Bohr Energies and Spectral Transitions

Introduction

The Bohr energy formula turns integer hydrogen levels into numerical photon predictions. Its sign convention and the distinction between atom energy change and photon energy are the main obstacles. A consistent calculation gives a positive photon energy, then a frequency or wavelength.

Core explanation

For isolated hydrogen in the simple Bohr treatment, Eₙ ≈ −13.6 eV/n² for positive integer n. The negative sign means a bound state is below the zero-energy reference of a separated proton and electron. As n increases, energy becomes less negative and approaches zero. Thus n = 1 is more tightly bound than n = 2.

For emission from nᵢ to n𝒇 with nᵢ > n𝒇, atom energy change ΔEatom = E𝒇 − Eᵢ is negative. The emitted photon carries positive energy Eγ = Eᵢ − E𝒇 = ΔEatom . For absorption, the atom gains positive energy E𝒇 − Eᵢ supplied by a photon. In both cases the photon frequency ν = Eγ/h and vacuum wavelength λ = hc/Eγ.

Keep energy units consistent. If the level formula gives eV, use h in eV s or convert energy to joules before using h = 6.62607015 × 10⁻³⁴ J s. One eV is 1.602176634 × 10⁻¹⁹ J. Mixing an eV energy directly with a joule-second value of h causes an enormous numerical error.

The energy needed to remove hydrogen's electron from level n is 0 − Eₙ. From ground state this is about 13.6 eV. From n = 2 it is about 3.40 eV. The lower energy required from an excited state follows because that state lies closer to zero. Ionization produces an unbound electron rather than a higher finite bound n state.

These calculations apply to an isolated one-electron hydrogen atom at introductory precision. In real samples, collisions, fields, fine structure, and reduced-mass effects can shift or split measured features. The simple formula captures the main scale and series structure, not every high-resolution detail.

Step-by-step reasoning

1. Calculate Eᵢ and E𝒇 using each positive integer n. 2. Subtract to find the atom's signed energy change. 3. Take the positive magnitude for photon energy. 4. Convert units consistently and use ν = Eγ/h or λ = hc/Eγ.

Visual explanation

Draw a vertical energy axis with zero at the top and negative levels approaching it from below. Downward arrows represent emission; upward arrows require absorption.

Real-world analogy

A person below sea level may climb toward zero elevation while still remaining below it. Hydrogen's negative bound-state energies similarly approach zero as n grows.

Real-world example

A spectrometer can measure the wavelength of light from an n = 3 to n = 2 hydrogen transition. The energy formula predicts its scale and links the observed red line with those two levels.

Why?

Why is a photon energy positive during emission when the atom's ΔE is negative? Conservation assigns the energy lost by the atom to the departing photon.

Common misconception

“A negative Eₙ means the electron has negative kinetic energy.” The negative value is total energy relative to a chosen separated-particle zero, not a negative kinetic-energy measurement.

Worked example

For n = 3 → n = 2, E₃ = −13.6/9 ≈ −1.51 eV and E₂ = −13.6/4 = −3.40 eV. The atom changes by −1.89 eV, so the photon carries 1.89 eV. In joules this is about 3.03 × 10⁻¹⁹ J. Wavelength λ = hc/Eγ ≈ 6.56 × 10⁻⁷ m, or 656 nm.

Quick check

1. How much energy ionizes ground-state hydrogen in the simple model? Answer: Approximately 13.6 eV to raise it from −13.6 eV to zero.

Exam focus

Draw the level order before subtracting. Convert eV to J if using SI h, and reserve positive values for photon energy.

Advanced insight

The same simple level formula yields the Rydberg reciprocal-wavelength relation because ΔE is proportional to 1/n𝒇² − 1/nᵢ² and Eγ = hc/λ. Two common hydrogen formulas therefore express one underlying energy pattern.

Summary

Hydrogen's approximate levels follow Eₙ = −13.6 eV/n². The positive difference between initial and final allowed energies becomes photon energy, which determines transition frequency and wavelength.

Practice questions

1. Which level is closer to ionization, n = 2 or n = 4? Answer: n = 4, because its energy is less negative and closer to zero. 2. Does hydrogen gain or lose energy in n = 4 → n = 2? Answer: It loses energy and emits a photon. 3. Why cannot 1.89 eV be divided directly by 6.626 × 10⁻³⁴ J s? Answer: The energy units differ; convert eV to joules or use h in eV s first.