PYQ Vault

MHT-CET Physics · Structure of Atoms and Nuclei

The Hydrogen Spectrum and its Series

Every line of the hydrogen spectrum is a jump down to a fixed lower level, with 1/λ = RZ²(1/n_f² − 1/n_i²); the lower level names the series — Lyman, Balmer, Paschen, Brackett — and within a series the first line is the longest wavelength and the series limit the shortest.

Why this matters

28 PYQs, 7 of them HARD. Sixteen are about the series themselves: which one is visible, the frequency or wave number of a series limit, and relations between limits and first lines. Twelve ask for the ratio of two named lines, sometimes in a hydrogen-like ion, or for the level a photon came from. Two cards.

Concept 1 of 2: The Series, Their First Lines and Their Limits

Fix the lower level n_f and let the upper level run: n_f = 1 gives the Lyman series (ultraviolet), 2 the Balmer (visible), 3 the Paschen, 4 the Brackett and 5 the Pfund (all infrared). The first line comes from the nearest level, n_f + 1, and has the smallest energy, so the LONGEST wavelength. The series limit comes from n = ∞, 1/λ = R/n_f², and has the SHORTEST wavelength. Frequency is c/λ and wave number is 1/λ, so the Balmer limit has wave number R/4 and frequency Rc/4. Because limits and first lines are differences of the same terms, they combine: the Lyman limit minus the first Lyman line is the Balmer limit. A source moving away stretches every line by Δλ/λ = v/c.

Definition

  • Lyman (n_f = 1, UV), Balmer (2, visible), Paschen (3), Brackett (4), Pfund (5) — infrared.
  • First line (nf+1→nf)(n_f + 1 \to n_f): longest λ. Series limit (∞→nf)(\infty \to n_f): shortest λ, 1λ=Rnf2\dfrac{1}{\lambda} = \dfrac{R}{n_f^2}.
  • Balmer limit: wave number R4\dfrac{R}{4}, frequency Rc4\dfrac{Rc}{4}. Lyman max/min λ = 4 : 3.
  • Limits in a ratio: λLymanλBalmer=14\dfrac{\lambda_{\text{Lyman}}}{\lambda_{\text{Balmer}}} = \dfrac{1}{4}; Paschen/Balmer = 9/4.
  • Combining terms: 1λL,∞−1λL,1=1λB,∞\dfrac{1}{\lambda_{L,\infty}} - \dfrac{1}{\lambda_{L,1}} = \dfrac{1}{\lambda_{B,\infty}}; νB,∞−νB,1=νP,∞\nu_{B,\infty} - \nu_{B,1} = \nu_{P,\infty}.
  • Photon from an energy gap: λ=hcΔE=1240ΔE (eV)\lambda = \dfrac{hc}{\Delta E} = \dfrac{1240}{\Delta E\,(\text{eV})} nm.
  • Receding source: Δλλ=vc\dfrac{\Delta\lambda}{\lambda} = \dfrac{v}{c}.

Series limit

1λ∞=Rnf2,ν∞=Rcnf2\frac{1}{\lambda_\infty} = \frac{R}{n_f^2}, \qquad \nu_\infty = \frac{Rc}{n_f^2}

Worked example

With R = 1.097 × 10⁷ m⁻¹, what is the wavelength of the Balmer series limit?
Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 12th May Shift 1 · Q16Moderate

Example 1 · Structure of Atoms and Nuclei · Hydrogen Spectrum and Spectral Series

In Lyman series, series limit of wavelength is λ1\lambda_{1}. The wavelength of first line of Lyman series is λ2\lambda_{2} and in Balmer series, the series limit of wavelength is λ3\lambda_{3}. Then the relation between λ1\lambda_{1}, λ2\lambda_{2} and λ3\lambda_{3} is

Calling the series limit the longest wavelength

The limit comes from n = ∞, the biggest jump into that level, so it has the most energy and the SHORTEST wavelength. The first line is the longest.

Using n_f = 1 for every series

Only Lyman ends on n = 1. The Balmer limit is R/4, not R; Paschen's is R/9. Set the lower level from the series name before substituting.

Concept 2 of 2: Ratios of Two Lines

Write 1/λ = RZ²(1/n_f² − 1/n_i²) for each line and divide; R cancels, and so does Z if both lines are in one atom. The wavelength ratio is the INVERSE of the ratio of the brackets, and the frequency ratio is the ratio of the brackets itself. The same transition in two hydrogen-like ions gives λ ∝ 1/Z². To find the level a photon came from, solve the formula for n_i. The highest-frequency emission is the one with the biggest bracket, which is usually a jump into n = 1: 2 → 1 beats 5 → 3 and 6 → 2. A jump UP is absorption, not emission.

Definition

  • 1λ=RZ2(1nf2−1ni2)\dfrac{1}{\lambda} = RZ^2\left(\dfrac{1}{n_f^2} - \dfrac{1}{n_i^2}\right); λ1λ2\dfrac{\lambda_1}{\lambda_2} = inverse ratio of the brackets.
  • Common brackets: 2→1:342\to1: \tfrac{3}{4}; 3→1:893\to1: \tfrac{8}{9}; 3→2:5363\to2: \tfrac{5}{36}; 4→2:3164\to2: \tfrac{3}{16}; 4→3:71444\to3: \tfrac{7}{144}; 5→4:94005\to4: \tfrac{9}{400}.
  • First Paschen to first Lyman: λPλL=3/47/144=1087\dfrac{\lambda_P}{\lambda_L} = \dfrac{3/4}{7/144} = \dfrac{108}{7}.
  • Same transition, two ions: λ∝1Z2\lambda \propto \dfrac{1}{Z^2} (He⁺⁺ : Li⁺⁺⁺ ⇒ 9 : 4).
  • Level of origin, photon λ into n = 1: n=λRλR−1n = \sqrt{\dfrac{\lambda R}{\lambda R - 1}}.

Rydberg formula

1λ=RZ2(1nf2−1ni2)\frac{1}{\lambda} = RZ^2\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)

Worked example

With R = 1.097 × 10⁷ m⁻¹, wavelength of the first Balmer line (3 → 2)?
Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 3rd May Shift 1 · Q49Hard

Example 2 · Structure of Atoms and Nuclei · Hydrogen Spectrum and Spectral Series

If 'λ1\lambda_1' and 'λ2\lambda_2' are the wavelengths of the first line of the Lyman and Paschen series respectively, then λ2:λ1\lambda_2:\lambda_1 is

Dividing wavelengths the way the brackets divide

The bracket is 1/λ. A larger bracket means a SHORTER wavelength, so the wavelength ratio is the brackets' ratio turned upside down.

Counting an upward jump as emission

n = 1 to n = 2 absorbs a photon. 'Which transition emits the highest frequency' excludes every jump upward, however large.

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

  • The Series, Their First Lines and Their Limits

    Series limit

    1λ∞=Rnf2,ν∞=Rcnf2\frac{1}{\lambda_\infty} = \frac{R}{n_f^2}, \qquad \nu_\infty = \frac{Rc}{n_f^2}
  • Ratios of Two Lines

    Rydberg formula

    1λ=RZ2(1nf2−1ni2)\frac{1}{\lambda} = RZ^2\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)

Watch out for (4)

Test yourself on Structure of Atoms and Nuclei

20 past MHT-CET questions from this chapter, timed at 18 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.