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Saha equation / hydrogen

  1. May 30, 2015 #1
    1. The problem statement, all variables and given/known data
    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.


    2. Relevant equations


    3. 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).
     
  2. jcsd
  3. May 30, 2015 #2

    mfb

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    The ionization energy is 13.6 eV for the transition H <-> H+, it is lower for H- <-> H.
     
  4. May 30, 2015 #3
    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.
     
  5. May 30, 2015 #4

    mfb

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    A large log means the ratio is large, but that means your H- are rare.
     
  6. May 31, 2015 #5
    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).
     
  7. May 31, 2015 #6
    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).
     
  8. May 31, 2015 #7

    mfb

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    "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.
     
  9. May 31, 2015 #8
    I thought it was all to the base 10. What do I have that is to the base e?
     
  10. May 31, 2015 #9

    mfb

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    log(2/1)
     
  11. May 31, 2015 #10
    How do I convert log(2/1) to base e, to the base 10?
     
  12. May 31, 2015 #11

    mfb

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    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.
     
  13. May 31, 2015 #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).
     
  14. May 31, 2015 #13

    mfb

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    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.
     
  15. May 31, 2015 #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.
     
  16. May 31, 2015 #15

    mfb

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    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

    10-6.272 = 5.35*10-7
     
  17. May 31, 2015 #16
    Is it okay to introduce a negative sign to the 6.272?
     
  18. May 31, 2015 #17

    mfb

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    You calculated the ratio of neutral to negative ions. The ratio of negative ions to neutral ions is the inverse of this.
     
  19. May 31, 2015 #18
    Okay, thanks for pointing that out.
     
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