Question Concerning Chemistry Model

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Homework Help Overview

The discussion revolves around the Bohr model of the hydrogen atom, specifically focusing on the energy levels of electrons and the transitions between these levels. Participants are examining the implications of energy approaching zero as the principal quantum number \( n \) approaches infinity, as well as the relationships between energy and wavelength in atomic transitions.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the behavior of energy levels as \( n \) increases, questioning the meaning of energy approaching zero. There are attempts to clarify the derivation of the energy transition formula and its implications for atomic spectra.

Discussion Status

The discussion is active, with participants providing insights into the formulas and their historical context. Some participants express confusion regarding the mathematical representation, while others seek clarification on the meaning of the equations and their applications.

Contextual Notes

There are mentions of potential issues with LaTeX formatting in the discussion, which may affect the clarity of mathematical expressions. Additionally, there is a caution against conflating the Bohr model with laser physics, indicating a need for careful distinction between concepts.

courtrigrad
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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 a lot
 
Last edited:
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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...

Daniel.
 
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)

[tex]\Delta{E}=E_{f}-E{i}=\frac{E_{1}}{n_{f}^{2}}-\frac{E_{1}}{n_{i}^{2}}=E_{1}\left({\frac{1}{n_{f}^{2}-\frac{1}{n_{i}^{2}}\right)[/tex]

[tex]\frac{1}{\lambda}=\frac{E_{1}}{hc}\left({\frac{1}{n_{f}^{2}-\frac{1}{n_{i}^{2}}\right)[/tex]

Hm...are you guys also not seeing the latex here? Something is weird...
 
Sirus said:
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)

[tex]\Delta{E}=E_{f}-E{i}=\frac{E_{1}}{n_{f}^{2}}-\frac{E_{1}}{n_{i}^{2}}=E_{1}\left({\frac{1}{n_{f}^{2}-\frac{1}{n_{i}^{2}}\right)[/tex]

[tex]\frac{1}{\lambda}=\frac{E_{1}}{hc}\left({\frac{1}{n_{f}^{2}-\frac{1}{n_{i}^{2}}\right)[/tex]

Hm...are you guys also not seeing the latex here? Something is weird...

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!
 
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:
Much thanks, Gokul. What happened/how did yours work?
 
Could someone please explain to me what the formula is acutally telling us?

Thanks
 
courtrigrad said:
Could someone please explain to me what the formula is acutally telling us?

Thanks

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...
 
plusaf said:
the deltaE is also the source of energy emitted by LASERS, you know...

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

Daniel.
 

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