2005 I 2

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-3 V

Drawing a tangent from the zero crossing to the ion saturation current level (-1 mA) gives roughly -2 V, so the electron temperature is 2 eV.

Argon ions will probably be singly charged, since the second ionization potential is significantly over 1.5 eV

The ion saturation current is -1 mA. since:

Isi = − 0.6neeApCs

One finds (assuming the mass of argon is 36):

C_{s}=9.79\times10^{5}\left(\gamma ZT_{e}/\mu\right)^{1/2}\mathrm{cm/s}\approx2\times10^{5}\mathrm{cm/s}
n_{e}=\frac{1\,\mathrm{mA}}{0.6e2\pi\left(0.2\,\mathrm{cm}\right)^{2}\left(2\times10^{5}\mathrm{cm/s}\right)}\approx2\times10^{11}\mathrm{cm}^{-3}

The electron saturation current is:

I_{se}=\frac{1}{4}eAn_{e}\bar{v}

We get:

\bar{v}=\sqrt{\frac{1}{2\pi}}v_{te}\approx\frac{1}{2.5}4.19\times10^{7}T_{e}^{1/2}\mathrm{cm/s}\approx2\times10^{7}\mathrm{cm/s}

So that the current is:

I_{se}=\frac{e}{4}2\pi\left(0.2\mathrm{cm}\right)^{2}\left(2\times10^{11}\mathrm{cm}^{-3}\right)\left(2\times10^{7}\mathrm{cm/s}\right)\approx40\,\mathrm{mA}

This beam electrons have velocity:

v_{b}=4.19\times10^{7}T_{e}^{1/2}\mathrm{cm/s}=1.3\times10^{8}\mathrm{cm/s}

The beam would increase the current for V > − 10V uniformly by:

I_{beam}=e\pi\left(0.2\mathrm{cm}\right)^{2}\left(2\times10^{8}\mathrm{cm}^{-3}\right)\left(1.3\times10^{7}\mathrm{cm/s}\right)=0.5\,\mathrm{mA}

This page was recovered in October 2009 from the Plasmagicians page on Generals_2005_I_2 dated 20:22, 15 May 2007.

Retrieved from "https://qed.princeton.edu/main/PlasmaWiki/Generals/2005_I_2"
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