Question Concerning Chemistry Model

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The discussion centers on Bohr's formula for calculating the energy levels of a hydrogen atom, specifically addressing whether the energy approaches zero as n approaches infinity. Participants confirm that as n increases, the energy indeed approaches zero, indicating that the energy required to transition to higher orbits diminishes. The formula also relates to the wavelengths of emitted radiation during electron transitions between energy levels, as originally described by Balmer and later refined by Bohr. It is noted that the energy change (ΔE) becomes smaller at higher energy levels, with implications for understanding laser emissions. The conversation highlights the distinction between Bohr's model and laser physics, emphasizing that they should not be conflated.
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Hello all

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

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

Thanks alot
 
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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:

E_{n}=\frac{E_{1}}{n^{2}} (Bohr)

\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)

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

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:

E_{n}=\frac{E_{1}}{n^{2}} (Bohr)

\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)

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

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 :

E_n=\frac{E_1}{n^2} (Bohr)

\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} )

\frac{1}{\lambda} = \frac {E_1}{hc} ( \frac{1}{n_f^2} - \frac{1}{n_i^2} )
 
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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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