# Electron energy levels

> Physics · CIE A-Level 9702
> Source: https://www.owlsprep.com/study/cie-9702-u26-electron-energy-levels/

This subtopic explains why atomic electrons can only have discrete fixed energies, how transitions between levels produce line spectra, and how to calculate photon energy for electron transitions.

**Prerequisites:** [Photon model and photon energy](https://www.owlsprep.com/study/cie-9702-u26-photon-model/); Basic atomic structure

## Learning objectives

- Describe discrete electron energy levels in atoms
- Explain the origin of emission and absorption line spectra
- Calculate photon energy and wavelength for electron transitions
- Solve problems involving excitation and ionisation of atoms

## Discrete Electron Energy Levels

**Discrete energy levels** — Electrons bound to an atom can only exist at specific fixed energies. Any energy between these values is not allowed for bound electrons.

*Notation:* E_n

*Example:* Hydrogen has energy levels at -13.6 eV, -3.4 eV, -1.51 eV, etc.

The negative sign for bound electron energy follows the convention that a free electron at rest has an energy of 0 eV. To remove a bound electron from the atom, you must add energy equal to the magnitude of its negative energy level: this value is the atom's ionisation energy.

**Worked example:** The ground state energy of a hydrogen atom is -13.6 eV. What is its ionisation energy in joules?

1. Ionisation energy is the energy required to move the electron from ground state to a free state (E = 0 eV):
2. $$E_{\text{ionisation}} = 0 - (-13.6 \text{ eV}) = 13.6 \text{ eV}$$
3. Convert from electron volts to joules, using 1 eV = 1.6 × 10⁻¹⁹ J:
4. $$E_{\text{ionisation}} = 13.6 \times 1.6 \times 10^{-19} = 2.18 \times 10^{-18} \text{ J}$$

## Transitions Between Energy Levels

When an electron moves between energy levels, it must gain or lose energy equal to the difference between the two levels. Moving from a lower to higher energy level requires absorption of energy (usually from a photon of the correct energy). Moving from a higher to lower energy level releases energy, usually as an emitted photon.

$$\Delta E = |E_{final} - E_{initial}| = hf = \frac{hc}{\lambda}$$

> **info**
>
> The photon energy always equals the absolute value of the energy difference between the two levels, regardless of whether the transition is emission or absorption.

**Worked example:** An electron transitions from $E_1 = -13.6$ eV to $E_2 = -3.4$ eV. Is a photon emitted or absorbed? Calculate the photon wavelength.

1. The electron moves from a lower energy to a higher energy, so it absorbs a photon. Calculate the energy difference:
2. $$\Delta E = (-3.4) - (-13.6) = 10.2 \text{ eV} = 10.2 \times 1.6 \times 10^{-19} = 1.632 \times 10^{-18} \text{ J}$$
3. Rearrange $\Delta E = hc/\lambda$ to solve for wavelength, with $h = 6.63 \times 10^{-34}$ J s, $c = 3.00 \times 10^8$ m/s:
4. $$\lambda = \frac{hc}{\Delta E} = \frac{(6.63 \times 10^{-34})(3.00 \times 10^8)}{1.632 \times 10^{-18}} \approx 1.22 \times 10^{-7} \text{ m} = 122 \text{ nm}$$

## Origin of Line Spectra

Each unique pair of energy levels produces a transition with a unique energy difference, which corresponds to a unique photon wavelength. When light from excited gas is separated by wavelength, it forms an emission line spectrum: bright lines at the wavelengths of emitted photons, on a dark background.

When white light (all wavelengths) passes through cool gas, photons of exactly the right energy are absorbed to excite electrons. This produces an absorption line spectrum: dark lines at the same wavelengths that the gas emits.

**Worked example:** How many different photon energies can be emitted if electrons are excited to the 3rd energy level (counting ground state as n=1) and can decay to lower levels?

1. Only downward transitions emit photons. List all possible transitions from n=3:
2. 1. 3 → 2, 2. 3 → 1, 3. 2 → 1
3. Each transition has a different energy difference, so there are 3 different possible photon energies.

## Excitation and Ionisation by Collision

Electrons can be excited not just by absorbing photons, but also by collisions with other energetic particles like free electrons. The incoming particle transfers exactly the energy needed for excitation, and retains the remaining kinetic energy.

If the incoming particle transfers enough energy to remove the bound electron completely, the process is ionisation, leaving a positively charged ion behind.

**Worked example:** A free electron with 12 eV kinetic energy collides with a hydrogen atom in its ground state (-13.6 eV). What is the maximum kinetic energy of the electron after the collision?

1. The energy needed to excite the ground state electron to the first excited state (-3.4 eV) is:
2. $$\Delta E = (-3.4) - (-13.6) = 10.2 \text{ eV}$$
3. The incoming electron can transfer up to 10.2 eV to excite the atom, so the remaining kinetic energy is:
4. $$12 \text{ eV} - 10.2 \text{ eV} = 1.8 \text{ eV}$$

## Common pitfalls

- **Wrong:** Ignoring negative signs when calculating ΔE, using absolute values of level energies.
  - Why it fails: This gives an incorrect energy difference when working with negative bound energy levels.
  - Correct: Always calculate ΔE as final energy minus initial energy, keeping the negative signs for bound levels.
- **Wrong:** Claims that moving to a higher energy level emits a photon.
  - Why it fails: An electron cannot gain energy by emitting energy; this reverses the physics of transitions.
  - Correct: Lower → higher energy = photon absorption; Higher → lower energy = photon emission.
- **Wrong:** Counting upward transitions when asked for the number of emitted photon energies.
  - Why it fails: Upward transitions absorb photons, they do not emit them, so they should not be counted.
  - Correct: Only count downward transitions when calculating the number of possible emitted photon energies.
- **Wrong:** Using energy in electron volts directly to calculate wavelength with SI values of h and c.
  - Why it fails: h and c are in SI units (J s and m/s), so ΔE must also be in joules to get a correct wavelength in meters.
  - Correct: Always convert ΔE from eV to joules by multiplying by $1.6 \times 10^{-19}$ before calculating wavelength.
- **Wrong:** Believes negative energy levels come from the electron's negative charge.
  - Why it fails: The negative sign is a convention for the reference energy, not related to the electron's charge.
  - Correct: Remember zero energy is defined for a free electron, so bound electrons have lower (negative) energy.

## Cheatsheet

| Concept | Key Rule/Formula |
| --- | --- |
| Discrete energy levels | Only specific fixed energies allowed for bound electrons |
| Energy difference | $\Delta E = \|E_{final} - E_{initial}\|$ |
| Emission transition | $E_{high} \to E_{low}$: photon emitted, $E_{photon} = \Delta E$ |
| Absorption transition | $E_{low} \to E_{high}$: photon absorbed, $E_{photon} = \Delta E$ |
| Photon energy relation | $E = hf = \frac{hc}{\lambda}$ |
| Ionisation energy | $IE = 0 - E_{ground} = \|E_{ground}\|$ |
| Unit conversion | $1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}$ |

## What's next

Electron energy levels are the foundation for understanding how light interacts with matter, and explain all observed atomic spectra. This topic is heavily tested in both multiple choice and structured questions in CIE A-Level Physics, with common problems asking for photon energy calculations and explanations for the origin of line spectra. Mastering this topic prepares you for further study of wave-particle duality, the photoelectric effect, and nuclear energy levels. It also underpins many practical applications of spectroscopy from astronomy to medical imaging.

- [Photoelectric effect](https://www.owlsprep.com/study/cie-9702-u26-photoelectric-effect/)
- [Emission and absorption spectra](https://www.owlsprep.com/study/cie-9702-u26-line-spectra/)
- [de Broglie wavelength](https://www.owlsprep.com/study/cie-9702-u26-de-broglie-wavelength/)

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