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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: Cavat on November 28, 2007, 07:56:10 PM

Title: Rydberg formula
Post by: Cavat on November 28, 2007, 07:56:10 PM

I tried searching for the answer to my question, but I couldn't seem to find it...

The one thing I did notice was that when people were using Rydberg's formula, they had it in the form (1/wavelength)=Rh {1/(ni^2) - 1/(nf^2)}

However, the way our professor showed us the equation was :

deltaE = R_h {1/(n_i²) - 1/(n_f²)}

So I just want to make sure, these equations or similar / why teacher showed it this way...

Also I actually do have a question pertaining to the equation... We were going over this in a review session, and no one was able to get the answer, and the professor just kind of left us on how to solve it...

So this is what I have:

Wavelength = 95.2 nm
N_f = 1
Find N_i

So first, I found energy:
E = H (c/wavelength)
E = (6.63 x 10^-34) * (3.00 x 10⁸ / 95.2)
E = 2.089x10^-27

So into Rydberg's equation:

2.089x10^-27 = R_h * [1/(n_i²) - 1/(n_f²)]

2.089x10^-27 = (2.18x10^-18) * [(1/(n_i^{^2}) - (1/1)]
2.089x10^-27= (2.18x10^-18) * [1/(n_i²) - 1]
9.582x10^-10 = [1/(n_i²)] - 1
1.0000000009583 = [1 / (n_i²)]
1.0000000009583(n_i²) = 1
n_i² = 1/1.0000000009583
n_i² = .9999
n_i= 1

And I forgot to mention, the answer in the back of the book is n_i=5
So, I'm not sure if its just calculation error, or what the problem.

Any help is appreciated!
Thanks.

Title: Re: Rydberg formula
Post by: Sev on November 28, 2007, 08:30:42 PM

Quote

deltaE = R_h {1/(n_i²) - 1/(n_f²)}

dE = -hcR_H (1/n_f²-1/n_i²)

dE = hc/lambda, so 1/lambda = -hcR_H/hc (1/n_f²-1/n_i²) = R_H(1/n_i² - 1/n_f²)

R_H is 1.097*10⁷, the value you quote as R_H is hcR_H.

I haven't checked your working, but if n_i = n_f nothing has happened.

Title: Re: Rydberg formula
Post by: Cavat on November 28, 2007, 10:00:14 PM

Quote from: Sev on November 28, 2007, 08:30:42 PM

Quote

deltaE = R_h {1/(n_i²) - 1/(n_f²)}

dE = -hcR_H (1/n_f²-1/n_i²)

dE = hc/lambda, so 1/lambda = -hcR_H/hc (1/n_f²-1/n_i²) = R_H(1/n_i² - 1/n_f²)

R_H is 1.097*10⁷, the value you quote as R_H is hcR_H.

I haven't checked your working, but if n_i = n_f nothing has happened.

I'm a bit confused, what is hcR_h?

My teacher never taught us this...  She only taught us that Rydberg's constant is 2.18x10¹⁸

Title: Re: Rydberg formula
Post by: Sev on November 28, 2007, 10:12:16 PM

hcR_H is a combination of constants: h is planck's constant, c is speed of light.  hcR_H is 2.18*10^-18

Title: Re: Rydberg formula
Post by: sjb on November 29, 2007, 03:47:50 AM

Quote from: Cavat on November 28, 2007, 07:56:10 PM

I tried searching for the answer to my question, but I couldn't seem to find it...

The one thing I did notice was that when people were using Rydberg's formula, they had it in the form (1/wavelength)=Rh {1/(ni^2) - 1/(nf^2)}

However, the way our professor showed us the equation was :

deltaE = R_h {1/(n_i²) - 1/(n_f²)}

The two equations are at least formally identical, less a constant factor. Recall that c = f.λ and E = h.f, so E = h.c / λ

S

Title: Re: Rydberg formula
Post by: Padfoot on November 29, 2007, 04:32:33 AM

Quote

E = H (c/wavelength)
E = (6.63 x 10^-34) * (3.00 x 10⁸ / 95.2)

Also, wavelength is in m, not nm  :)

Title: Re: Rydberg formula
Post by: Cavat on December 03, 2007, 11:19:38 AM

haha, alright well thank you very much for the help.   :)

I finally ended up getting the answer after many tries...  Thankfully my teacher didn't end up putting one of these on the exam...