http://branespace.blogspot.com/2012/04/simplesolarradiativetransfer1.html
I examined the basic conditions for radiative transfer in a stellar medium or atmosphere. Of course, this is barely one half the story. Ultimately, the solar physicist examines radiative transfer to obtain first clues in how to understand emission and absorption processes which can occur in that atmosphere.
In this post we examine details of emission and absorption in terms of what are called transition probabilities. Basically one can consider and evaluate three possible cases:
i) A stimulated emission probability
ii) A spontaneous emission probability
iii) An absorption (or negative emission) probability
We will use a condensed notation in order to not have too much notation clutter within a limited blog post space To do this we will apply the following symbols to the differing probabilities:
Of critical importance in considering transitions between energy states are the statistical weights of the states, which we denote by g. Then given statistical weights[1], say g_{1} and g _{2}, we first wish to show that:
g_{1 } B _{12.} = g _{2 }B _{21}
In other words, the absorption probability for state g_{1} must be equal to the stimulated emission probability for state g_{2.}. There can’t be more transitions than the numbers of electrons available for them in those states. For thermal equilibrium, especially, we have detailed balancing. i.e. the number of transitions from level one to level two must be equal to the number of transitions from level two to level one.
From level 2 to level 1 we write: N_{2} (A_{21 } + I _{u }B _{12.})
For detailed balancing we require:
N_{1} I _{u }B _{12.} = N_{2} (A_{21 } + I _{u }B _{12.})
We note here that in thermodynamic equilibrium Boltzmann’s equation applies:
N_{2 }/ N_{1 } = I _{u }B _{12.}/ (A_{21 } + I _{u }B _{12.}) =
[g_{2} / g_{1 }]_{ }exp ( E2 – E1) / kT
Where E2, E1 designate the respective energy levels.
Then: I _{u} = 2h u ^{3} / c ^{2} [1/ exp (hc/lkT]
As T ® ¥ and I _{u }® ¥
Then:
B _{12.}/ B _{21} = g_{2} / g_{1 } or g_{1 } B _{12.} = g _{2 }B _{2}
We also need to show, for detailed balancing:
A _{21.}/ B _{12} = 2h u ^{3} / c ^{2} [g_{2} / g_{1 }]
We use:
I _{u }B _{12.} =
(A_{21 } + I _{u }B _{21.}) [g_{2} / g_{1 }]_{ }exp ( E2 – E1) / kT
And:
I _{u} [B _{12.} B _{21.} (g_{2} / g_{1 })_{ }exp ( E2 – E1) / kT]
= (A_{21}) (g_{2} / g_{1 })_{ }exp ( E2 – E1) / kT
® I _{u} [B _{12.} _{.} (g_{2} / g_{1 })( g_{1} / g_{2})_{ }B _{12 }exp ( E2 – E1) / kT
I _{u} = A _{21.}/ B _{12} (g_{2}) e ^{( E2 – E1) / kT}/ g_{1} [1  e ^{( E2 – E1) / kT}]
= A _{21.}/ B _{12} (g_{2} / g_{1 }) [e ^{(E2 – E1) / kT}  1]
= 2h u ^{3} / c ^{2} [1 / e ^{h}^{u}^{ / kT}  1_{ }]
And:
A _{21.}/ B _{12} (g_{2} / g_{1 }) = 2h u ^{3} / c ^{2}
Then: A _{21.}/ B _{12} = 2h u ^{3} / c ^{2} [g_{2} / g_{1 }]
Thus, we see that in thermodynamic equilibrium, the ratio of populations in upper and lower levels is given by the Boltzmann formula. In most solar applications of interest, the stimulated emission is negligible compared to the spontaneous emission.
Note also that Einstein showed that the spectral line transition probabilities are related by:
A _{21.}= 8 p h u ^{3} / c ^{2} [B _{21}] = 8 p h u ^{3} / c ^{2} (g_{1} / g_{2 }) B _{12} =
6.67 x 10 ^{16} [g f/g2 l^{2} Å]
Where g is the Gaunt factor, of order unity and f is the oscillator strength. The latter generally has specific values for discrete transitions. (For the strongest spectral line from a level one in an atom one can usually use f = 1.)
The radiation density is defined:
u _{u} = 4 p B _{u }/ c = 8 p h u ^{3} / c ^{2} [1 / e ^{h}^{u}^{ / kT}  1]
With the Planck function:
B _{u } du = 2h u ^{3} / c ^{2} [1 / e ^{h}^{u}^{ / kT}  1_{ }] du
This is for the frequency domain, but can also be expressed in the wavelength domain:
B _{l}_{ } dl = 2 p h c ^{2 }/l^{5 } [1 / e ^{hc / k}^{l}^{ T}  1_{ }] dl
(Be careful treating the differential for the frequency!)
du = ( c/ l^{2}) dl which must be used when transferring from the frequency scale Hz ^{1} to the wavelength scale
(cm ^{1} or m^{1})
For hydrogenic atoms (Z » 1) the absorption crosssection, a _{u }, plays a critical role, defined:
a _{u }= 2.815 x 10 ^{29} Z ^{4}/ n ^{5} (g /u ^{3})
Or:
a _{u }= 7.91 x 10 ^{16} Z ^{4}/ n ^{5} (R_{y }/ h u)^{3 } g
(cm ^{2} / bound electron in state n.)
Where R_{y }is the modified Rydberg constant for atomic physics. In terms of the standard Rydberg constant, R_{∞} = 1.0974 × 10^{7} m^{−}^{1}, it is: R_{y }= hc R_{∞} = 13.605 eV.^{}
When dealing with complex atoms one needs to allow for the number of electrons in the absorbing state:
a _{u }= 4 p/ c (B _{n}_{e}) h u
where B _{n}_{e} is the Einstein coefficient or “continuum fvalue”. Thus, it may also be posed:
a _{u }= 8.067 × 10^{18} (df/de)
(cm ^{2} / bound electron in state n.)
A useful table that will come in handy for spectral line computations is the following:
Quantum Numbers and Energies For Hydrogen Atom:
Ground state s1
n ℓ m _{ℓ }m _{s }

Energy E1 (eV)

1 0 0 +½
1 0 0 ½

  13.6
  13.6

First Excited States
n ℓ m _{ℓ }m _{s }

Energy E1 (eV)

2 0 0 +½

 3.40

2 0 0 ½

 3.40

2 1 1 +½

 3.40

2 1 1 ½

 3.40

2 1 0 +½

 3.40

2 1 0 ½

 3.40

2 1 1 +½

 3.40

2 1 1 ½

 3.40

Inspection of the table above shows two quantum states with the same energy (13.6 eV) and eight states with (3.40 eV). Thus, two states are degenerate for the n=1 level and eight states are degenerate for the n=2 level. Since g _{n }= 2n^{2}, then:
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
Problems for Budding Astrophysicists:
1)Consider a gas of neutral hydrogen. Using the Boltzmann equation and the information in the table above, compute the temperature at which one will expect equal numbers of atoms in the ground state and the first excited state.
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 specific intensity from:
I _{u} = 2h u ^{3} / c ^{2} [1/ exp (hc/lkT]
3)Calculate the transition probability ( A _{21} ) you get using the Einstein equation:
_{} If: A _{21.}= 6.67 x 10 ^{16} [g f/g2 l^{2} Å]
What possible errors might cause the values to diverge? (Take g = f » 1)
[1] The “statistical weight” or degeneracy is just the number of different atomic substates included in the state being considered. As we saw each atomic state of angular momenta L,S leading to total angular momentum J can be split by magnetic field into 2J + 1 states. Then for all J levels of a term LS there are:
g(L.S) = (2L + 1)(2S + 1) = S _{j} (2J + 1)
different M_{ j} sublevels possible. For a hydrogenic shell (n) there are 2n ^{2} sublevels possible.
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