Emission and Relaxation

Introduction

An excited atom can release light as it settles to a lower energy state. In an energy-level diagram the arrow points downward, while the emitted photon carries the positive energy difference. An atom need not return to the ground state in one step; a sequence of transitions can release several photons whose energies add to the total energy drop.

Core explanation

Suppose an atom begins in an allowed state with energy E high and ends in one with lower energy E low. Its own energy change is ΔE atom = E low − E high, a negative number. An emitted photon carries E photon = E high − E low = ΔE atom , a positive quantity. Using E photon = hν = hc/λ connects the level difference to the spectral line's frequency and wavelength. A larger gap produces a higher-frequency, shorter-wavelength photon.

For hydrogen in a rounded Bohr diagram, take E₂ ≈ −5.45 × 10⁻¹⁹ J and E₁ ≈ −2.18 × 10⁻¹⁸ J. The atom going from n = 2 to n = 1 loses about 1.64 × 10⁻¹⁸ J. The photon carries that positive energy, corresponding to ultraviolet light. Reversing the transition requires absorption of a matching photon. Saying “the photon has negative energy because the atom loses energy” confuses two different energy changes.

Now imagine an atom initially at n = 3. It may emit directly to n = 1, releasing one photon with energy E₃ − E₁. Alternatively, if both steps occur, it may go n = 3 → n = 2 and then n = 2 → n = 1, releasing two photons with energies E₃ − E₂ and E₂ − E₁. Adding the two gives E₃ − E₁: the intermediate E₂ terms cancel. The total released energy is the same for these endpoints, but the photon wavelengths and number differ. Which pathways occur depends on transition probabilities and conditions.

An excited atom needs some earlier source of energy. Electrical discharge, collision, light absorption or heat-related processes can populate excited states. The later emission is not energy appearing from nowhere. The atom releases energy it gained or retained above a lower state. In a collection of atoms, many transitions produce a pattern of lines, and intensity at a line depends on how many photons of that transition reach the detector.

The term “relaxation” does not mean every atomic change produces visible light. A photon may be ultraviolet or infrared, and some energy can be transferred in collisions to nearby particles rather than emitted as light under certain conditions. A simple isolated-atom diagram focuses on radiative transitions. Real gas pressure and environment affect line shapes and intensities.

The wavelength of an emitted line identifies an energy gap, not a unique starting and ending level without additional context. Different systems may have gaps of similar magnitude, and one element has many possible transitions. A measured set of lines and a model together support an assignment. It is unsafe to claim one colour alone proves one element or one particular electron path.

In modern language, the atom changes quantum state; an electron is not a glowing bead visibly falling through space from one drawn ring to another. Bohr's arrows are energy bookkeeping diagrams. This distinction preserves the model's predictive use without turning its drawing into a literal movie.

Step-by-step reasoning

1. Identify initial higher and final lower atomic energies. 2. Compute positive photon energy as E initial − E final. 3. If needed, use E = hν or hc/λ to obtain frequency or wavelength. 4. For a cascade, sum all emitted photon energies and check that the intermediate levels cancel.

Visual explanation

Draw three horizontal levels E₃, E₂ and E₁. Show one direct arrow E₃ → E₁ and a two-arrow path E₃ → E₂ → E₁. Label each photon gap and write their sum equals the direct gap.

Real-world analogy

A ball descending a staircase may drop from the top to the bottom in one move or pause at a middle step, releasing the same total change in gravitational energy across the full descent. The analogy illustrates energy accounting, not the quantum rules that determine atomic transitions.

Real-world example

A hydrogen discharge tube emits a pinkish overall glow built from several spectral contributions. A spectrometer separates its light into individual lines, allowing each detected wavelength to be linked to a particular energy difference in an atomic model.

Why?

Why can one excited atom lead to more than one emitted photon? It may relax through an intermediate allowed state. Each downward step releases a photon, and the sum of their energies equals the difference between the starting and final states.

Common misconception

“A downward transition creates a negative-energy photon.” The atom's energy change is negative, but the emitted photon's energy is positive and equals the magnitude of that loss.

Worked example

Let E₃ = −2.42 × 10⁻¹⁹ J, E₂ = −5.45 × 10⁻¹⁹ J and E₁ = −2.18 × 10⁻¹⁸ J. A direct n = 3 → n = 1 photon has energy E₃ − E₁ = 1.938 × 10⁻¹⁸ J. Via n = 2, the first photon has E₃ − E₂ = 3.03 × 10⁻¹⁹ J and the second E₂ − E₁ = 1.635 × 10⁻¹⁸ J. Their sum is 1.938 × 10⁻¹⁸ J, matching the direct gap within rounding.

Quick check

1. For emission, which is positive: the atom's energy change or the photon's energy? Answer: The photon energy is positive; the atom's own energy change is negative.

Exam focus

Draw downward arrows for emission and use initial minus final to get positive photon energy. For a multi-step path, show that the sum of photon energies equals the overall level drop. Keep intensity separate from photon energy.

Advanced insight

Selection rules and state lifetimes influence which transitions are strong or weak. Some excited states are metastable, meaning radiative relaxation can be relatively slow. Collisions can compete with light emission, so spectra also depend on the gas environment.

Summary

Emission occurs when an excited atom moves to a lower energy state and releases a positive-energy photon equal to the atomic energy lost. A cascade can release multiple photons whose energies add to the same overall gap. Spectral-line positions reflect gaps, while intensities depend on populations and probabilities.

Practice questions

1. Write the photon energy for a downward transition from E₃ to E₂. Answer: E photon = E₃ − E₂, a positive value when E₃ is higher. 2. What is the atom's signed energy change during emission? Answer: E final − E initial, which is negative. 3. If a cascade releases 2 × 10⁻¹⁹ J and 5 × 10⁻¹⁹ J photons, what total energy was lost? Answer: 7 × 10⁻¹⁹ J. 4. Does every emitted photon have visible wavelength? Answer: No. Depending on the level gap, photons can be ultraviolet, infrared or elsewhere.