recall and use hf = E1 – E2

Energy Levels in Atoms and Line Spectra

Learning Objective

Recall and use the relationship $hf = E_1 - E_2$ to analyse atomic line spectra.

Key Concepts

• Atoms have discrete energy levels $E_n$.
• When an electron moves between two levels, a photon is emitted or absorbed.
• The energy of the photon is $hf$, where $f$ is the frequency and $h = 6.626\times10^{-34}\,\text{J·s}$.
• Using $c = \lambda f$, the wavelength $\lambda$ can be found: $\lambda = \dfrac{hc}{E_1 - E_2}$.

Derivation of the Formula

Starting from the conservation of energy for a transition from level $E_1$ (higher) to $E_2$ (lower):

$$ E_1 = E_2 + hf $$

Rearranging gives the working equation:

$$ hf = E_1 - E_2 $$

Typical Energy Level Data (Hydrogen Atom)

Level $n$	Energy $E_n$ (eV)
1	-13.6
2	-3.40
3	-1.51
4	-0.85
5	-0.54

Worked Example

1. Identify the transition: $n = 3 \to n = 2$.
2. Read the energies: $E_3 = -1.51\ \text{eV}$, $E_2 = -3.40\ \text{eV}$.
3. Calculate the energy difference: $$\Delta E = E_2 - E_3 = (-3.40) - (-1.51) = -1.89\ \text{eV}$$ Since the electron drops to a lower level, the photon energy is $|\Delta E| = 1.89\ \text{eV}$.
4. Convert to joules: $1\ \text{eV} = 1.602\times10^{-19}\ \text{J}$, $$\Delta E = 1.89 \times 1.602\times10^{-19}\ \text{J} = 3.03\times10^{-19}\ \text{J}$$
5. Find the frequency: $$f = \frac{\Delta E}{h} = \frac{3.03\times10^{-19}}{6.626\times10^{-34}} \approx 4.57\times10^{14}\ \text{Hz}$$
6. Find the wavelength: $$\lambda = \frac{c}{f} = \frac{3.00\times10^{8}}{4.57\times10^{14}} \approx 6.56\times10^{-7}\ \text{m} = 656\ \text{nm}$$

The 656 nm line is the well‑known H‑α line in the Balmer series.

Common Pitfalls

• Mixing up $E_1$ and $E_2$ – $E_1$ must be the higher (initial) energy for emission.
• For absorption, the photon energy is still $hf = E_{\text{higher}} - E_{\text{lower}}$, but the electron moves upward.
• Remember to convert electron‑volts to joules before using $h$.
• Use $c = 2.998\times10^{8}\ \text{m s}^{-1}$ for accurate wavelength values.

Practice Questions

1. Calculate the wavelength of the photon emitted when an electron in a hydrogen atom falls from $n = 5$ to $n = 2$.
2. What is the frequency of the photon required to excite an electron from $n = 1$ to $n = 3$ in hydrogen?
3. Explain why the Balmer series appears in the visible region while the Lyman series is ultraviolet.

Summary

• Atoms possess quantised energy levels.
• Transitions between levels produce photons whose energy is given by $hf = E_{\text{initial}} - E_{\text{final}}$.
• Using $c = \lambda f$, the wavelength of the emitted or absorbed light can be calculated.
• Mastery of this relationship allows prediction and interpretation of line spectra, a cornerstone of atomic physics.