Exercises — Hydrogen emission spectrum — Lyman, Balmer, Paschen, Brackett, Pfund

The series is fixed by the lower level $n_1$. Here $n_1=2$ → Balmer series. Balmer sits in the visible region. (Look at the green arrows in the staircase figure: all Balmer lines land on step 2.)

Read off $n_1$ each time.

• (a) $n_1=1$ → Lyman (UV)
• (b) $n_1=3$ → Paschen (IR)
• (c) $n_1=5$ → Pfund (IR)
• (d) $n_1=2$ → Balmer (visible)

True. The first line of a series ($n_2=n_1+1$) is the smallest energy jump, and smallest $\Delta E$ ⇒ largest $\lambda$ (because $\lambda \propto 1/\Delta E$). So $3\to 2$ (H-α, red, 656 nm) is the longest-wavelength Balmer line.

Balmer ⇒ $n_1=2$; second line ⇒ $n_2=4$. $\frac{1}{\lambda }=R_H\left(\frac{1}{2^2}-\frac{1}{4^2}\right)=1.097\times 10^7\left(\frac{1}{4}-\frac{1}{16}\right)$ $\frac{1}{4}-\frac{1}{16}=\frac{4-1}{16}=\frac{3}{16}=0.1875$. $\frac{1}{\lambda }=1.097\times 10^7\times 0.1875=2.057\times 10^6{\text{m}}^{-1}$. $\lambda =1/2.057\times 10^6=4.86\times 10^{-7}\text{m}$. Convert with $1\text{nm}=10^{-9}\text{m}$: $4.86\times 10^{-7}\text{m}\times 10^9=486\mathbf{nm}$ → blue-green, exactly the observed H-β. ✓

$\frac{1}{\lambda }=1.097\times 10^7\left(\frac{1}{1^2}-\frac{1}{2^2}\right)=1.097\times 10^7\times \frac{3}{4}=8.228\times 10^6{\text{m}}^{-1}$ $\lambda =1.215\times 10^{-7}\text{m}$. Convert with $1\text{nm}=10^{-9}\text{m}$: $1.215\times 10^{-7}\text{m}\times 10^9=121.5\mathbf{nm}$ → far UV (Lyman-α). ✓

At the limit $1/n_2^2\to 0$, so only the first term survives: $\frac{1}{\lambda }=\frac{R_H}{n_1^2}=\frac{1.097\times 10^7}{9}=1.219\times 10^6{\text{m}}^{-1}$ $\lambda =8.21\times 10^{-7}\text{m}$. Convert with $1\text{nm}=10^{-9}\text{m}$: $8.21\times 10^{-7}\text{m}\times 10^9=821\mathbf{nm}$ (near-IR). This is the shortest wavelength of the Paschen series. ✓

Energy of a photon: $\Delta E=hc/\lambda$. First put the wavelength in metres: $121.5\text{nm}=121.5\times 10^{-9}\text{m}=1.215\times 10^{-7}\text{m}$. $\Delta E=\frac{(6.626\times 10^{-34})(3.00\times 10^8)}{1.215\times 10^{-7}}=1.636\times 10^{-18}\text{J}$ Convert to eV: $\frac{1.636\times 10^{-18}}{1.602\times 10^{-19}}=10.2\mathbf{eV}$. Bohr check: $E_2-E_1=-\frac{13.6}{4}-\left(-\frac{13.6}{1}\right)=-3.4+13.6=10.2$ eV. ✓ They match — the photon carries exactly the electronic energy lost.

Wavenumber $\bar{\nu }\propto \Delta E$, so compare the brackets.

• Paschen $4\to 3$: $\frac{1}{9}-\frac{1}{16}=\frac{16-9}{144}=\frac{7}{144}=0.04861$.
• Brackett $5\to 4$: $\frac{1}{16}-\frac{1}{25}=\frac{25-16}{400}=\frac{9}{400}=0.02250$. $0.04861>0.02250$, so Paschen $4\to 3$ is more energetic. Deeper landing (smaller $n_1$) means a deeper part of the energy well, hence bigger drops. ✓

First line: $\frac{1}{{\lambda }_1}=R_H\left(\frac{1}{4}-\frac{1}{9}\right)=R_H\cdot \frac{5}{36}$. Series limit: $\frac{1}{{\lambda }_{\infty }}=R_H\cdot \frac{1}{4}$. Ratio of wavelengths is the inverse ratio of wavenumbers: $\frac{{\lambda }_1}{{\lambda }_{\infty }}=\frac{1/{\lambda }_{\infty }}{1/{\lambda }_1}=\frac{R_H\cdot \frac{1}{4}}{R_H\cdot \frac{5}{36}}=\frac{\frac{1}{4}}{\frac{5}{36}}=\frac{1}{4}\times \frac{36}{5}=\frac{36}{20}=\frac{9}{5}=1.8$ So ${\lambda }_1/{\lambda }_{\infty }=9/5=1.8$. The first line's wavelength is $1.8\times$ the limit wavelength, i.e. longer, exactly as expected since the first line is the smallest jump of the series. ✓

The series limit of series $n_1$ obeys $\frac{1}{{\lambda }_{\infty }}=\frac{R_H}{n_1^2}$, i.e. ${\lambda }_{\infty }=\frac{n_1^2}{R_H}$. So ${\lambda }_{\infty }\propto n_1^2$. Comparing Lyman ($n_1=1$) to Balmer ($n_1=2$): $\frac{{\lambda }_{\infty }^{\text{Lyman}}}{{\lambda }_{\infty }^{\text{Balmer}}}=\frac{1^2}{2^2}=\frac{1}{4}$ ${\lambda }_{\infty }^{\text{Lyman}}=\frac{1}{4}\times 364.6\text{nm}=91.2\mathbf{nm}$ The whole scale factor is just $n_1^2$ — Lyman's limit is a quarter of Balmer's, sitting in the far UV. ✓ (Matches the parent note's Lyman limit.)

Energy is a state function — it depends only on the endpoints, so any route must give the same total. Verify with $E_n=-13.6/n^2$ eV. Direct $5\to 1$: $E_1-E_5=-13.6-(-13.6/25)=-13.6+0.544=-13.056$ eV emitted, i.e. $13.056$ eV. Cascade $5\to 2$: $E_2-E_5=-3.40-(-0.544)=-2.856$ eV → $2.856$ eV emitted. Cascade $2\to 1$: $E_1-E_2=-13.6-(-3.40)=-10.2$ eV → $10.2$ eV emitted. Sum of cascade: $2.856+10.2=13.056$ eV. ✓ Identical to the direct fall. Each individual step is still a legal spectral line (one Paschen-region, one Lyman-α), so a real gas shows all of them.

See Rydberg Constant and Spectra of Hydrogen-like Ions. The only change is the factor $Z^2=4$. $\frac{1}{\lambda }=1.097\times 10^7\times 4\times \left(\frac{1}{1^2}-\frac{1}{2^2}\right)=1.097\times 10^7\times 4\times 0.75=3.291\times 10^7{\text{m}}^{-1}$ $\lambda =3.04\times 10^{-8}\text{m}$. Convert with $1\text{nm}=10^{-9}\text{m}$: $3.04\times 10^{-8}\text{m}\times 10^9=30.4\mathbf{nm}$. Compared to H's Lyman-α at 121.5 nm, the He⁺ line is exactly $\frac{1}{Z^2}=\frac{1}{4}$ as long: $121.5/4=30.4$ nm. Higher nuclear charge pulls levels deeper, so bigger jumps and shorter wavelengths. ✓

Absorbed energy raises the electron: $E_n-E_1=12.75$ eV. $E_1=-13.6$ eV, so $E_n=-13.6+12.75=-0.85$ eV. Solve $-13.6/n^2=-0.85\Rightarrow n^2=13.6/0.85=16\Rightarrow n=4$. Number of distinct downward lines from level $n$ back toward level 1 is the number of ways to pick 2 levels out of $\{1,2,3,4\}$: $\left(\genfrac{}{}{0ex}{}{4}{2}\right)=\frac{4\cdot 3}{2}=6$ lines. (Those 6: $4\to 3,4\to 2,4\to 1,3\to 2,3\to 1,2\to 1$ — spanning Paschen, Balmer and Lyman regions.) ✓

$\frac{1}{\lambda }=1.097\times 10^7\left(\frac{1}{1}-\frac{1}{9}\right)=1.097\times 10^7\times \frac{8}{9}=9.751\times 10^6{\text{m}}^{-1}$ $\lambda =1.026\times 10^{-7}\text{m}$. Convert with $1\text{nm}=10^{-9}\text{m}$: $1.026\times 10^{-7}\text{m}\times 10^9=102.6\mathbf{nm}$. Ordering: $91.2<102.6<121.5$ nm. ✓ As $n_2$ climbs $2\to 3\to \infty$, the Lyman wavelengths shrink from 121.5 → 102.6 → 91.2 nm, converging on the limit. This matches the "lines crowd together toward the series limit" pattern in the figure.

A wavelength is shortest when $\bar{\nu }=R_H\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$ is largest. To maximise the bracket: make the positive term $\frac{1}{n_1^2}$ as large as possible ⇒ smallest $n_1=1$; and make the subtracted term $\frac{1}{n_2^2}$ as small as possible ⇒ largest $n_2\to \infty$ (term $\to 0$). Then ${\bar{\nu }}_{\max }=R_H(1-0)=R_H=1.097\times 10^7{\text{m}}^{-1}$, giving ${\lambda }_{\min }=\frac{1}{R_H}=9.12\times 10^{-8}\text{m}=91.2\mathbf{nm}(\text{using }1\text{nm}=10^{-9}\text{m}).$ No emission line can be shorter because $\frac{1}{n_1^2}\le 1$ always (since $n_1\ge 1$) and $\frac{1}{n_2^2}\ge 0$ always. This shortest wavelength equals the energy to fully ionize from the ground state — see Ionization Energy of Hydrogen (13.6 eV). ✓

$4\to 2$)? $\approx 486$ nm (blue-green), from $\bar{\nu }=R_H\cdot \frac{3}{16}$.

The Lyman limit, $\lambda =1/R_H\approx 91.2$ nm — maximise $\frac{1}{n_1^2}-\frac{1}{n_2^2}$ with $n_1=1,n_2\to \infty$.

$n$? $\left(\genfrac{}{}{0ex}{}{n}{2}\right)=\frac{n(n-1)}{2}$ distinct downward transitions.

$Z^2=4$ on the wavenumber, so wavelengths are $1/4$ of hydrogen's.