Question on the Rydberg - Ritz Fromula The question is to determine which of the following wavelengths
$\lambda_1=7460nm$  $\lambda_2 = 4654nm$ $\lambda_3 = 4103nm$ $\lambda_4 = 3741 nm  $ 
does not belong to hydrogen. My guess is that the Rydberg - Ritz Formula $$\frac{1}{\lambda} = R(\frac{1}{m^2}-\frac{1}{n^2})$$ where $R_H = 1.097*10^7\frac{1}{m}$  is what will be used to solve this problem. My question is what determines the integers m and n? The textbook could not be more vague on the issue. I know that the wavelength of hydrogen must have integer values for n and m am I suppose to guess or what am I missing?
 A: m and n are natural numbers and shell number of atom, and the condition that wavelength is positive applies, and btw formula isn't complete there is a z^2 on right side too for general 1 electron species
set z=1 and find maximum and minimum value of wavelength and check if any of the given are out of that range
Hint : try and use values like 1 and infinity to maximise or minimise it
extra : it is not really a formula but is using energy difference between 2 shells and equating it with the energy of photon emmited I.e h*frequency
and there are specific name series for values of m 1,2,3,4 are Lyman, balmer, paschen, brakett,pfund
A: $n$ and $m$ are indeed integers. Whenever a photon is emitted or absorbed, the electron transitions from one energy level to another. $n$ and $m$ are the numbers associated with these energy levels. That is, if the electron transitions from level $6$ to level $3$, you would use the formula with $n = 6$ and $m = 3$ to find the wavelength of the emitted photon. Thus $n$ and $m$ can be any positive integers (or $\infty$ in the case of bound-free transitions).
By the way, for fixed $m$ and varying $n$ you get all the members of each "series" of hydrogen lines. For example, $m = 2$ corresponds to the Balmer series.
A: The values of $1/(\lambda R)$ are $0.0122,\,0.0196,\,0.0222,\,0.0244$. The closest one can get to $m^{-2}$ with values of $n^{-2}$ for $n>m$ is to take $n=m+1$, giving a difference of $\frac{2m+1}{m(m+1)}$. This does not become as small as desired until we reach $m=4$, which unfortunately doesn't allow us to reproduce any of these values. But for $m=5$ we get the desired values for $n=6,\,7,\,?,\,8$. The question mark denotes our not achieving $1/(\lambda_3 R)$ when $m=5$. 
