recall and use hf = E1 – E2

Energy Levels in Atoms and Line Spectra

Learning Objective

Recall and use the relationship \$hf = E1 - E2\$ 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}{E1 - E2}\$.

Derivation of the Formula

Starting from the conservation of energy for a transition from level \$E1\$ (higher) to \$E2\$ (lower):

\$\$

E1 = E2 + hf

\$\$

Rearranging gives the working equation:

\$\$

hf = E1 - E2

\$\$

Typical Energy Level Data (Hydrogen Atom)

Worked Example

1. Identify the transition: \$n = 3 \to n = 2\$.
2. Read the energies: \$E3 = -1.51\ \text{eV}\$, \$E2 = -3.40\ \text{eV}\$.
3. Calculate the energy difference:

   \$\Delta E = E2 - E3 = (-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 \$E1\$ and \$E2\$ – \$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.