S1.3.2 Hydrogen emission spectrum

Emission Line Spectra of Hydrogen and Energy Transitions

The Emission Line Spectrum of Hydrogen: Evidence for Discrete Energy Levels

When a pure gaseous element like hydrogen is subjected to high voltage under reduced pressure, it emits light.
Passing this light through a prism produces a line emission spectrum—a series of distinct lines, each corresponding to a specific wavelength of light.
Unlike a continuous spectrum, which includes all wavelengths (like a rainbow), the line spectrum has clear gaps. But why do these gaps occur?

Quantization of Energy Levels

The hydrogen line spectrum arises because electrons in hydrogen atoms can only occupy specific, discrete energy levels.

Analogy

Think of these energy levels as rungs on a ladder: an electron cannot exist between the rungs.

When an electron absorbs energy, it "jumps" to a higher energy level (excited state).
When it falls back to a lower energy level, it releases energy in the form of a photon.
The energy of this photon corresponds precisely to the difference between the two energy levels.

Mathematically, the energy of an electron in a hydrogen atom is described by the equation:

$E_{n} = - R_{H} \frac{1}{n^{2}}$

Where:

$E_{n}$ = energy of the electron in the $n$ -th energy level (in joules, J)
$R_{H} = 2.18 \times 10^{- 18}, J$ (Rydberg constant)
$n$ = principal quantum number (1, 2, 3, ...)

Hint

The negative sign indicates that the energy is relative to the ionized state, where the electron is completely removed from the atom ( $n = \infty$ ).

Analogy

Think of the energy levels like steps on a staircase. You can stand on any step, but you can’t hover between them. Similarly, electrons can only "stand" on specific energy levels.

Convergence of Lines at Higher Energies

If you examine the hydrogen spectrum closely, you’ll notice that the lines become closer together (converge) at higher energy levels.
This convergence occurs because the energy levels themselves get closer as $n$ increases.
At the limit of convergence, the energy difference between levels approaches zero, corresponding to the ionization of the atom.

Energy Transitions: Linking Energy Levels to Spectral Lines

Each line in the hydrogen spectrum corresponds to an electron transitioning between two energy levels.
The energy of the emitted photon is directly related to the difference in energy between these levels:

$Δ E = E_{higher} - E_{lower}$

Using Planck’s equation, the energy of the photon can also be expressed as:

$E = h \cdot f = \frac{h \cdot c}{λ}$

Where:

$h = 6.63 \times 10^{- 34} Js$ (Planck’s constant)
$f$ = frequency of the photon (in Hz)
$c = 3.00 \times 10^{8} {ms}^{- 1}$ (speed of light)
$λ$ = wavelength of the photon (in meters)

The Balmer, Lyman, and Paschen Series

The hydrogen spectrum is divided into series based on the final energy level ( $n_{f}$ ) to which the electron transitions:

Lyman Series: Transitions to $n_{f} = 1$ (UV region)
Balmer Series: Transitions to $n_{f} = 2$ (visible region)
Paschen Series: Transitions to $n_{f} = 3$ (infrared region)

Note

You are not required to know the names of the series by heart.

Example

The red line in the Balmer series corresponds to an electron falling from $n = 3$ to $n = 2$ , emitting a photon with a wavelength of 656 nm.
The violet line corresponds to a transition from $n = 6$ to $n = 2$ , with a shorter wavelength of 410 nm.

Example question

Calculate the wavelength of light emitted when an electron in a hydrogen atom transitions from $n = 4$ to $n = 2$ .

Solution

Using the energy formula for hydrogen:
$E_{n} = - R_{H} \frac{1}{n^{2}}$
Energy at $n = 4$ : $E_{4} = - 2.18 \times 10^{- 18} \frac{1}{4^{2}} = - 1.36 \times 10^{- 19} J$ Energy at $n = 2$ : $E_{2} = - 2.18 \times 10^{- 18} \frac{1}{2^{2}} = - 5.45 \times 10^{- 19} J$
Calculate the energy difference:
$Δ E = E_{2} - E_{4} = - 5.45 \times 10^{- 19} - (- 1.36 \times 10^{- 19}) = 4.09 \times 10^{- 19} J$
Use Planck’s equation to find the wavelength:
$λ = \frac{h \cdot c}{Δ E}$
Substituting values:
$λ = \frac{6.63 \times 10^{- 34} \cdot 3.00 \times 10^{8}}{4.09 \times 10^{- 19}} = 486 nm$ This corresponds to the cyan line in the Balmer series.

Common Mistake

Students often forget to convert units when solving problems involving energy and wavelength. Always ensure that energy is in joules (J), wavelength in meters (m), and frequency in hertz (Hz).

Electron transitions to different energy levels.

Reflection and Practice

Self review

Why do the lines in the hydrogen spectrum converge at higher energies?
If an electron transitions from $n = 5$ to $n = 3$ , in which region of the electromagnetic spectrum would the emitted photon lie?
How does the energy difference between levels change as $n$ increases?

Theory of Knowledge

How does the quantization of energy challenge classical physics? What does this suggest about the limitations of human intuition in understanding atomic-scale phenomena?

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