Energy levels in atoms and line spectra

1. Overview

The energy of an electron bound to an isolated atom is quantized, meaning it can only exist in specific, discrete energy levels. Electrons transition between these levels by absorbing or emitting a single photon of electromagnetic radiation. The energy of this photon must exactly match the difference between the two energy levels. Because these energy levels are unique to each element, the resulting emission and absorption line spectra act as definitive "fingerprints" for identifying chemical elements. This phenomenon provides direct experimental evidence for the existence of discrete energy states within atoms, moving away from classical models that predicted continuous energy ranges.

---

Key Definitions

• Discrete Energy Level: One of a specific set of quantized energy values that an electron can occupy while bound to an atom. Electrons cannot exist between these levels.
• Ground State: The lowest energy state (most negative value) that an electron can occupy in an atom. It is the most stable configuration.
• Excited State: Any energy level with a higher energy than the ground state. Electrons in these states are unstable.
• Excitation: The process where an electron absorbs energy (from a photon or a colliding electron) to move from a lower energy level to a higher energy level.
• De-excitation: The process where an electron releases energy by moving from a higher energy level to a lower energy level, emitting a single photon in the process.
• Photon: A quantum (discrete packet) of electromagnetic radiation. Its energy is proportional to its frequency ($E = hf$).
• Ionization Energy: The minimum energy required to remove an electron completely from its ground state to an infinite distance from the nucleus (where its energy is defined as zero).
• Emission Line Spectrum: A spectrum consisting of bright, distinct colored lines on a dark background, produced when excited atoms de-excite and emit photons of specific frequencies.
• Absorption Line Spectrum: A spectrum consisting of dark lines at specific frequencies superimposed on a continuous "rainbow" background, produced when a cool gas absorbs specific frequencies from a continuous light source.

---

Content

3.1 The Nature of Discrete Energy Levels

In isolated atoms, such as those in a low-pressure gas, the electrons are confined by the electrostatic attraction of the nucleus. This confinement leads to quantization.

• The Zero Reference: By convention, the energy of an electron is defined as zero when it is at an infinite distance from the nucleus (the ionization limit).
• Negative Energy Values: Because work must be done to move an electron from the atom to infinity, all bound energy levels have negative values. A more negative value (e.g., -13.6 eV) represents a "deeper" or lower energy state than a less negative value (e.g., -3.4 eV).
• Isolated Atoms: Quantization is most clearly observed in isolated atoms. In solids or high-pressure gases, the proximity of neighboring atoms causes energy levels to shift and overlap, forming "energy bands" rather than distinct lines.

3.2 The Mechanism of Photon Emission and Absorption

The interaction between light and matter is governed by the principle of conservation of energy.

1. Photon Emission (Formation of Emission Spectra)

• Excitation: Atoms in a gas are excited, usually by heating or by passing an electric discharge through them. This provides energy to electrons, moving them to higher, unstable energy levels.
• Transition: The excited electrons spontaneously "fall" to lower energy levels (de-excitation).
• Photon Release: For every transition, a single photon is emitted. The energy of this photon ($hf$) is exactly equal to the difference in energy between the initial high level ($E_1$) and the final lower level ($E_2$).
• Observation: When this light is passed through a diffraction grating or prism, each specific transition appears as a distinct, bright line of a specific color (frequency).

2. Photon Absorption (Formation of Absorption Spectra)

• Continuous Source: White light (containing all visible frequencies) is passed through a cool gas.
• Selective Absorption: An electron in the gas will only absorb a photon if the photon's energy is exactly equal to the difference between the electron's current level and a higher available level.
• Re-emission: The excited electrons quickly de-excite. However, they emit the new photons in all possible directions.
• Observation: In the original direction of the light beam, the intensity of those specific frequencies is greatly reduced, appearing as dark lines in the continuous spectrum.

3.3 Mathematical Relationship: $hf = E_1 - E_2$

The fundamental equation for atomic transitions relates the wave properties of light to the energy states of the atom.

$$hf = E_1 - E_2$$ (This equation must be memorised and is the core of most calculations in this topic.)

Where:

• $h$ is the Planck constant ($6.63 \times 10^{-34}$ J s).
• $f$ is the frequency of the emitted or absorbed photon (Hz).
• $E_1$ is the energy of the higher (less negative) level (J).
• $E_2$ is the energy of the lower (more negative) level (J).

Since $c = f\lambda$, we can substitute $f = \frac{c}{\lambda}$ to find the wavelength:

$$\frac{hc}{\lambda} = E_1 - E_2$$

3.4 Energy Units: Joules vs. Electronvolts

In atomic physics, the Joule is often too large a unit for convenience. We use the electronvolt (eV).

• Definition: 1 eV is the energy gained by an electron accelerated through a potential difference of 1 Volt.
• Conversion: $1 \text{ eV} = 1.60 \times 10^{-19} \text{ J}$
• Crucial Step: When using $hf = \Delta E$, the energy difference $\Delta E$ must be in Joules.

---

4. Worked Examples

Worked Example 1 — Calculating Emission Frequency

An electron in a mercury atom de-excites from an energy level of $-3.71 \times 10^{-19}$ J to a level of $-8.82 \times 10^{-19}$ J. Calculate the frequency of the photon emitted.

1. Identify the energy difference ($\Delta E$): $$\Delta E = E_{high} - E_{low}$$ $$\Delta E = (-3.71 \times 10^{-19}) - (-8.82 \times 10^{-19})$$ $$\Delta E = 5.11 \times 10^{-19} \text{ J}$$
2. Use the photon energy equation: $$hf = \Delta E$$ $$(6.63 \times 10^{-34}) \times f = 5.11 \times 10^{-19}$$
3. Solve for $f$: $$f = \frac{5.11 \times 10^{-19}}{6.63 \times 10^{-34}}$$ $$f = 7.71 \times 10^{14} \text{ Hz}$$

Worked Example 2 — Wavelength from eV Levels

A gas atom has energy levels at $-0.85$ eV, $-1.51$ eV, and $-3.40$ eV. Calculate the wavelength of the photon emitted when an electron transitions from the $-1.51$ eV level to the $-3.40$ eV level.

1. Calculate the energy difference in eV: $$\Delta E = (-1.51) - (-3.40) = 1.89 \text{ eV}$$
2. Convert the energy difference to Joules: $$\Delta E = 1.89 \times 1.60 \times 10^{-19} = 3.024 \times 10^{-19} \text{ J}$$
3. Use the wavelength equation: $$\frac{hc}{\lambda} = \Delta E$$ $$\lambda = \frac{hc}{\Delta E}$$ $$\lambda = \frac{(6.63 \times 10^{-34}) \times (3.00 \times 10^8)}{3.024 \times 10^{-19}}$$
4. Final Answer: $$\lambda = 6.58 \times 10^{-7} \text{ m (or 658 nm)}$$

Worked Example 3 — Identifying Transitions

A hypothetical atom has four discrete energy levels. How many distinct spectral lines can be produced in its emission spectrum?

1. Method: A spectral line is produced for every possible downward transition between any two levels.
2. List the transitions (Labeling levels 1, 2, 3, 4 from bottom to top):
   • From level 4: $4 \rightarrow 3, 4 \rightarrow 2, 4 \rightarrow 1$ (3 lines)
   • From level 3: $3 \rightarrow 2, 3 \rightarrow 1$ (2 lines)
   • From level 2: $2 \rightarrow 1$ (1 line)
3. Total: $3 + 2 + 1 = 6$ lines. (General formula: $\frac{n(n-1)}{2}$ where $n$ is the number of levels.)

---

Key Equations

Equation	Description	Status
$E = hf$	Energy of a photon	Data Sheet
$c = f\lambda$	Wave equation for light	Data Sheet
$hf = E_1 - E_2$	Energy difference between levels	Memorise
$\frac{hc}{\lambda} = E_1 - E_2$	Wavelength from energy levels	Memorise
$1 \text{ eV} = 1.60 \times 10^{-19} \text{ J}$	Energy unit conversion	Data Sheet

Standard Constants:

• Planck constant ($h$): $6.63 \times 10^{-34}$ J s
• Speed of light ($c$): $3.00 \times 10^8$ m s⁻¹

---

Common Mistakes to Avoid

• ❌ Wrong: Forgetting to convert eV to Joules before using $h$.
  • ✅ Right: Always perform the conversion $\text{eV} \rightarrow \text{J}$ as your first step in any calculation involving $h$ or $c$.
• ❌ Wrong: Getting a negative frequency or wavelength because of the negative energy values.
  • ✅ Right: Photon energy is a scalar quantity and must be positive. Use $\Delta E = |E_{initial} - E_{final}|$.
• ❌ Wrong: Thinking that an absorption spectrum is produced by a hot gas.
  • ✅ Right: Absorption requires a cool gas to be placed in front of a continuous (hot) source. If the gas itself is hot, it will produce an emission spectrum.
• ❌ Wrong: Confusing the number of energy levels with the number of spectral lines.
  • ✅ Right: The number of lines is the number of possible transitions between levels.
• ❌ Wrong: Stating that electrons "move between levels" without mentioning photons.
  • ✅ Right: In the context of spectra, you must explicitly state that a photon is emitted or absorbed during the transition.

---

Exam Tips

1. The "Explain" Question: If asked to explain how a line spectrum provides evidence for discrete energy levels, your answer must include:
   • Photons have energy $E = hf$.
   • Photons are emitted/absorbed when electrons move between levels.
   • $\Delta E = hf$, so the frequency depends on the energy difference.
   • Since only discrete frequencies (lines) are observed, the energy differences must be discrete.
   • Therefore, the energy levels themselves must be discrete.
2. Longest vs. Shortest Wavelength:
   • Longest wavelength ($\lambda$) = Smallest energy change ($\Delta E$). This is usually a transition between the two highest (closest) energy levels.
   • Shortest wavelength ($\lambda$) = Largest energy change ($\Delta E$). This is usually a transition from a high level down to the ground state ($n=1$).
3. Ionization Limit: If a question mentions "ionization," it refers to a transition to the $E = 0$ level. The energy required is simply the absolute value of the ground state energy.
4. Precision: Use the values for $h$ and $c$ exactly as provided on the 9702 Data Sheet. Round your final answer to the same number of significant figures as the least precise data given in the question (usually 2 or 3 s.f.).
5. Direction of Arrows: In energy level diagrams, ensure arrows for emission point downwards and arrows for absorption point upwards.