Approximating the Voigt function as where a = A/4xAvp , the equivalent width can be expressed as: . Solve analytically for the equivalent width of the line. Then show that the HI Lyman a line at A 1215 A has the following relation between column density and equivalent width: ( N(HI, Lya) = 1.867 x 101 W?(cm=3) where, WA is equivalent width in Angstrom. The Einstein A coefficient for this transitions is 6.27 x 10°s- and the oscillator strength is 0.4162. Hints: 1) 2) To = Nơo where me Avp is the absorption cross-section at line center.
Approximating the Voigt function as where a = A/4xAvp , the equivalent width can be expressed as: . Solve analytically for the equivalent width of the line. Then show that the HI Lyman a line at A 1215 A has the following relation between column density and equivalent width: ( N(HI, Lya) = 1.867 x 101 W?(cm=3) where, WA is equivalent width in Angstrom. The Einstein A coefficient for this transitions is 6.27 x 10°s- and the oscillator strength is 0.4162. Hints: 1) 2) To = Nơo where me Avp is the absorption cross-section at line center.
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Transcribed Image Text:Approximating the Voigt function as
where
A/4n Avp
a =
, the equivalent width can be expressed as:
Wa = /cAvp / (1 – e TuVEz²;
. Solve analytically for the equivalent width of the line. Then show that the HI
Lyman a line at
A 1215 A has the following relation between column density and equivalent
width: (
N(HI, L,a) = 1.867 × 1018W3(cm-2)
where,
WA
is equivalent width in Angstrom. The Einstein A coefficient for this transitions is
6.27 x 10°s-1
and the oscillator strength is 0.4162.
Hints: 1)
(1 – e")y-/² dy = 2V
2)
To = Noo
, where
VTe f
me Avp
is the absorption cross-section at line center.
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