Q.Starting from the formula for energy of an electron in the nth orbit of hydrogen atom, derive the formula for the wavelengths of Lyman and Balmer series spectral lines and determine the shortest wavelengths of lines in both these series.

[!TLDR] Photon energy on an $m\rightarrow n$ transition is $13.6(1/n^2-1/m^2)$ eV, giving $1/\lambda=R(1/n^2-1/m^2)$; shortest wavelengths (series limits, $m\rightarrow\infty$) are $1/R=911.6$ Å for Lyman (n=1) and $4/R=3646$ Å for Balmer (n=2). [!ANSWER] Lyman shortest $\lambda\approx911.6$ Å; Balmer shortest $\lambda\approx3646$ Å.

Starting from $E_n=-13.6/n^2$ eV, a transition from a higher orbit m down to a lower orbit n releases a photon of energy $\Delta E=E_m-E_n=13.6\left(\frac{1}{n^2}-\frac{1}{m^2}\right)$ eV. Using $\Delta E=hc/\lambda$ and converting eV to joules, this becomes the wavelength (Rydberg) formula $\frac{1}{\lambda}=R\left(\frac{1}{n^2}-\frac{1}{m^2}\right)$, with $R=1.097\times10^7$ m$^{-1}$. For the Lyman series, n=1 and m=2,3,4,...; for the Balmer series, n=2 and m=3,4,5,.... In each series, wavelength DECREASES as m increases, so the SHORTEST wavelength in a series is reached in the limit $m\rightarrow\infty$, where $1/m^2\rightarrow0$. For Lyman: $\frac{1}{\lambda_{min}}=R\left(1-0\right)=R$, so $\lambda_{min}=\frac{1}{R}=\frac{1}{1.097\times10^7}\approx9.116\times10^{-8}$ m = 911.6 Å. For Balmer: $\frac{1}{\lambda_{min}}=R\left(\frac{1}{4}-0\right)=\frac{R}{4}$, so $\lambda_{min}=\frac{4}{R}\approx3.646\times10^{-7}$ m = 3646 Å. [!ANSWER] Lyman shortest $\lambda\approx911.6$ Å; Balmer shortest $\lambda\approx3646$ Å.

Derive $1/\lambda=R(1/n^2-1/m^2)$ from $E_n=-13.6/n^2$ eV and $\Delta E=hc/\lambda$, then take the $m\rightarrow\infty$ limit (series limit = shortest wavelength) for each series.

Looking for the shortest wavelength at the smallest m (m=n+1, the FIRST line of the series) instead of the largest m ($m\rightarrow\infty$) -- wavelength actually decreases as m increases within a series, so the shortest wavelength is the series LIMIT, not the first line.