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

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The Saha equation is used to estimate the ratio of ionized to neutral hydrogen in the solar photosphere, with specific calculations provided for a temperature of 6,000 K and a pressure of log Pe = 2.7. Initial calculations incorrectly used an ionization energy of 13.6 eV for H-, leading to a much larger ratio than expected. Adjusting the ionization energy to 3.4 eV and correcting logarithmic calculations still did not yield the expected result of 5x10^(-7). Further discussion suggested using a binding energy of around 0.7 eV, which brought the results closer to the expected value. The conversation highlights the importance of accurate values for ionization energy and the distinction between neutral and negative hydrogen ions in these calculations.
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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.
 

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