# Density of electrons

1. Feb 23, 2006

### stunner5000pt

A planet with an atmosphere entirely consisting of O2 and atomic oxygen, O, is exposed to solar EUV radiation at a wavelength of λ = 80 nm. The O2 does not absorb this radiation but the atomic oxygen is ionized by it with an absorption and ionization cross section of σ = 1x10-18 cm2. The O+ ions rapidly undergo charge exchange with O2 to form O2+ ions, and the only electron and ion loss process is dissociative recombination of the O2+ ions
O2+ e→O+O k1 = 1x10-8 cm3 s-1
If the solar flux is 1x10^11 photons cm-2 s-1 at 80 nm, the atomic oxygen scale height is 50 km and the O density is 1.6x10^12 cm-3 at 100 km, calculate the peak electron density of the resulting ionospheric layer for a solar zenith angle of 0°. What is the altitude of the electron density maximum?

thing is we know the solar flux anbd the nergy of each photon thus we know the totla energy coming into the planet

is this the formula i have to be using
$$n_{e}(h,\chi) = \sqrt{\frac{q_{max}}{keff}} \exp[0.5{1-z-\sec \chi \exp(-z)}]$$ where
$$z = \frac{h-h^0_{max}}{H_{O}}$$
where ne is the electron density at some height h and zenith angle chi

q max is the ion production rate
h is the height
h0 is the maximum height at where the maximum occurs

Last edited: Feb 23, 2006
2. Feb 26, 2006

### SpaceTiger

Staff Emeritus
You will need that equation, but it doesn't describe the electron density. Did you learn anything in class about detailed balance and/or equilibrium calculations?

3. Feb 26, 2006

### stunner5000pt

we never actually did any calculations in class per se

but from term equilibrium i gather that it has something to do iwth hte amount of energy being deposited

$$q(h) = I_{t} \sigma n_{0} \exp\left(\frac{h}{H} - \sigma n_{0} H \sec(\chi) \exp(-\frac{h}{H}) \right)$$

from here we can geta ratio between qmax and hmax
not sure where to proceed from there

could we assume that hte max depsotion occurs at 200 km? From this we can say z (from the first post) is zero. We can also calculate qmax using this asumption.
Could we then differentiate this function wrt height h and get what we need?

$$n_{e}(h,\chi) = \sqrt{\frac{q_{max}}{keff}} \exp[0.5{1-z-\sec \chi \exp(-z)}]$$
?

Last edited: Feb 26, 2006
4. Feb 26, 2006

### SpaceTiger

Staff Emeritus
It is related to this. What are the two processes being balanced? In other words, if one assumes that the rate at which electrons are released (ionized) is the same as being captured, what are the processes causing release and capture? Why would there be a balance in the first place?

5. Feb 26, 2006

### stunner5000pt

wouldnt simly the forward reaction O2+ e→O+O k1 = 1x10-8 cm3 s-1 and the corresponding backward reaction be the balancing processes?
there would be balance because at these levls there is failry consatnt concentration of the constituents?

6. Feb 27, 2006

### SpaceTiger

Staff Emeritus
This is indeed the reaction that removes electrons from the plasma.

No, how do you suppose the photons fit into this?