How Does the Saha Equation Estimate Ion Populations in the Solar Photosphere?

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Homework Help Overview

The discussion revolves around the application of the Saha equation to estimate the relative populations of negative hydrogen ions (H-) and neutral hydrogen (H) in the solar photosphere, specifically at a temperature of 6,000 K and a pressure of log Pe = 2.7. Participants are exploring the implications of ionization energy values and the calculations involved in determining the ratio of ion populations.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants are attempting to calculate the ratio of H- to H using the Saha equation, questioning the appropriate value for the ionization energy (χion) and its impact on the results. There is discussion about the logarithmic calculations and potential errors in previous attempts.

Discussion Status

The discussion is ongoing, with participants providing various calculations and questioning the assumptions made regarding ionization energy. Some participants suggest alternative values for χion and explore how these changes affect the calculated ratios. There is recognition of the complexity involved in the calculations and the need for consistency in logarithmic bases.

Contextual Notes

Participants note discrepancies in the expected results, with some suggesting that the electron partial pressure may need to be considered. There is also mention of differing values for the binding energy of H-, indicating a lack of consensus on this parameter.

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Homework Statement


The Saha equation for the hydrogen atom can be written as

log(N+/N) = log(u+/u) + (5/2)logT - log(Pe) - χionӨ - 0.18

where Ө=5040/T
χion is measured in electron volts (eV).

Calculate the number of negative hydrogen ions (H-) in the solar photosphere relative to neutral hydrogen (H) for a temperature of T = 6,000 K and a pressure of log Pe = 2.7.

Homework Equations

The Attempt at a Solution



log(N+/N) = log(u+/u) + (5/2)logT - log(Pe) - χionӨ - 0.18
log(N+/N) = log(2/1) + (5/2)log6000 - 2.7 - (13.6)x(5040/6000) - 0.18
log(N+/N) = 0.693 + 21.748 -2.7 - 11.42 - 0.18
log(N+/N) = 8.14

I chose to input 13.6 for χion since the ionisation energy for the hydrogen atom in the ground state is χion = 13.6eV.

But apparently the correct solution should be 5x10(-7).
 
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The ionization energy is 13.6 eV for the transition H <-> H+, it is lower for H- <-> H.
 
mfb said:
The ionization energy is 13.6 eV for the transition H <-> H+, it is lower for H- <-> H.

If you mean the value for χion should instead be 3.4eV, then this would have the effect of making my solution even larger than it is. Whereas apparently the correct solution is a lot smaller.
 
A large log means the ratio is large, but that means your H- are rare.
 
log(N+/N) = log(u+/u) + (5/2)logT - log(Pe) - χionӨ - 0.18
log(N+/N) = log(2/1) + (5/2)log6000 - 2.7 - (3.4)x(5040/6000) - 0.18
log(N+/N) = 0.693 + 21.748 -2.7 - 2.86 - 0.18
log(N+/N) = 16.70
(N+/N) = 10^16.70
(N+/N) = 5.01x10^16

Here I have changed I have changed χion to 3.4eV. This gives a value for the ratio of 5.01x10^16

But apparently the correct solution should be 5x10^(-7).
 
I discovered one mistake, (5/2)log6000 should of course equal 9.44 (not 21.748)!

log(N+/N) = log(u+/u) + (5/2)logT - log(Pe) - χionӨ - 0.18
log(N+/N) = log(2/1) + (5/2)log6000 - 2.7 - (3.4)x(5040/6000) - 0.18
log(N+/N) = 0.693 + 9.44 -2.7 - 2.86 - 0.18
log(N+/N) = 4.39
(N+/N) = 10^4.39
(N+/N) = 2.45x10^4

Here χion is 3.4eV. Which gives a value for the ratio of 2.45x10^4 which is still a long way from 5x10^(-7).
 
"N+" are your neutral hydrogen atoms here. Your target is 2*106, the inverse of 2*10-7.
You mix logarithms to base 10 and e here. It does not matter which one you use but you have to be consistent.
 
I thought it was all to the base 10. What do I have that is to the base e?
 
  • #10
mfb said:
log(2/1)
How do I convert log(2/1) to base e, to the base 10?
 
  • #11
You can calculate the logarithm of 2 in base 10 in the same way you calculated all other logarithms.
0.693 is the logarithm of 2 in base e.
 
  • #12
So you mean:

log(N+/N) = log(u+/u) + (5/2)logT - log(Pe) - χionӨ - 0.18
log(N+/N) = log(2/1) + (5/2)log6000 - 2.7 - (3.4)x(5040/6000) - 0.18
log(N+/N) = 0.30 + 9.44 -2.7 - 2.86 - 0.18
log(N+/N) = 4.00
(N+/N) = 10000
(N+/N) = 1x10^4

Though I'm still not getting 5x10^(-7).
 
  • #13
Where does the number of 3.4 eV binding energy come from? I see sources giving 0.75 eV (e.g. here).
Using that value I get a result close to the given value.

But now I think that calculation cannot work like this.
The negative sign for the exponential suppresses higher-energetic states, like the neutral H plus electron here. The lower-energetic state H- has to be rare, because we don't have enough electrons around to form many of them. Simply taking the pressure is not sufficient, the number of hydrogen atoms and the number of free electrons are completely different things. The given value would have to be the electron partial pressure.
 
  • #14
I'd got 3.4 by taking the 13.8eV I originally and dividing by 2^2. Though I have head other students mention they had used 0.7 for this value. So you're 0.7-0.75 could well be correct.

When I re-do the calculation with 0.7eV I get a result of 6.272, which is a lot better, though I'm not sure how 5x10^(-7) was obtained.
 
  • #15
ZedCar said:
I'd got 3.4 by taking the 13.8eV I originally and dividing by 2^2.
That does not work:
(1) the second electron will also take the n=1 orbital (but with opposite spin) and
(2) electron repulsion is important

Though I have head other students mention they had used 0.7 for this value. So you're 0.7-0.75 could well be correct.

When I re-do the calculation with 0.7eV I get a result of 6.272, which is a lot better, though I'm not sure how 5x10^(-7) was obtained.
10-6.272 = 5.35*10-7
 
  • #16
Is it okay to introduce a negative sign to the 6.272?
 
  • #17
You calculated the ratio of neutral to negative ions. The ratio of negative ions to neutral ions is the inverse of this.
 
  • #18
Okay, thanks for pointing that out.