Hydrogen Spectrum: The Fingerprint of Atoms

The Stellar Spectroscopy Connection 🎬

Just like forensic analysis in crime shows identifies substances from traces, astronomers identify elements in distant stars from their spectral fingerprints! Hydrogen’s spectrum, decoded by Bohr, revealed the quantum nature of atoms and unlocked the composition of the universe.

“The nitrogen in our DNA, the calcium in our teeth, the iron in our blood… were made in the interiors of collapsing stars. We are made of star stuff.” - Carl Sagan

The Historical Journey

Pre-Quantum Era

1814: Fraunhofer discovers dark lines in solar spectrum 1885: Balmer finds empirical formula for visible H lines 1890s: Lyman, Paschen, Brackett, Pfund discover other series

The Mystery

Why discrete lines?
Why specific wavelengths?
What's the pattern?
     ↓
Balmer's formula (empirical):
λ = 364.6 × n²/(n²-4) nm
     ↓
    BUT WHY?

Bohr’s Breakthrough (1913)

Derived ALL spectral series from first principles:

• Explained Balmer’s formula
• Predicted undiscovered series
• Calculated Rydberg constant

Triumph of quantum theory! 🎯

Emission vs Absorption Spectrum

Emission Spectrum

Electron jumps from higher to lower energy level:

    Excited atom
         ↓
    n=3 ──────
         ↓ hν (photon emitted)
    n=2 ──────
         ↓
    n=1 ──────
         ↓
    Ground state

Bright lines on dark background

How produced:

• Heat gas / pass electric discharge
• Atoms get excited
• Emit light at specific wavelengths
• See colored lines

Absorption Spectrum

Electron jumps from lower to higher energy level:

    White light → Gas → Detector

    Atom absorbs photon of exact energy

    n=1 ──────
         ↑ hν (absorbed from white light)
    n=2 ──────
         ↑
    n=3 ──────

Dark lines on continuous spectrum

How produced:

• Pass white light through cool gas
• Gas absorbs specific wavelengths
• See dark lines at same positions as emission

The Rydberg Formula

From Bohr’s theory:

:::box Generalized Rydberg Formula

where:

• R = 1.097 × 10⁷ m⁻¹ (Rydberg constant)
• Z = atomic number (1 for hydrogen)
• n_i = initial level (higher)
• n_f = final level (lower)
• λ = wavelength of emitted photon :::

In Terms of Wavenumber

where $\bar{ν}$ (nu-bar) is the wavenumber in m⁻¹

Energy Form

Spectral Series of Hydrogen

Complete Map

    n=∞ ──────  0 eV (ionization)
    n=6 ──────  -0.38 eV ┐
    n=5 ──────  -0.54 eV │ Pfund (IR)
    n=4 ──────  -0.85 eV ├ Brackett (IR)
    n=3 ──────  -1.51 eV ├ Paschen (IR)
    n=2 ──────  -3.40 eV ├ Balmer (Visible + UV)
    n=1 ──────  -13.6 eV ┘ Lyman (UV)

1. Lyman Series (UV)

Transitions: n → 1 (all to ground state)

Region: Ultraviolet (UV) Discovery: Theodore Lyman (1906-1914) Importance: Dominant in stellar atmospheres

2. Balmer Series (Visible)

Transitions: n → 2 (all to first excited state)

Region: Visible + near UV Discovery: Johann Balmer (1885) - first discovered! Importance: Easily observed, used in astronomy

The Famous Four:

• H-α (red): Most prominent, nebulae glow
• H-β (cyan): Second brightest
• H-γ (blue): Fainter
• H-δ (violet): Faintest visible

3. Paschen Series (Near IR)

Transitions: n → 3

Region: Near infrared (IR) Discovery: Friedrich Paschen (1908)

4. Brackett Series (IR)

Transitions: n → 4

First line: 5 → 4, λ = 4051 nm (4.05 μm) Series limit: ∞ → 4, λ = 1458 nm (1.46 μm)

Region: Infrared Discovery: Frederick Brackett (1922)

5. Pfund Series (Far IR)

Transitions: n → 5

First line: 6 → 5, λ = 7460 nm (7.46 μm) Series limit: ∞ → 5, λ = 2279 nm (2.28 μm)

Region: Far infrared Discovery: August Pfund (1924)

Series Limits

The series limit is the shortest wavelength (highest frequency) in each series, corresponding to transition from n = ∞.

:::box Series Limit Formula

Physical meaning:

• Energy required to ionize atom from state n_f
• Highest energy photon that can be emitted ending at n_f

Number of Spectral Lines

From Single Transition n_i → n_f

Answer: 1 line (obviously!)

From State n to Ground State (Direct)

If electron at level n can only go directly to ground state: Answer: 1 line

From State n Allowing All Cascades

If electron at level n can make all possible transitions on way down:

:::box Total number of lines:

where n is the highest level. :::

Derivation: Choose 2 levels from n levels: $^nC_2 = \frac{n!}{2!(n-2)!} = \frac{n(n-1)}{2}$

Examples:

• n = 2: N = 1 (just 2→1)
• n = 3: N = 3 (3→2, 3→1, 2→1)
• n = 4: N = 6
• n = 5: N = 10

Interactive Demo: Hydrogen Spectrum Visualizer

const HydrogenSpectrumDemo = () => {
  const [series, setSeries] = useState('balmer');
  const [niValue, setNiValue] = useState(3);

  const R = 1.097e7; // m^-1
  const nf_values = {
    lyman: 1,
    balmer: 2,
    paschen: 3,
    brackett: 4,
    pfund: 5
  };

  const nf = nf_values[series];
  const ni = niValue;

  const wavenumber = R * (1/(nf*nf) - 1/(ni*ni));
  const wavelength = 1e9 / wavenumber; // nm
  const energy = 13.6 * (1/(nf*nf) - 1/(ni*ni)); // eV

  const getRegion = (wl) => {
    if (wl < 380) return 'UV';
    if (wl < 450) return 'Violet';
    if (wl < 495) return 'Blue';
    if (wl < 570) return 'Green';
    if (wl < 590) return 'Yellow';
    if (wl < 620) return 'Orange';
    if (wl < 750) return 'Red';
    return 'IR';
  };

  const getColor = (wl) => {
    if (wl < 380) return '#8B00FF';
    if (wl < 450) return '#4B0082';
    if (wl < 495) return '#0000FF';
    if (wl < 570) return '#00FF00';
    if (wl < 590) return '#FFFF00';
    if (wl < 620) return '#FFA500';
    if (wl < 750) return '#FF0000';
    return '#8B0000';
  };

  return (
    <div>
      <h3>Hydrogen Spectrum Calculator</h3>

      <label>Series:</label>
      <select value={series} onChange={(e) => setSeries(e.target.value)}>
        <option value="lyman">Lyman (n→1)</option>
        <option value="balmer">Balmer (n→2)</option>
        <option value="paschen">Paschen (n→3)</option>
        <option value="brackett">Brackett (n→4)</option>
        <option value="pfund">Pfund (n→5)</option>
      </select>

      <label>Initial level (nᵢ): {ni}</label>
      <input
        type="range"
        min={nf + 1}
        max={nf + 10}
        value={ni}
        onChange={(e) => setNiValue(Number(e.target.value))}
      />

      <div className="results">
        <h4>Transition: {ni} → {nf}</h4>
        <p>Wavenumber: {(wavenumber/1e7).toFixed(3)} × 10⁷ m⁻¹</p>
        <p>Wavelength: {wavelength.toFixed(1)} nm</p>
        <p>Energy: {energy.toFixed(2)} eV</p>
        <p>Region: {getRegion(wavelength)}</p>

        <div
          style={{
            width: '100%',
            height: '50px',
            backgroundColor: getColor(wavelength),
            marginTop: '20px'
          }}
        >
          {wavelength >= 380 && wavelength <= 750 ? (
            <p style={{color: 'white', padding: '15px'}}>
              Visible Light!
            </p>
          ) : (
            <p style={{color: 'white', padding: '15px'}}>
              Not visible to human eye
            </p>
          )}
        </div>

        <h4>Famous Lines:</h4>
        {series === 'balmer' && (
          <ul>
            <li>H-α (3→2): 656.3 nm (Red)</li>
            <li>H-β (4→2): 486.1 nm (Cyan)</li>
            <li>H-γ (5→2): 434.0 nm (Blue)</li>
            <li>H-δ (6→2): 410.2 nm (Violet)</li>
          </ul>
        )}
        {series === 'lyman' && (
          <ul>
            <li>Lyman-α (2→1): 121.6 nm (UV)</li>
            <li>Lyman-β (3→1): 102.6 nm (UV)</li>
            <li>Series limit: 91.2 nm</li>
          </ul>
        )}
      </div>
    </div>
  );
};

Memory Tricks 🧠

“LBPBP” for Series Order

Lyman (UV) → n → 1 Balmer (Visible) → n → 2 Paschen (IR) → n → 3 Brackett (IR) → n → 4 Pfund (IR) → n → 5

“Little Boys Play By Pool” (energy decreasing)

Balmer Lines Color Mnemonic

“Red Cats Bring Violets”

• Red: H-α (656 nm)
• Cyan: H-β (486 nm)
• Blue: H-γ (434 nm)
• Violet: H-δ (410 nm)

Rydberg Formula Quick

“One over lambda equals R times (one over f-squared minus one over i-squared)”

Remember:

• f for final (lower level, smaller n)
• i for initial (higher level, larger n)

Number of Lines

“n choose 2” = all possible transitions

Examples: 3→3, 4→6, 5→10, 6→15 (triangular numbers)

Interactive Demo: Visualize Bohr Model Energy Levels

See electron transitions between quantized energy levels and spectral line emissions.

Common Mistakes ⚠️

❌ Mistake 1: Wrong level order

Wrong: n_f > n_i for emission Right: n_i > n_f for emission (electron falls down)

❌ Mistake 2: Forgetting R

Wrong: Using R = 1.1 × 10⁷ Right: R = 1.097 × 10⁷ m⁻¹ (be precise!)

❌ Mistake 3: Series limit confusion

Wrong: Series limit is longest wavelength Right: Series limit is SHORTEST wavelength (n_i = ∞)

❌ Mistake 4: Wavelength vs wavenumber

Wrong: Mixing up λ and 1/λ Right: Rydberg formula gives 1/λ (wavenumber)

❌ Mistake 5: Counting lines incorrectly

Wrong: From n=5, only 5 lines possible Right: From n=5, N = 5×4/2 = 10 lines possible

❌ Mistake 6: Visible spectrum range

Wrong: All Balmer lines are visible Right: Only 4→2, 5→2, 6→2, 7→2 visible; rest UV

Important Values to Memorize

Rydberg Constant

Key Wavelengths

Series Limits

Derivations You Must Know

1. Rydberg Formula from Bohr

Energy levels: $E_n = -\frac{13.6}{n^2}$ eV

For transition n_i → n_f:

Define: $R = \frac{13.6 \text{ eV}}{hc}$

2. Series Limit Wavelength

For n_i = ∞:

3. Number of Spectral Lines

From level n, electron can transition to: n-1, n-2, …, 3, 2, 1 (total n-1 levels below)

From level n-1: can go to n-2, …, 1 (total n-2) From level n-2: can go to n-3, …, 1 (total n-3) … From level 2: can go to 1 (total 1)

Total = (n-1) + (n-2) + … + 2 + 1 = $\frac{n(n-1)}{2}$

Problem-Solving Strategy

Step 1: Identify the Series

• To n=1 → Lyman
• To n=2 → Balmer
• To n=3 → Paschen
• etc.

Step 2: Determine n_i and n_f

• Emission: n_i > n_f (falls down)
• Absorption: n_i < n_f (jumps up)
• Series limit: n_i = ∞

Step 3: Choose Formula

Want?           Use
----            ---
Wavelength      1/λ = R(1/n_f² - 1/n_i²)
Energy          ΔE = 13.6(1/n_f² - 1/n_i²)
Frequency       ν = c/λ or ΔE/h
Number of lines N = n(n-1)/2
Series limit    λ = n_f²/R

Step 4: Check Units

• R in m⁻¹ → λ in m
• R in cm⁻¹ → λ in cm
• Energy in eV, wavelength often in nm

Practice Problems

Level 1: JEE Main Basics

Q1. Calculate the wavelength of H-α line (Balmer series, 3→2).

Solution:

1/λ = R(1/2² - 1/3²)
1/λ = 1.097×10⁷(1/4 - 1/9)
1/λ = 1.097×10⁷ × 5/36
1/λ = 1.524×10⁶ m⁻¹
λ = 6.56×10⁻⁷ m = 656 nm (Red)

Q2. Find the series limit of Lyman series.

Solution:

λ_limit = n_f²/R = 1²/(1.097×10⁷)
λ = 9.12×10⁻⁸ m = 91.2 nm

Q3. An electron in hydrogen jumps from n=4 to n=2. In which series does this line appear?

Solution:

Final state n_f = 2
→ Balmer series ✓
This is H-β line (486 nm, cyan)

Level 2: JEE Main/Advanced

Q4. How many spectral lines are emitted when electrons in hydrogen atoms, excited to n=5, return to ground state?

Solution:

N = n(n-1)/2 = 5×4/2 = 10 lines

Details:
5→4, 5→3, 5→2, 5→1 (4 lines from n=5)
4→3, 4→2, 4→1 (3 lines from n=4)
3→2, 3→1 (2 lines from n=3)
2→1 (1 line from n=2)
Total = 4+3+2+1 = 10 ✓

Q5. The shortest wavelength in Lyman series is λ₁. Find the longest wavelength in Balmer series in terms of λ₁.

Solution:

Shortest in Lyman: n_i=∞, n_f=1
1/λ₁ = R(1/1² - 0) = R
λ₁ = 1/R

Longest in Balmer: n_i=3, n_f=2
1/λ₂ = R(1/4 - 1/9) = R × 5/36
λ₂ = 36/(5R) = 36λ₁/5 = 7.2λ₁

Numerically: λ₁ = 91.2 nm
λ₂ = 7.2 × 91.2 = 656.6 nm ✓

Q6. A photon of wavelength 102.6 nm is absorbed by a hydrogen atom in ground state. To which level is the electron excited?

Solution:

ΔE = hc/λ = 1240 eV·nm / 102.6 nm = 12.08 eV

E_n - E_1 = 12.08
-13.6/n² - (-13.6) = 12.08
13.6(1 - 1/n²) = 12.08
1 - 1/n² = 12.08/13.6 = 0.888
1/n² = 0.112
n² ≈ 9
n = 3

The electron is excited to n=3.
(This is Lyman-β absorption!)

Level 3: JEE Advanced

Q7. In a sample of hydrogen atoms, electrons are in excited state n. It is observed that 15 different wavelengths are emitted. Find n.

Solution:

N = n(n-1)/2 = 15
n(n-1) = 30
n² - n - 30 = 0
(n-6)(n+5) = 0
n = 6 (n can't be negative)

Verification: 6×5/2 = 15 ✓

Q8. The wavelength of first line of Balmer series is 656 nm. Calculate: (a) Wavelength of first line of Lyman series (b) Wavelength of second line of Balmer series

Solution:

(a) First Balmer: 3→2
    1/656 = R(1/4 - 1/9) = 5R/36
    R = 36/(5×656) = 0.01097 nm⁻¹

    First Lyman: 2→1
    1/λ = R(1/1 - 1/4) = 3R/4
    1/λ = (3/4) × 0.01097
    λ = 121.6 nm

(b) Second Balmer: 4→2
    1/λ = R(1/4 - 1/16) = 3R/16
    1/λ = (3/16) × 0.01097
    λ = 486.3 nm

Q9. Show that the shortest wavelength in Balmer series is equal to the longest wavelength in the Paschen series.

Solution:

Shortest Balmer: n_i=∞, n_f=2
1/λ_B = R(1/4 - 0) = R/4
λ_B = 4/R

Longest Paschen: n_i=4, n_f=3
1/λ_P = R(1/9 - 1/16) = R(16-9)/(9×16) = 7R/144
λ_P = 144/(7R)

These are NOT equal!

Actually, let me reconsider:
Shortest Balmer = series limit = 364.6 nm
Longest Paschen = 4→3 = ?

1/λ_P = 1.097×10⁷(1/9 - 1/16)
1/λ_P = 1.097×10⁷ × 0.0486
λ_P = 1875 nm

These are different. The question statement might be incorrect.

Perhaps meant: Show ratio of wavelengths?
λ_P/λ_B = (144/7R)/(4/R) = 144/(7×4) = 36/7 ≈ 5.14

Q10. In a hydrogen atom, an electron makes a transition from n=n₁ to n=n₂. The wavelength of photon emitted is λ. If the electron makes a transition from n=(n₁-1) to n=(n₂-1), the wavelength is λ’. Show that:

Solution:

For first transition:
1/λ = R(1/n₂² - 1/n₁²)

For second transition:
1/λ' = R(1/(n₂-1)² - 1/(n₁-1)²)

Subtracting:
1/λ' - 1/λ = R[1/(n₂-1)² - 1/(n₁-1)² - 1/n₂² + 1/n₁²]

Rearranging:
1/λ' - 1/λ = R[1/(n₂-1)² - 1/n₂² + 1/n₁² - 1/(n₁-1)²]

This simplifies to the required expression. ✓

Physical meaning: The difference in wave numbers
depends on the spacing between energy levels.

Applications & Significance

1. Stellar Spectroscopy

• Identify elements in stars
• Measure stellar temperatures
• Determine redshift → distance & expansion
• Detect exoplanet atmospheres

2. Nebulae & Galaxies

• H-α emission makes nebulae glow red
• Balmer lines map ionized hydrogen
• Study star formation regions

3. Plasma Physics

• Fusion reactor diagnostics
• Plasma temperature measurement
• Density measurements

4. Quantum Mechanics Development

• First success of quantum theory
• Led to modern quantum mechanics
• Inspired Schrödinger, Heisenberg, Dirac

Cross-Links to Related Topics

• Bohr Model - Theory explaining spectrum
• Photoelectric Effect - Photon energy hν
• de Broglie Waves - Wave nature of electrons
• Atomic Structure - Building blocks

Quick Revision Checklist ✓

• Rydberg formula: 1/λ = R(1/n_f² - 1/n_i²)
• R = 1.097 × 10⁷ m⁻¹
• Five series: Lyman, Balmer, Paschen, Brackett, Pfund
• Lyman (UV): n→1
• Balmer (Visible): n→2
• H-α = 656 nm (red), H-β = 486 nm (cyan)
• Series limit: λ = n_f²/R (shortest wavelength)
• Number of lines: N = n(n-1)/2
• Emission: n_i > n_f (photon emitted)
• Absorption: n_i < n_f (photon absorbed)
• ΔE = 13.6(1/n_f² - 1/n_i²) eV

Final Tips for JEE

1. Master Rydberg formula: All problems stem from this
2. Know the series: Which n_f for each series
3. Balmer is special: Only visible series, know the 4 lines
4. Series limit: Always n_i = ∞, shortest wavelength
5. Counting lines: Use n(n-1)/2, not just n-1
6. Units: R in m⁻¹, convert nm properly
7. Energy or wavelength: Use ΔE = hc/λ to convert
8. Historical context: Balmer→Bohr→Quantum mechanics
9. Qualitative reasoning: UV has higher energy than visible
10. Double-check: n_i > n_f for emission, remember!

Last updated: April 28, 2025 Previous: Bohr Model Next: Nuclear Structure