to voltage:

\[ I\simeq 2e\beta\sqrt{\frac{Z_{\mathrm{ext}}}{r}}\, \frac SL\,U. \]

Comparing this with

\[ I=\frac{S}{\rho_{\mathrm{gas}}L}\,U \]

gives

\[ \boxed{\rho_{\mathrm{gas}} =\frac{1}{2e\beta}\sqrt{\frac{r}{Z_{\mathrm{ext}}}}.} \]

The length \(L\) cancels. With SI quantities, the result is in \(\Omega\,\mathrm m\).

B1. Electron-current equation

In Part B, recombination is neglected. Take \(I_e\) and \(I_i\) as nonnegative current magnitudes, so that

\[ I=I_e(x)+I_i(x). \]

Consider a thin slice of thickness \(dx\). The external ionizer creates

\[ Z_{\mathrm{ext}}S\,dx \]

electrons per unit time in the slice. Avalanche multiplication creates an additional

\[ \alpha\frac{I_e(x)}e\,dx \]

electrons per unit time. Each new electron adds \(e\) times its production rate to the electron current leaving the slice. Therefore,

\[ I_e(x+dx)-I_e(x) =\left[eSZ_{\mathrm{ext}}+\alpha I_e(x)\right]dx. \]

In differential form,

\[ \frac{dI_e}{dx}=\alpha I_e+eSZ_{\mathrm{ext}}. \]

For constant \(\alpha>0\), its general solution is

\[ I_e(x)=C_1e^{\alpha x}-\frac{eSZ_{\mathrm{ext}}}{\alpha}. \]

Comparison with the prescribed expression gives

\[ \boxed{A_1=\alpha,\qquad A_2=-\frac{eSZ_{\mathrm{ext}}}{\alpha}.} \]

B2. Ion-current equation

Every ionization event also produces one ion. Ions move toward decreasing \(x\), so the ion current leaving the slice through its left face exceeds that entering through its right face:

\[ I_i(x)-I_i(x+dx) =\left[eSZ_{\mathrm{ext}}+\alpha I_e(x)\right]dx. \]

Thus

\[ \frac{dI_i}{dx} =-eSZ_{\mathrm{ext}}-\alpha I_e(x). \]

Using the electron-current solution,

\[ \frac{dI_i}{dx}=-\alpha C_1e^{\alpha x}, \]

and integration gives

\[ I_i(x)=C_2-C_1e^{\alpha x}. \]

Consequently,

\[ \boxed{B_1=-C_1,\qquad B_2=\alpha.} \]

The two differential equations also show that

\[ \frac{d}{dx}(I_e+I_i)=0, \]

which expresses stationary charge conservation.

B3. Boundary condition at the anode

The anode is at \(x=L\). No ions enter the gas from this electrode, and ions created inside the gas drift away from it. Therefore,

\[ \boxed{I_i(L)=0.} \]

Equivalently,

\[ C_2=C_1e^{\alpha L}. \]

B4. Boundary condition at the cathode

At \(x=0\), the ion arrival rate is \(I_i(0)/e\). Each arriving ion releases, on average, \(\gamma\) electrons. Hence

\[ \frac{I_e(0)}e=\gamma\frac{I_i(0)}e, \]

or

\[ \boxed{I_e(0)=\gamma I_i(0).} \]

B5. Total stationary current

For brevity, define

\[ K=\frac{eSZ_{\mathrm{ext}}}{\alpha}, \qquad T=e^{\alpha L}. \]

Then

\[ I_e(x)=C_1e^{\alpha x}-K, \qquad I_i(x)=C_2-C_1e^{\alpha x}. \]

The anode condition gives

\[ C_2=C_1T. \]

The cathode condition becomes

\[ C_1-K=\gamma(C_2-C_1) =\gamma C_1(T-1). \]

Therefore,

\[ C_1=\frac{K}{1-\gamma(T-1)}. \]

Adding the electron and ion currents,

\[ I=C_2-K=C_1T-K. \]

Substituting \(C_1\) and simplifying,

\[ I =K\frac{T-1+\gamma(T-1)} {1-\gamma(T-1)}. \]

Thus

\[ \boxed{ I=\frac{eSZ_{\mathrm{ext}}}{\alpha} \frac{(1+\gamma)(e^{\alpha L}-1)} {1-\gamma(e^{\alpha L}-1)}. } \]

With positive external ionization, a physical finite stationary current requires

\[ \boxed{1-\gamma(e^{\alpha L}-1)>0.} \]

At zero denominator, the linear stationary model predicts divergence; a negative denominator would give an unphysical negative current magnitude.

B6. Critical length for self-sustained discharge

One electron emitted from the cathode grows into

\[ e^{\alpha L} \]

electrons by the time the avalanche reaches the anode. Since the original electron was already present, the avalanche creates

\[ e^{\alpha L}-1 \]

new electron–ion pairs.

With recombination neglected, all these ions reach the cathode. They release, on average,

\[ \mathcal R=\gamma(e^{\alpha L}-1) \]

secondary electrons. This is the number of cathode electrons produced by one preceding cathode electron.

If \(\mathcal R<1\), successive generations decrease and external ionization is necessary. The threshold occurs when each generation exactly replaces itself:

\[ \gamma(e^{\alpha L_{\mathrm{cr}}}-1)=1. \]

Solving,

\[ \boxed{ L_{\mathrm{cr}} =\frac{1}{\alpha} \ln\!\left(1+\frac{1}{\gamma}\right). } \]

For \(L>L_{\mathrm{cr}}\), the feedback amplifies a seed discharge even after the external ionizer is removed. The present linear model determines the ignition threshold; additional physical effects are needed to determine a finite operating current above threshold.

Neither \(Z_{\mathrm{ext}}\), \(e\), nor \(S\) enters the threshold. If \(\alpha=0\) or \(\gamma=0\), this mechanism cannot sustain a discharge at any finite length.

The solution was checked against the official solution.