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IB DP Chemistry – S1.3.2 Hydrogen emission spectrum – Study Notes

IB DP Chemistry - S1.3.2 Hydrogen emission spectrum - Study Notes - New Syllabus - 2026, 2027 & 2028

IB DP Chemistry – S1.3.2 Hydrogen emission spectrum – Study Notes – New Syllabus

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 Hydrogen Emission Spectrum

 Hydrogen Emission Spectrum

The hydrogen atom has only one electron. When it absorbs energy, the electron moves to a higher energy level (excited state). When the electron returns to a lower energy level, it emits energy in the form of light (a photon).

This emitted light appears as a line spectrum a series of discrete lines at specific wavelengths, providing evidence that electrons exist in quantised energy levels.

Spectral Series in Hydrogen:

  • Lyman series: transitions to \( n = 1 \) (ultraviolet region)
  • Balmer series: transitions to \( n = 2 \) (visible region)
  • Paschen series: transitions to \( n = 3 \) (infrared region)

Concepts:

  • Each line in the emission spectrum corresponds to an electron transition from a higher to a lower energy level.
  • Lines converge (get closer together) at higher energies – this is due to energy levels getting closer as \( n \) increases.
  • The shortest wavelength (highest frequency) line corresponds to the largest energy transition.

Interpretation:

  • Lines in the UV region (Lyman series) involve transitions with larger energy differences → higher energy photons.
  • Lines in the visible region (Balmer series) correspond to transitions like \( n = 3 \rightarrow 2 \), \( n = 4 \rightarrow 2 \), etc.
  • Lines in the infrared region (Paschen series) involve smaller energy changes → lower energy photons.

Evidence for Quantisation:

  • The presence of discrete lines (not a continuous spectrum) confirms that electrons can only occupy certain allowed energy levels.
  • As \( n \rightarrow \infty \), the levels converge indicating ionisation limit (electron completely removed).

Formula for Energy Transitions:

\( \Delta E = E_{\text{high}} – E_{\text{low}} = hf = \frac{hc}{\lambda} \)

Where \( h \) is Planck’s constant and \( c \) is the speed of light.

Example

Which transition in the hydrogen atom would emit a photon of the highest energy?

▶️Answer/Explanation

Energy is highest when the difference between energy levels is greatest. The largest difference occurs when the electron falls from \( n = \infty \) to \( n = 1 \).

Hence, this transition emits the highest-energy photon, which lies in the ultraviolet (UV) region of the spectrum (Lyman series).

Example

In which region of the electromagnetic spectrum does the Balmer series of hydrogen appear, and what type of transitions cause it?

▶️Answer/Explanation

The Balmer series appears in the visible region of the spectrum.

These spectral lines are produced when electrons fall from higher levels (\( n \geq 3 \)) down to the \( n = 2 \) energy level.

For example: \( n = 3 \rightarrow 2 \), \( 4 \rightarrow 2 \), \( 5 \rightarrow 2 \), etc.

Example

Explain why the lines in the emission spectrum of hydrogen converge at higher frequencies.

▶️Answer/Explanation

As the principal quantum number \( n \) increases, the energy levels in the hydrogen atom become closer together.

Therefore, transitions from very high levels (e.g., \( n = 7 \rightarrow 1 \), \( n = 8 \rightarrow 1 \)) release photons with increasingly similar energies.

This causes the spectral lines to appear closer together — a phenomenon called line convergence — especially visible in the UV region (Lyman series).

Example

A spectral line in the Lyman series of hydrogen has a wavelength of 121.6 nm. Calculate the energy of the photon emitted in joules and in electronvolts (eV).

▶️Answer/Explanation

\( E = \frac{hc}{\lambda} \)

  • \( h = 6.63 \times 10^{-34} \, \text{J·s} \)
  • \( c = 3.00 \times 10^8 \, \text{m/s} \)
  • \( \lambda = 121.6 \, \text{nm} = 1.216 \times 10^{-7} \, \text{m} \)

Calculate energy in joules
\( E = \frac{(6.63 \times 10^{-34})(3.00 \times 10^8)}{1.216 \times 10^{-7}} \approx 1.63 \times 10^{-18} \, \text{J} \)

Convert to electronvolts (eV)
\( 1 \, \text{eV} = 1.60 \times 10^{-19} \, \text{J} \)
\( E = \frac{1.63 \times 10^{-18}}{1.60 \times 10^{-19}} \approx 10.2 \, \text{eV} \)

Final Answer: Energy = \( 1.63 \times 10^{-18} \, \text{J} \) or 10.2 eV

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