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Homework Help: Question Concerning Chemistry Model

  1. Jan 25, 2005 #1
    Hello all

    Bohr calculated the energies of each orbit of a hydrogen atom using the following forumula: [tex] (-2.18 \times \frac{1}{10^1^8} \frac{1}{n^2}) [/tex]
    Is it true that [tex] n \rightarrow \infty [/tex] the energy appraches 0?

    Also [tex] \Delta E = hv = \frac{hc}{\lambda} = (-2.18 \times \frac{1}{10^1^8})(\frac {1}{n^2_f} - \frac {1}{n^2_i}) [/tex] how do we get this? What is it saying?

    Thanks alot
    Last edited: Jan 25, 2005
  2. jcsd
  3. Jan 25, 2005 #2


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    1.Of course it approaches zero...

    2.It's the formula that gives the H atom's spectrum (the wavelengths of the radiation/photons emmited in a transition from one energy level to another)discovered empirically by Balmer in 1885 and refound by Bohr in 1913...

  4. Jan 25, 2005 #3
    I'm not sure how Balmer developed it, but it can be done as follows using the Bohr model of the atom:

    [tex]E_{n}=\frac{E_{1}}{n^{2}}[/tex] (Bohr)



    Hm...are you guys also not seeing the latex here? Something is weird...
  5. Jan 25, 2005 #4
    both showed up as errors to me. can tex have embedded spaces? clicking on the error msgs showed extra spaces.... but i'm worse than naive about this!
  6. Jan 25, 2005 #5


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    Sirus' post :

    [tex]E_n=\frac{E_1}{n^2}[/tex] (Bohr)

    [tex]\Delta E=E_f - E_i=\frac{E_1}{n_f^2}-\frac{E_1}{n_i^2} = E_1 ( \frac {1}{n_f^2} - \frac {1}{n_i^2} ) [/tex]

    [tex]\frac{1}{\lambda} = \frac {E_1}{hc} ( \frac{1}{n_f^2} - \frac{1}{n_i^2} )[/tex]
    Last edited: Jan 25, 2005
  7. Jan 26, 2005 #6
    Much thanks, Gokul. What happened/how did yours work?
  8. Jan 30, 2005 #7
    Could someone please explain to me what the formula is acutally telling us?

  9. Feb 1, 2005 #8
    the equation basically says that as an electron is raised to higher and higher orbits (higher and higher energy levels) around its nucleus, the amount of energy to raise it to the NEXT higher energy level (orbit) gets smaller and smaller.

    as n=>infinity, delta E goes to zero.

    for small values of n (lower orbits), the deltaE is large and pretty granular (not continuous); as n gets really high the energy change is less and less. the deltaE is also the source of energy emitted by LASERS, you know.....
  10. Feb 1, 2005 #9


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    Let's not mix Bohr's model with LASER effect,as the former cannot account for the latter.Okay??

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