Solving for n_1 and n_2 in Hydrogen Line Spectra

AI Thread Summary
To solve for the quantum numbers n_1 and n_2 in the hydrogen line spectra, the Rydberg Equation is used: 1/λ = R_H(1/n_2^2 - 1/n_1^2). The wavelengths provided correspond to the Balmer series, with n_1 being less than n_2. To find n_1 and n_2, start by assuming values for n_1, such as 1 or 2, and solve for n_2 as an integer. It's also important to ensure the correct value for Rydberg's constant is used in calculations. Understanding the selection rules and ionization energy can further aid in solving the problem.
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I need help on this problem that asks me to solve for n_1 and n_2 (the initial and final quantum numbers).

This is the given information (the line spectra for Hydrogen):

color red known wavelength: 656.4 nm
color turquoise known wavelength: 486.3 nm
purple wavelength: 434.2 nm
purple wavelength: 410.3 nm

n_1 and n_2 are unknown for each one.

I've tried using this equation, the Rydberg Equation, to solve for n_1 and n_2:

\frac {1}{\lambda} = (R_H)(\frac {1}{n^2_2} - \frac {1}{n^2_2})

where \lambda is the wavelength, n_1 and n_2 are the initial and final principal quantum numbers, with the initial one being larger than the final one. R_H is Ryberg's constant.

I've plugged in the numbers and (for the color red) I got
\frac {1}{656.4} = \frac {R_H}{n^2_1} - \frac{R_H}{n^2_2}

I still can't find n1 and n2. Am I using the right formula? I just couldn't understand how to solve a problem with 2 variables.
 
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Well n2 > n1, so pick n1 = 1, then solve for n2, which must be an integer.

If that doesn't work, then try n1 = 2, and solve for n2.

Alternatively, one can select n1=1, and then using n2 = 2, 3, 4, . . . solve for the wave lengths.

What value is one using for Rydberg's constant.

Try this reference - http://hyperphysics.phy-astr.gsu.edu/Hbase/hyde.html

Hint - the visible lines are in the Balmer series.
 
Also, consider the selection rules as well as the determination of the ionization energy.
 
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