Chapman
α-layer
for vertically incident solar radiation
- Given:
n0 = 2 ∙ 1025 m-3 (density of neutrals to be ionized, at z=0)
H = 10 km (scale heights of neutrals)
= 2 ∙ 10-21
m2
(average ionization cross section for relevant wavelength range)
I∞ = 2 ∙ 10-3 W m-2 (solar spectrum until 100 nm, above the ionosphere)
- Using the equations given in the lecture, calculate:
a) The height z0 of the maximum electron density of the Chapman α layer
b) The light intensity I(z) and the absorption rate q(z) for the altitude range [ z0 – 2 H ; z0 + 2 H ]. Plot a graph of both curves.
c) The recombination coefficient α under the assumption of a stationary situation, and that α-recombination is the only loss process. For this, the electron density at the layer maximum z0 be 2 ∙ 1011 m-3 (E-layer, daytime, low latitudes), and the parameter γ (number of produced ion/electron pairs per energy unit) be 2 ∙ 1017 J-1.
d) With the value for α as calculated in part c), use the balance equation (55), inserting L for α-recombination, to calculate ne(t) at z=z0 for the case that q(z0) is instantly “switched on” at t=0, and the initial electron density is zero. That is,
ne(t=0) = 0 ; q(t) = 0 for t<0 ; q(t) = q(z0) for t≥0, with the value for q(z0) from b)
How long does it take until ne(t) increases from zero to half of its stationary value?
Plot a graph of ne(t) for the time interval from zero to the time as calculated above.
Note: Task d) can be completed either analytically or computationally!