**equivalent width**”. This provides the total strength of a line, yielding

*the same area of the line*(for a rectangular facsimile) provided the depth is complete from the continuum to zero brightness.

We can express the equivalent width in two
ways, based on frequency of wavelength, or wavelength: itself, e.g.

W =

**ò**^{¥}**(I**_{0 }_{c}– I_{u }/ I_{c})du =**ò**^{¥}**(F**_{0 }_{c}– F_{l }/ F_{c}) dl
The left side defines W in terms of the
intensity from the continuous spectrum outside the spectral line where the
quantity (I

_{c}– I_{u }/ I_{c}) is referred to as the “depth of the line”, the analogous quantity to (F_{c}– F_{l }/ F_{c}) on the right side where we have radiant flux units. Technically, the integral should be taken only from one side of the line to the other but the limits can be as shown provided I_{c}(or F_{c}) is kept constant in the neighborhood of the line.*The Curve of Growth*:

Having obtained the equivalent width, the next logical step is
to generate the “curve of growth”.
Recall from the previous section that W, the equivalent width,
represents the line strength. Then the curve of growth describes how the latter
increases as the optical depth, τ
increases.

To fix ideas, we use a model slab of finite thickness ds and over which the
optical depth increases by d τ. The
incident intensity is I

_{c}_{ }for frequencies in the neighborhood of the line, and the intensity emerging from the opposite side is I_{u }which we seek to find.Starting with the original basic transfer equation based on wavelength:

dI(l)/ds = -k(l) I(l) + k(l) S(l)

We can derive the frequency form:

d I

**/ dτ**_{u }**= - I**_{u }**– S**_{u}_{u}_{ }
We have (from the original equation) after adjusting for
frequency:

d I

**/ ds**_{u }**= -k**_{ }**I**_{u}**+ k**_{u}**S**_{u}_{u}
Where: k

**= d τ**_{u}**/ ds**_{u }
So:

d I

**/ ds**_{u }**= -( d τ**_{ }**/ ds) I**_{u }**+ (d τ**_{u}**/ ds) S**_{u }_{u}
d I

**= - d τ**_{u }**I**_{u }**+ d τ**_{u}**S**_{u }**= - d τ**_{u }**(I**_{u }**– S**_{u}**)**_{u}_{}
whence:

d I

**/ dτ**_{u }**= - I**_{u }**– S**_{u}_{u}_{ }
For which the solution is found (on integration):

I

**= I**_{u}**exp (-τ**_{ c}**)**_{u}*Where:*

*τ*

_{u}*= N L*

*a*

_{ u}*= N L*

*a*

_{o}*f*

_{u}

Where ‘N’ is the number of absorbing atoms per unit volume
and we already saw that:

a

**= a**_{o }**/**_{u }**f**_{ }_{u}**= a**_{ }**[1 - e**_{u}^{- h }^{u o}^{ / kT}] (**p**e^{2}/ mc) f
In terms of this new information, the equivalent width of the
line can then be written as:

W

**=**_{ u}**ò**^{¥}**[1 - exp (- N L a**_{0 }**f**_{o}_{u}) ] du
For small x we may use the approximation that:

exp (-x)

*much greater than*1 – x
So rewrite
the equation for W

**:**_{ u}
W

**»**_{ u}**ò**^{¥}**[1 - (1 - N L a**_{0 }**f**_{o}_{u}) du
»

**ò**^{¥}**N L a**_{0 }**f**_{o}_{u}du » N L a_{o}**ò**^{¥}**f**_{0 }_{u}du
So
we find W

**depends only on the form for the broadening function. For very weak lines, for example:**_{ u}**ò**

^{¥}**f**

_{0 }_{u}du

*much greater than*1 so that W

**» N L a**

_{ u}

_{o}
Or simply proportional to N, the
number of absorbing atoms. For “strong” lines the absorption near the center is
very large so we can expect: N L a

**f**_{o}_{u}>> 1 and with sufficient working this can be shown to give the corresponding equivalent width: W**= 2**_{ u}**D**u_{D}
And for the

*moderately strong*lines:
W

**= 2**_{ u}**D**u_{D}**{ln (N L a****/Ö**_{o}**p****D**u**)}**_{D}^{½}
While
very strong lines yield:

W

**= 1/**_{ u}**p**(N L a**g)**_{o}^{½}
Where
g is the damping constant.

Assembling
all the diverse W

**and plotting log W**_{ u}**vs. log (N L a**_{ u}**) one gets the curve of growth shown below:**_{o}
Here, each contribution can be
separately considered, in terms of the “linear”, “saturated” and “damping”
segments. The first then is for the weakest lines such that W

**is simply proportional to the number of absorbing atoms. But as more atoms are added this approximation becomes invalid and so we are in the saturated part. In effect, the near zero slope shows the degree to which the equivalent width increases at an almost uniform rate. Finally, we have the damping segment wherein the line under consideration develops “damping wings” because the number of absorbing atoms is so large. (Bear in mind again that a**_{ u}**is proportional to the transition probability)**_{o}
Problems:

1)Consider
the ionization of helium, with

*three stages of ionization*for which the partition functions are: Z_{I}= Z_{3}= 1, and Z_{2}= 2
With
ionization energies:

c

_{ }_{ 1 }= 24.58 eV and c_{ }_{ 2 }= 13.6 eV
For
a temperature of T = 5040 K write out two versions of the Saha equation for:

log(
P

_{e}N_{2}/N_{1}) and log( P_{e}N_{3}/N_{2})
And
hence, compute: (P

_{e}N_{2}/N_{1}) and (P_{e}N_{3}/N_{2})
2)The
diagram below was sketched by a first year astrophysics student. It represents part of the emission spectrum of atomic hydrogen. It
contains a series of lines, and the wavelengths of some (in nm) are marked.
'

There are no lines in the series less than 91.2 nm

There are no lines in the series less than 91.2 nm

(a) In which region of
the E-M spectrum would these lines occur?

(b)
Obtain a relation between

*E and***and thence find the***l**photon**energies*equivalent for all the wavelengths marked.
(c)Use
this information to map a partial energy level diagram for hydrogen. Show and
label clearly the electronic transitions responsible for the emission lines
labeled.

3)a)
A stellar atmosphere is composed of pure hydrogen, at a temperature of 9600 K
and with electron pressure, P

that are ionized in relation to the total, i.e. find: N

_{e }= 200 dyne cm^{-2}. Use the Saha equation to compute the fraction of atomsthat are ionized in relation to the total, i.e. find: N

_{II}/( N_{I }+ N_{II})
The
partition functions are: Z

_{II}= 1 and Z_{I}= 2
Hint:
Check to see whether (E

_{2}- E_{1}) >> kT with
k
= 8.6174 x 10

^{-5}eV/K
So
the Boltzman factor exp (- E2 – E1) / kT
<< 1

b)Repeat
the exercise again for a temperature of 5040 K and 20,000 K and thence account
for why the Balmer lines attain maximum intensity at a temperature of 9520 K.

(4)
Calculate how far you could see through the Earth’s atmosphere if it had the
same opacity as the solar atmosphere, where:

k

**= 0.264 cm**_{l }^{2}/g.
Take
the density of Earth’s atmosphere as:

**r**= 0.0012 g cm_{ }^{-3}
The
density of the solar atmosphere is taken as :

**r**

_{s}

**= 2.5 x 10**

_{ }^{-7}g cm

^{-3}

And we
know the intensity declines over a characteristic distance (length scale) such that:

ℓ = 1/ k

_{l }**r**
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