# Balmer & Rydberg equations

1. Jun 21, 2007

### bluestar

I think I am correct in saying that the Balmer & Rydberg formulas provided the foundation for quantum physics and if not they were significant in contribution. Anyway, I have been examining the original of both formulas and wondered if someone knew and can show me how Rydberg rewrote Balmers equation in such a manner that he was able to determine Balmers equation was a special case of his more generalized equation. I see the end result but don't know how he got there.

Balmer original equation: λ = hm2/(m2 − 4),….OK

Rydberg original equation: n = no − No/(m+m')2,….OK

Rydberg rewrite of Balmer equation: n = no − 4no/m2 ???

The reference link states: “This shows that hydrogen is a special case with m'=0 and No=4no. No is a universal constant common to all elements. Now this constant is known as the Rydberg constant, and m' is known as the quantum defect. Rydberg noted that m' is approximately the same different "diffuse" or different "sharp" series, but that diffuse and sharp series of the same order have essentially the same value of no. “
Note, n = 1/ λ
Also, n0 = n sub 0, and N0 = N sub 0
Likewise, m2 is m squared

http://w3.msi.vxu.se/~pku/Rydberg/LifeWork.html

Thanks a bunch,

2. Jun 21, 2007

### Staff: Mentor

If you invert this equation:
You get this equation, now in terms of n instead of λ:

3. Jun 21, 2007

### bluestar

Yeah, making the inversion was the first operation I performed.
Then I substituted n for 1/. But afterwards the derivation didn’t get any closer. I tried working it from the Balmer equation down and also from the Rydberg equation up. Unfortunately, both ways were unsuccessful.

I was able to make the constant substitutions into Rydberg’s original equation and come up the answer, but no luck when trying to derive the Balmer equation into the Rydberg solution.

Is it possible to show some of the math so I can get over this hump?

4. Jun 21, 2007

### Staff: Mentor

Starting with this:
$$\lambda = h m^2/(m^2 - 4)$$

Invert to get this:
$$n = (m^2 - 4)/(h m^2) = (1/h)(1 - 4/m^2)$$

But as m goes to infinity:
$$n_0 = 1/h$$

Thus:
$$n = n_0 (1 - 4/m^2) = n_0 - 4n_0/m^2$$

That's all there is to it. Make sense?

5. Jun 21, 2007

### bluestar

OK Great,
Step 3 was the ringer.
How did you know that n0=1/h?
By definition? By inference?
I didn’t notice it in the reference link?
I only noted “no is the series limit when the ordinal number m approaches infinity”
I can see the last operation in step two is similar to the second operation in the last step. Likewise when the limits of m go to infinity in both equations then they become equal.
Never mind. If my logic is correct I think I just answered my question.

Hey, Thanks a bunch!

6. Jun 22, 2007

### Staff: Mentor

I think you've got it.
That means that n_0 is n as we let m approach infinity in the second equation in post #4:
$$n_0 = \lim_{m\rightarrow \infty} n = \lim_{m\rightarrow \infty} (1/h)(1 - 4/m^2) = 1/h$$

Last edited: Jun 22, 2007
7. Jun 22, 2007

### bluestar

It should show that I had/have a weakness in determining when to apply limits and in resolving proofs.