Ion density in sheath of Langmuir probe in plasma

[goodphy]

Hello.

I'm studying principle of Langmuir probe and got several questions.

1st, the textbook suddenly tells that ion density within sheath is

n_i(x) = n_is(V_s/V(x))^1/2

where n_is, V_s are ion density and plasma potential at sheath edge.

I found some document which shows that it is obtained by using

(1/2)m_iV² = -eV (Conservation of energy)
n_iv = const (Particle conservation) where v is velocity of the particle.

but I still don't get how to combine them to get desired result.

1. Could you please tell me how to get the expression above?And I have impression during studying that in derivation of I-V characteristic curve of probe in plasma, electrons are all in thermal equilibrium thus Maxwell distribution and Boltzmann factor are used for their distribution and density while ions are not.

2. Why are ions not in equilibrium and why electrons are treated as equilibrium even within sheath where they're repelling or attracting depending on probe voltage?

Please help me to get deeper understanding.

 

[goodphy]

I've found personal conclusion.

First, ion density distribution is obtained by using
1. Current conservation; v_i(x)n_i(x) = n_isv_is.
2. Energy conservation; v_is = (2eV_s/m_i)^1/2, v_i(x) = (2eV(x)/m_i)^1/2.
Combining these two gives ion density in my question.

Second, electron has light mass, which leads
1. Electron heat flux (n_ev_th where v_th = (k_BT_e/m_e)^1/2) is large so fast thermalization time.
2. Thermal velocity v_th is normally exceeding drift velocity by E-field within sheath. If this doesn't holds, for example, high bias voltage of probe than plasma potential V_p, Electron can follow similar density distribution of ions above.