Hydrogen Spectrum

Introduction

The hydrogen spectrum was a major puzzle before Bohr’s model. When hydrogen gas is excited, it emits light at specific wavelengths—not a continuous spectrum. Bohr’s model beautifully explains these discrete spectral lines.

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Origin of Spectral Lines

When an electron transitions between energy levels, it emits or absorbs a photon with energy equal to the difference:

$$\Delta E = h\nu = E_{n_2} - E_{n_1}$$

Using Bohr’s energy formula:

$$h\nu = -13.6 Z^2 \left(\frac{1}{n_2^2} - \frac{1}{n_1^2}\right) \text{ eV}$$

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Rydberg Formula

For Wavelength

$$\boxed{\frac{1}{\lambda} = R_H Z^2 \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)}$$

where:

• $R_H = 1.097 \times 10^7$ m⁻¹ (Rydberg constant)
• $n_1$ = lower energy level (final state)
• $n_2$ = higher energy level (initial state)
• $n_2 > n_1$ for emission

For Wave Number

$$\bar{\nu} = \frac{1}{\lambda} = R_H Z^2 \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$$

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Spectral Series

Each series is named after its discoverer and corresponds to transitions ending at a specific level:

graph TD
    A[n=6] --> B[Pfund n₁=5]
    A --> C[Brackett n₁=4]
    A --> D[Paschen n₁=3]
    A --> E[Balmer n₁=2]
    A --> F[Lyman n₁=1]
    
    style F fill:#9b59b6
    style E fill:#e74c3c
    style D fill:#e67e22

Series	n₁	n₂	Region	Formula
Lyman	1	2, 3, 4…	UV	$\frac{1}{\lambda} = R_H\left(\frac{1}{1^2} - \frac{1}{n^2}\right)$
Balmer	2	3, 4, 5…	Visible	$\frac{1}{\lambda} = R_H\left(\frac{1}{2^2} - \frac{1}{n^2}\right)$
Paschen	3	4, 5, 6…	Infrared	$\frac{1}{\lambda} = R_H\left(\frac{1}{3^2} - \frac{1}{n^2}\right)$
Brackett	4	5, 6, 7…	Far IR	$\frac{1}{\lambda} = R_H\left(\frac{1}{4^2} - \frac{1}{n^2}\right)$
Pfund	5	6, 7, 8…	Far IR	$\frac{1}{\lambda} = R_H\left(\frac{1}{5^2} - \frac{1}{n^2}\right)$

Interactive Demo: Explore Hydrogen Spectral Series

Watch electron transitions between energy levels and see how different spectral series are formed. Observe the Lyman, Balmer, and Paschen series.

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Series Limits

Definition

The series limit is the shortest wavelength (highest energy) transition in each series, occurring when $n_2 → ∞$.

$$\frac{1}{\lambda_{limit}} = R_H Z^2 \times \frac{1}{n_1^2}$$

Series Limits for Hydrogen

Series	λ_limit (nm)
Lyman	91.2
Balmer	364.6
Paschen	820.4

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Number of Spectral Lines

When an electron is excited to level $n$, the total number of possible emission lines is:

$$\boxed{N = \frac{n(n-1)}{2}}$$

For transition from $n_2$ to $n_1$:

$$N = \frac{(n_2 - n_1)(n_2 - n_1 + 1)}{2}$$

Example:

From n=4 to n=1:

$$N = \frac{4 \times 3}{2} = 6 \text{ lines}$$

These are: 4→3, 4→2, 4→1, 3→2, 3→1, 2→1

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Calculations

Example 1: First Line of Lyman Series

Problem: Calculate the wavelength of the first line of the Lyman series.

Solution: First line: n₁ = 1, n₂ = 2

$$\frac{1}{\lambda} = R_H\left(1 - \frac{1}{4}\right) = R_H \times \frac{3}{4}$$ $$\frac{1}{\lambda} = 1.097 \times 10^7 \times \frac{3}{4} = 8.228 \times 10^6 \text{ m}^{-1}$$ $$\lambda = 121.6 \text{ nm (UV region)}$$

Example 2: Balmer Series Limit

Problem: Find the series limit of the Balmer series.

Solution: Series limit: n₁ = 2, n₂ → ∞

$$\frac{1}{\lambda} = R_H\left(\frac{1}{4} - 0\right) = \frac{R_H}{4}$$ $$\lambda = \frac{4}{R_H} = \frac{4}{1.097 \times 10^7} = 364.6 \text{ nm}$$

Example 3: Identifying a Transition

Problem: A hydrogen line has wavelength 656.3 nm. Identify the transition.

Solution: 656.3 nm is in the visible region → Balmer series (n₁ = 2)

$$\frac{1}{656.3 \times 10^{-9}} = 1.097 \times 10^7 \left(\frac{1}{4} - \frac{1}{n_2^2}\right)$$

Solving: $\frac{1}{n_2^2} = \frac{1}{4} - \frac{1}{8} = \frac{1}{8}$

Wait, let me recalculate:

$$1.524 \times 10^6 = 1.097 \times 10^7 \left(0.25 - \frac{1}{n_2^2}\right)$$ $$\frac{1}{n_2^2} = 0.25 - 0.139 = 0.111 ≈ \frac{1}{9}$$

So $n_2 = 3$. This is the Hα line (first line of Balmer series).

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Limitations of Bohr’s Model

What’s Next?

These limitations led to the development of:

• de Broglie’s wave-particle duality
• Heisenberg’s uncertainty principle
• Schrödinger’s wave equation

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JEE Important Points

1. Lyman series has the shortest wavelengths (highest energy)
2. First line of any series has the longest wavelength in that series
3. Series limit has the shortest wavelength in each series
4. For He⁺, multiply $R_H$ by $Z^2 = 4$
5. Balmer series is the only one in visible region for hydrogen

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Practice Problems

1. Calculate the wavelength of the third line of the Balmer series.

2. In which series does the line with λ = 1875 nm belong?

3. Find the ratio of longest wavelengths in Lyman and Balmer series.

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