IPhO 2014, Теория — Задача 3. Simplest model of gas discharge

Problem 3: Simplest model of gas discharge · 10 т.

Внимание: Бележка към темата

The question prints b in the prose defining mobility after v=βE; that distinction is retained. The official solution’s (B1.3) middle expression and B6 reference to (B5.1) are retained as printed. The extra B7 belongs only to the presentation, not the scored competition paper. The archive item “3 copy.pdf” is not a readable PDF; the complete 3.pdf and 3_sol.pdf were used.

Theoretical competition

Tuesday, 15 July 2014 — 45th IPhO, Kazakhstan–Astana.

Условие

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 (4,8 т.)

In this part of the problem the so-called non-self-sustained gas discharge is studied. To maintain it a permanent operation an external ionizer is needed, which creates ZextZ_{\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 recombines with an ion to form a neutral atom. The number of recombining events ZrecZ_{\mathrm{rec}} that occurs in the gas per unit volume and per unit time is given by

Zrec=rneni,Z_{\mathrm{rec}}=rn_en_i,

where rr is a constant called the recombination coefficient, and ne,nin_e,n_i denote the electron and ion number densities, respectively.

Suppose that at time t=0t=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 ne(t)n_e(t) depends on time tt as follows:

ne(t)=n0+atanh⁡bt,n_e(t)=n_0+a\tanh bt,

where n0,an_0,a and bb are some constants, and tanh⁡x\tanh x stands for the hyperbolic tangent.

A1 Find n0,a,bn_0,a,b and express them in terms of ZextZ_{\mathrm{ext}} and rr. [1,8 т.]

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 ne1=12⋅1010 cm−3n_{e1}=12\cdot10^{10}\ \mathrm{cm^{-3}}. When the second external ionizer is switched on, the electron number density reaches its equilibrium value of ne2=16⋅1010 cm−3n_{e2}=16\cdot10^{10}\ \mathrm{cm^{-3}}.

A2 Find the electron number density nen_e at equilibrium when both external ionizers are switched on simultaneously. [0,6 т.]

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 SS separated by the distance L≪SL\ll\sqrt S from each other. The voltage UU 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.

Assume that both the electrons (denoted by the subscript ee) and the ions (denoted by the subscript ii) acquire the same ordered speed vv due to the electric field strength EE found as

v=βE,v=\beta E,

where bb is a constant called charge mobility.

A3 Express the electric current II in the tube in terms of U,β,L,S,Zext,rU,\beta,L,S,Z_{\mathrm{ext}},r and ee which is the elementary charge. [1,7 т.]

A4 Find the resistivity ρgas\rho_{\mathrm{gas}} of the gas at sufficiently small values of the voltage applied and express it in terms of β,L,Zext,r\beta,L,Z_{\mathrm{ext}},r and ee. [0,7 т.]

Part B. Self-sustained gas discharge (5,2 т.)

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

Attention! In the sequel assume that the external ionizer continues to operate with the same ZextZ_{\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 is a secondary electron emission, and the second one is a 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 N˙e\dot N_e per unit time to the number of ions N˙i\dot N_i hitting the cathode per unit time is called the coefficient of the secondary electron emission, γ=N˙e/N˙i\gamma=\dot N_e/\dot N_i.

The gas discharge tube: cathode at x=0, anode at x=L, electric field and charge-carrier directions, and a thin layer between x and x+dx.

The formation of the electron avalanche is explained 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 anode significantly increases. This process is described by the Townsend coefficient α\alpha, which characterizes an increase in the number of electrons dNedN_e due to moving NeN_e electrons that have passed the distance dldl, i.e.

dNedl=αNe.\frac{dN_e}{dl}=\alpha N_e.

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

Ie(x)=C1eA1x+A2,I_e(x)=C_1e^{A_1x}+A_2,

where A1,A2,C1A_1,A_2,C_1 are some constants.

B1 Find A1,A2A_1,A_2 and express them in terms of Zext,α,e,L,SZ_{\mathrm{ext}},\alpha,e,L,S. [2 т.]

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

Ii(x)=C2+B1eB2x,I_i(x)=C_2+B_1e^{B_2x},

where B1,B2,C2B_1,B_2,C_2 are some constants.

B2 Find B1,B2B_1,B_2 and express them in terms of Zext,α,e,L,S,C1Z_{\mathrm{ext}},\alpha,e,L,S,C_1. [0,6 т.]

B3 Write down the condition for Ii(x)I_i(x) at x=Lx=L. [0,3 т.]

B4 Write down the condition for Ii(x)I_i(x) and Ie(x)I_e(x) at x=0x=0. [0,6 т.]

B5 Find the total current II and express it in terms of Zext,α,γ,e,L,SZ_{\mathrm{ext}},\alpha,\gamma,e,L,S. Assume that it remains finite. [1,2 т.]

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

B6 Find LcrL_{\mathrm{cr}} and express it in terms of Zext,α,γ,e,L,SZ_{\mathrm{ext}},\alpha,\gamma,e,L,S. [0,5 т.]

Supplementary presentation: B7

The archived presentation adds the following extension on slide 11; it is absent from the two-page competition paper and carries no stated score. The presentation gives

α=α1pexp⁡(−α2pE).\alpha=\alpha_1p\exp\left(-\frac{\alpha_2p}{E}\right).

B7. Find and evaluate the minimum voltage Umin⁡U_{\min} applied across the plates, at which the self-sustained gas discharge can still appear in neon.

The slide does not give numerical values of α1,α2\alpha_1,\alpha_2 or γ\gamma.

Решение

Archived presentation

The accompanying 11-slide presentation dedicates this problem to Prof. F. Baimbetov. It illustrates the same derivations and adds an unscored B7 extension, included below. The full presentation remains available as an original supplementary source.


Оригинал в Архива: 3.pdf · официални решения: 3_sol.pdf · допълнителни данни: 3_pres.pdf