Q.Show that the frequency of the first line in Lyman series is equal to the difference between the limiting frequencies of Lyman and Balmer series.

[!TLDR] The Lyman series limit frequency is $Rc$ and the Balmer series limit frequency is $Rc/4$; their difference, $Rc-Rc/4=3Rc/4$, is exactly the frequency of Lyman's first line (n=1 to m=2), $\nu=Rc\left(1-\frac14\right)=\frac{3Rc}{4}$. [!ANSWER] Both equal $3 …

The frequency of any spectral line is $\nu=c/\lambda=Rc\left(\frac{1}{n^2}-\frac{1}{m^2}\right)$. The first line of the Lyman series is the n=1, m=2 transition, giving $\nu_{L1}=Rc\left(\frac{1}{1}-\frac{1}{4}\right)=\frac{3Rc}{4}$. The Lyman series LIMIT is reached as $m\rightarrow\infty$ with n=1, giving $\nu_{L,\text{limit}}=Rc\left(1-0\right)=Rc$. The Balmer series limit is reached as $m\rightarrow\infty$ with n=2, giving $\nu_{B,\text{limit}}=Rc\left(\frac{1}{4}-0\right)=\frac{Rc}{4}$. Subtracting the two limiting frequencies gives $\nu_{L,\text{limit}}-\nu_{B,\text{limit}}=Rc-\frac{Rc}{4}=\frac{3Rc}{4}$ -- exactly equal to $\nu_{L1}$, the frequency of the very first Lyman line, computed above. This is not a coincidence: physically, the Lyman limit represents the energy to ionize an electron directly from n=1, while the …

Write each of the three frequencies (first Lyman line, Lyman limit, Balmer limit) using $\nu=Rc(1/n^2-1/m^2)$ with the appropriate n and m (including $m\rightarrow\infty$ f …

Confusing a series 'limit' (the m to infinity, shortest-wavelength convergence point) with its FIRST line (the m=n+1 transition, its longest-wavelength, least energetic member) -- the pro …