Bohr Model Formula
Quantized energy levels in hydrogen-like atoms
Understanding the Bohr Model
In 1913, Danish physicist Niels Bohr revolutionized atomic theory by proposing a quantized model of the hydrogen atom that successfully explained the discrete emission spectra observed in laboratory experiments. Bohr's model introduced the radical concept that electrons orbit the nucleus only at specific allowed energy levels, rather than at any arbitrary distance. This groundbreaking work earned him the Nobel Prize in Physics in 1922 and laid the foundation for modern quantum mechanics.
The Bohr model elegantly combines classical mechanics with quantum postulates to predict electron energies and orbital radii in hydrogen-like atoms (atoms with a single electron, such as H, Heâº, Li²âº). The model's greatest success was explaining the Rydberg formula for hydrogen spectral lines and calculating the Rydberg constant from first principles. While later replaced by the more complete quantum mechanical model using wave functions, the Bohr model remains invaluable for understanding atomic structure and electronic transitions, particularly in introductory chemistry and physics courses.
The model's key insight—that angular momentum is quantized as L = n℠where ℠is the reduced Planck constant—leads directly to discrete energy levels and orbital radii. Although the Bohr model fails for multi-electron atoms and doesn't account for electron spin or wave properties, it provides accurate predictions for hydrogen-like systems and offers an intuitive framework for visualizing atomic transitions, photon emission, and absorption processes.
Energy Levels Formula
En = -13.6 eV × Z² / n²
Energy of electron in level n (hydrogen-like atoms)
En = energy of level n (eV or J)
Z = atomic number (1 for H, 2 for Heâº)
n = principal quantum number (1, 2, 3...)
13.6 eV = Rydberg energy constant
Key Points:
- Negative sign indicates electron is bound to nucleus (energy required to remove)
- Energy increases (becomes less negative) as n increases
- Ground state (n = 1) has lowest energy: E₠= -13.6 Z² eV
- As n → ∞, E → 0 (electron escapes, ionization)
- Higher Z (more protons) increases binding energy proportionally to Z²
Orbital Radius Formula
rn = a₀ n² / Z
where aâ‚€ = 0.529 Ã… = 5.29 × 10â»Â¹Â¹ m (Bohr radius)
The orbital radius increases with the square of the principal quantum number (n²), meaning higher energy levels are exponentially farther from the nucleus. For hydrogen (Z = 1), the electron in n = 2 orbits at 4 times the Bohr radius, while n = 3 orbits at 9 times the Bohr radius. This quantization explains why electrons don't spiral into the nucleus—they can only exist at specific discrete distances.
Detailed Example 1: Hydrogen First Excited State (n = 2)
Problem: Calculate the energy and orbital radius for a hydrogen atom electron in the n = 2 state.
Step 1: Calculate Energy
Given: Z = 1 (hydrogen), n = 2
E₂ = -13.6 eV × (1)² / (2)²
Eâ‚‚ = -13.6 eV / 4
Eâ‚‚ = -3.4 eV
Step 2: Calculate Orbital Radius
r₂ = 0.529 Å × (2)² / (1)
r₂ = 0.529 Å × 4
râ‚‚ = 2.116 Ã…
Answer: Eâ‚‚ = -3.4 eV, râ‚‚ = 2.12 Ã…
The n = 2 level has 1/4 the binding energy and 4 times the radius of the ground state.
Detailed Example 2: He⺠Ion Ground State
Problem: Calculate the ground state energy and radius for a He⺠ion (one electron, Z = 2).
Energy: E₠= -13.6 eV × (2)² / (1)² = -13.6 × 4 = -54.4 eV
Radius: r₠= 0.529 Å × (1)² / (2) = 0.265 Å
Interpretation:
He⺠has 4 times stronger binding energy than H due to doubled nuclear charge (Z² effect). The orbital is half the size, pulling the electron closer to the nucleus.
Electronic Transitions & Photon Energy
ΔE = Efinal - Einitial
ΔE = -13.6 Z² eV × (1/nf² - 1/ni²)
Emission (nf < ni)
Electron drops to lower level, photon emitted. ΔE is negative (energy released), photon energy Ephoton = |ΔE| = hν = hc/λ.
Absorption (nf > ni)
Electron jumps to higher level, photon absorbed. ΔE is positive, matching the energy of the absorbed photon.
Example: Balmer Series Transition (n = 3 → n = 2) in Hydrogen
ΔE = -13.6 eV × [(1/2²) - (1/3²)]
ΔE = -13.6 eV × [(1/4) - (1/9)]
ΔE = -13.6 eV × [(9 - 4)/36] = -13.6 × (5/36)
ΔE = -1.89 eV (negative = emission)
Photon energy: Ephoton = 1.89 eV
Wavelength: λ = hc/E = (1240 eV·nm) / 1.89 eV = 656 nm (red light)
Key Applications & Concepts
1. Spectroscopy & Emission Lines
The Bohr model explains hydrogen's discrete spectral lines (Lyman, Balmer, Paschen series). Each series corresponds to transitions ending at a specific n level: Lyman (n → 1, UV), Balmer (n → 2, visible), Paschen (n → 3, IR).
2. Ionization Energy
The energy required to remove an electron (ionization) equals |En|. For hydrogen ground state: 13.6 eV. For Heâº: 54.4 eV. This Z² dependence explains periodic trends in ionization energy.
3. Astrophysics & Stellar Spectra
Astronomers use Bohr model calculations to identify elements in distant stars by analyzing absorption and emission lines. The presence of hydrogen Balmer lines indicates stellar temperatures around 10,000 K.
4. Quantum Number Foundation
The principal quantum number n in Bohr's model became the primary quantum number in modern quantum mechanics, describing electron shells (K, L, M shells for n = 1, 2, 3).
Common Mistakes & Tips
Forgetting the Negative Sign
Energy levels are negative because they represent bound states. E = 0 means the electron is free (ionized). More negative = more stable/bound.
Applying to Multi-Electron Atoms
Bohr model only works accurately for hydrogen-like atoms (one electron). For helium, lithium, etc., electron-electron repulsion requires quantum mechanical treatment.
Confusing Emission vs Absorption
Emission: electron drops (ni > nf), ΔE negative, photon emitted. Absorption: electron rises (ni < nf), ΔE positive, photon absorbed.
Pro Tip: Energy Conversion
Remember: 1 eV = 1.602 × 10â»Â¹â¹ J. For photon wavelength: λ (nm) = 1240 eV·nm / E (eV). This shortcut saves time in spectroscopy problems.