What are spectral lines and how are they formed? It is first necessary to understand that a “line” is really an idealization. It exists only to the extent that it is distinctive and is only distinctive to the degree it corresponds to a specific wavelength (or frequency) for which the absorption coefficient, i.e.

a ** _{u }**= 7.91 x 10

^{16}Z

^{4}/ n

^{5}(R

_{y }/ h u

_{ })

^{3 }g

is greater
than usual. (Else it would not have a very good chance of escaping before
absorption). There are also two
distinct processes by which lines can be formed:

1)
Bound-bound transitions

2)
Bound-free transitions

In the first case, the photon goes from one bound atom to another, such as depicted in the diagram below for the transitions between energy levels in the hydrogen atom.

*Emission* occurs when an electron
in the atom makes a transition from a higher to a lower energy
level. This is always accompanied by the
emission of a photon with a defined energy E = hf = h (c/ l). Consider for example, a
transition from the n = 2 to the n = 1 level, as depicted in the lower left of the preceding diagram, i.e.* the first line
of the Lyman series*.

The energy at the n= 2 level is:

E(n=2)
= - 13.6/** **n^{2 } ** ^{ }**= - 13.6/

**(2)**

^{2 }

**= - 13.6/4 (eV)**

^{ }Now,
1 eV = 1.6 x 10^{-19} J so:

E(n=2)
= - 13.6/4 (eV) =
-(3.4) x 1.6 x 10^{-19} J
=

-5.4 x 10^{-19} J

** **The n= 1 level has energy**:**

E(n=1)
= - 13.6/** **n^{2 } ** ^{ }**= - 13.6/ (1)

^{2 }

**= - 13.6 (eV)**

^{ }E(n=1)
= -(13.6) x 1.6 x 10^{-19} J = -21.8 x 10^{-19} J

Then the energy difference is:

E2
– E1 = [- 5.4 – (-21.8)] x 10^{-19}
J = 16.4 x 10 ^{-19} J

From this, the wavelength of the photon emitted can be found.

Since E = hf = h (c/ l): l = hc/ (E2 – E1)

l = (6.626069 x 10^{- 34} J-s)(3 x 10 ^{8}
m/s)/ 16.4 x 10^{-19} J

l = 1.21 x 10 ^{-7} m

The frequency can then be found from:

f
= (c/ l) = (3 x 10 ^{8}
m/s) / 1.21 x 10 ^{-7} m = 2.47
x 10 ^{15} Hz** **

The Balmer series emission lines for hydrogen are shown in the graphic below:

Bound-bound
transitions (such as those shown) differ in significant ways from bound-free transitions. In the
first case, the transitions are also affected by a broadening function which is
not so important for continuous emission. If we write out the equation for
absorption in more detail we get:

a
** _{u }**= [1 - e

^{- h }

^{u }^{o}

^{ / kT}] (

**p**e

^{2}/ mc) f f

_{u}

which yields units in cm^{2} / atoms at lower level. Two
other absorption derivative values are possible from the preceding:

(i)The absorption coefficient per unit
length (cm^{-1})

(ii)The mass absorption coefficient k _{u} .

The value for (i) is just a ** _{u
}**multiplied
by the number of absorbing atoms per unit volume. The value for k

_{u}is just multiplied by the number of absorbing atoms. The value for a

_{o }is just:

a _{o }= a _{u }/** _{ }**f

_{u. }

Sometimes referred to as a “fudge factor”,
f is known as the oscillator strength or f-value of the line. It is basically
the transition probability for the line and is to be computed by quantum
mechanics or measured in the laboratory.

The *broadening
function* f _{u } is:

f _{u
}= 1/ Öp [exp (u
-u _{o})_{ }**/**D** **u
_{D}** **]^{2} D u_{D}_{}

Which can also be rewritten as: f _{u }du =

1/ Ö**p** [exp (u - u _{o})_{
}/D u_{D} ]^{2} du / D u_{D}_{}

This would be the probability that the
absorbed photon lies between u and u + du, assuming equal intensities for all frequencies. Thus the
integral:

ò f _{u
}du
= 1

Where du is over all frequencies.
The value of f _{u} is larger near u _{o } the frequency of the line
center.

This may be deduced from the line profile diagram
below (for the sodium 5889.95 A ** **line):

Note here that D u_{D} is the *Doppler half-width* of the
line. As can be seen on
inspection, the broadening function, f _{u} is very large at line center
and falls off in the “wings”, i.e.
at larger and smaller frequencies. (Called ‘wings’ because of the visual similarity to wings in the line profile.)

The three important types of broadening are due to Doppler
effect, natural and pressure broadening. We confine our attention here to the
first type which is given by the broadening probability equation, provided the
velocity is Maxwellian and that the frequency at line center u _{o} is also
observed for some known frequency u .

At the same time we note in passing that
the natural broadening is a consequence of the energy-time uncertainty
principle: ΔE
Δ t ³ h/2π where we define the probability for a
specific decay as: P = 10^{-6 } sec.

The study of spectral lines is
facilitated by using what is called the “**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 (u) *or* wavelength (l):

W = ò** **^{¥}_{0 } (I _{c} – I _{u
}/ I _{c} )du =
ò ^{¥}_{0 } (F _{c} – F _{l
}/ F _{c} ) dl

The left side defines W in terms of the intensity e.g. from the continuous
spectrum outside of the spectral line where the quantity (I _{c}
– I _{u}/ I _{c} ) is referred to as
the “*depth of the line*”. This is 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.

* Suggested Problems*:

1) Find the frequency of the spectral line for which the hydrogen electron transits from the n=3 to the n= 2 energy level.

2)For the Balmer a line (called H- alpha), we know:

E3 – E2 = - 13.6 eV ( 1/** **3 ^{2 } ^{ }-
1/ 2 ^{2 }) = 1.88 eV

a) From this information calculate the
ratio N_{2}/ N_{1} for
T = 10 ^{4 }K

b) Obtain the frequency of the spectral line for this
transition

---------------

At the n=1 level the statistical weight is: g _{1 }=
2(1)^{2} = 2

At the n=2 level the statistical weight is: g _{1 }=
2(2)^{2} = 8

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