Exercises — Rydberg formula 1 - λ = R(1 - n₁² − 1 - n₂²)

The electron falls from level 5 to level 2, so it loses energy → emission. The lower (landing) level is $n_1=2$; the higher (start) level is $n_2=5$. Slots: $n_1=2$, $n_2=5$. The rule $n_2>n_1$ is satisfied ($5>2$). ✓

$n_1=3$ is the Paschen series, lying in the infrared. (Landing on $1$ = Lyman/UV, $2$ = Balmer/visible, $3$ = Paschen/IR.)

$\frac{1}{\lambda }=R_H\left(\frac{1}{2^2}-\frac{1}{4^2}\right)=1.097\times 10^7\left(\frac{1}{4}-\frac{1}{16}\right)$ $\frac{1}{4}-\frac{1}{16}=\frac{4-1}{16}=\frac{3}{16}=0.1875$ $\frac{1}{\lambda }=1.097\times 10^7\times 0.1875=2.057\times 10^6{\text{m}}^{-1}$ $\lambda =\frac{1}{2.057\times 10^6}=4.86\times 10^{-7}\text{m}=486\text{nm (blue-green)}.$

$\frac{1}{\lambda }=R_H\left(\frac{1}{1^2}-\frac{1}{2^2}\right)=1.097\times 10^7\left(1-\frac{1}{4}\right)=1.097\times 10^7\times \frac{3}{4}$ $=8.228\times 10^6{\text{m}}^{-1}\Rightarrow \lambda =1.215\times 10^{-7}\text{m}=121.5\text{nm (UV)}.$

Way 1 — from the wavelength, using $E=hc/\lambda$: $E=\frac{hc}{\lambda }=\frac{(6.626\times 10^{-34})(3.00\times 10^8)}{656\times 10^{-9}}=3.03\times 10^{-19}\text{J}=\frac{3.03\times 10^{-19}}{1.602\times 10^{-19}}=1.89\text{eV}.$ Way 2 — from the Bohr ladder (the why of that number). H-alpha is the $3\to 2$ drop, so the photon carries the gap between those rungs. Using $E_n=-\frac{13.6\text{eV}}{n^2}$ from the Bohr model: $E_2=-\frac{13.6}{2^2}=-3.40\text{eV},E_3=-\frac{13.6}{3^2}=-1.51\text{eV}.$ The photon energy is the size of the fall $E_3-E_2$: $E_{\text{photon}}=E_3-E_2=(-1.51)-(-3.40)=+1.89\text{eV}.✓$ Both roads land on $1.89$ eV — the wavelength route and the energy-ladder route are the same physics (that is exactly what the Rydberg formula encodes).

Shortest $\lambda$ = biggest energy gap = largest bracket, which happens as $n_2\to \infty$: $\frac{1}{{\lambda }_{\min }}=R_H\left(\frac{1}{2^2}-0\right)=\frac{R_H}{4}=\frac{1.097\times 10^7}{4}=2.7425\times 10^6{\text{m}}^{-1}$ ${\lambda }_{\min }=3.646\times 10^{-7}\text{m}=364.6\text{nm}.$ Physically: an electron from $n=\infty$ (a free electron, just barely unbound) drops into $n=2$. Equivalently it is the threshold for ionising an electron that sits in $n=2$. Go back to the staircase figure in the intro: this is the tallest arrow that still lands on rung 2, and every finite $n_2\to 2$ line sits just below it — which is why the lines pile up at this edge.

Bluer = larger bracket. Compare: $3\to 1:1-\frac{1}{9}=\frac{8}{9}=0.889,4\to 2:\frac{1}{4}-\frac{1}{16}=\frac{3}{16}=\mathrm{0.1875.}$ $0.889>0.1875$, so $3\to 1$ has the bigger gap → bluer (in fact deep UV, Lyman-beta). Numbers: $3\to 1$ gives $1/\lambda =1.097\times 10^7\times 0.889=9.751\times 10^6\Rightarrow \lambda =102.6\text{nm}$; $4\to 2$ gives $486\text{nm}$. Indeed $102.6<486$. ✓

$\frac{1}{\lambda }=\frac{1}{434\times 10^{-9}}=2.304\times 10^6{\text{m}}^{-1}$ $\frac{1}{\lambda R_H}=\frac{2.304\times 10^6}{1.097\times 10^7}=0.2100=\frac{1}{4}-\frac{1}{n_2^2}$ $\frac{1}{n_2^2}=0.25-0.2100=0.0400\Rightarrow n_2^2=25\Rightarrow n_2=5.$ So it is the $5\to 2$ (H-gamma) line. ✓

Absorption vs emission changes only the direction of energy flow, never the wavelength. Sort the levels first: the lower is $1$, the higher is $3$, so $n_1=1$, $n_2=3$. $\frac{1}{\lambda }=R_H\left(\frac{1}{1^2}-\frac{1}{3^2}\right)=1.097\times 10^7\left(1-\frac{1}{9}\right)=1.097\times 10^7\times \frac{8}{9}$ $=9.751\times 10^6{\text{m}}^{-1}\Rightarrow \lambda =1.026\times 10^{-7}\text{m}=102.6\text{nm (UV)}.$ This is exactly the same number as the $3\to 1$ emission line in L3·Q2 — the atom absorbs precisely the colour it would later emit. Now the wrong order $n_1=3,n_2=1$: $\frac{1}{\lambda }=R_H\left(\frac{1}{3^2}-\frac{1}{1^2}\right)=1.097\times 10^7\left(\frac{1}{9}-1\right)=1.097\times 10^7\times \left(-\frac{8}{9}\right)<0,$ so $\lambda <0$ — a negative wavelength, which is physically impossible. That impossible sign is the alarm bell telling you the slots were swapped.

For a one-electron ion the whole formula scales by $Z^2$: $\frac{1}{\lambda }=R_H{Z}^2\left(\frac{1}{2^2}-\frac{1}{3^2}\right)=1.097\times 10^7\times 4\times \frac{5}{36}$ $=1.097\times 10^7\times 4\times 0.1389=6.095\times 10^6{\text{m}}^{-1}$ $\lambda =1.641\times 10^{-7}\text{m}=164.1\text{nm}.$ This is exactly $656/4=164$ nm: because every wavenumber is multiplied by $Z^2=4$, every wavelength is divided by 4. He⁺ pushes hydrogen's red line into the UV.

For the same transition, $\lambda \propto 1/R$. So the fractional shift is $\frac{{\lambda }_H-{\lambda }_D}{{\lambda }_H}=\frac{1/R_H-1/R_D}{1/R_H}=1-\frac{R_H}{R_D}=1-\frac{1.09678}{1.09707}=1-0.999736=2.64\times 10^{-4}.$ A shift of about $0.0264\%$. On the $656.28$ nm line that is $\Delta \lambda \approx 656.28\times 2.64\times 10^{-4}\approx 0.173\text{nm}$ — small, but this isotope shift is exactly how deuterium was first discovered.

The series limit is the photon that just frees an electron from $n=1$: $E=\frac{hc}{{\lambda }_{\min }}=\frac{(6.626\times 10^{-34})(3.00\times 10^8)}{91.13\times 10^{-9}}=2.181\times 10^{-18}\text{J}$ $E=\frac{2.181\times 10^{-18}}{1.602\times 10^{-19}}=13.62\text{eV}\approx 13.6\text{eV}.✓$ The energy needed to lift an electron from $n=1$ to $n=\infty$ is exactly the energy of the photon emitted going the other way — i.e. $0-E_1=0-(-13.6)=13.6$ eV on the Bohr ladder.

Longest Lyman line = smallest gap in Lyman = $2\to 1$: $\frac{1}{{\lambda }_{\text{Ly,max}}}=R_H\left(1-\frac{1}{4}\right)=\frac{3}{4}{R}_H=8.228\times 10^6\Rightarrow \lambda =121.5\text{nm}.$ Shortest Balmer line = Balmer series limit = $\infty \to 2$: $\frac{1}{{\lambda }_{\text{Ba,min}}}=R_H\left(\frac{1}{4}-0\right)=\frac{1}{4}{R}_H=2.743\times 10^6\Rightarrow \lambda =364.6\text{nm}.$ The entire Lyman series lives at $\lambda \le 121.5$ nm; the entire Balmer series lives at $\lambda \ge 364.6$ nm. Since $121.5<364.6$, they never overlap — there is a clean gap (121.5–364.6 nm) with no lines from either. The figure below draws both bands and the empty gap between them.

$\frac{1}{\lambda }=R_H{Z}^2\left(1-\frac{1}{4}\right)=\frac{3}{4}{R}_H{Z}^2$ $\frac{1}{30.4\times 10^{-9}}=3.289\times 10^7{\text{m}}^{-1}=\frac{3}{4}(1.097\times 10^7)Z^2=8.228\times 10^6{Z}^2$ $Z^2=\frac{3.289\times 10^7}{8.228\times 10^6}=3.997\approx 4\Rightarrow Z=2.$ $Z=2$, one electron → He⁺. (Its Lyman-alpha is at hydrogen's $121.5/4=30.4$ nm.) ✓

Since $\lambda \propto 1/(\text{bracket})$, the ratio is the inverse ratio of brackets: ${\text{Bracket}}_A=1-\frac{1}{9}=\frac{8}{9},{\text{Bracket}}_B=\frac{1}{4}-\frac{1}{36}=\frac{9-1}{36}=\frac{8}{36}=\frac{2}{9}.$ $\frac{{\lambda }_A}{{\lambda }_B}=\frac{{\text{Bracket}}_B}{{\text{Bracket}}_A}=\frac{2/9}{8/9}=\frac{2}{8}=\frac{1}{4}.$ Line A is one-quarter the wavelength of line B — the $R_H$ cancels entirely. Notice $6\to 2$ is the "$Z$-scaled twin" of $3\to 1$ (double both levels), a neat self-similarity of the $1/n^2$ ladder.

Sort levels so $n_1<n_2$; the bracket $(1/n_1^2-1/n_2^2)$ is proportional to energy and to $1/\lambda$; multiply by $Z^2$ for one-electron ions; take the reciprocal last to get $\lambda$; series limits give finite shortest wavelengths, and ratios let $R_H$ cancel.