Hydrogen Orbitals Calculator
Calculate hydrogen orbitals with our free science calculator. Uses standard scientific formulas with unit conversions and explanations.
Formula
E_n = -13.6 * Z^2 / n^2 eV
Where E_n is the energy of level n in electron volts, Z is the atomic number, and n is the principal quantum number. The orbital radius is r = a0 * n^2 / Z where a0 = 52.9 pm is the Bohr radius. Angular momentum magnitude is L = hbar * sqrt(l*(l+1)).
Worked Examples
Example 1: Hydrogen 2p Orbital Properties
Problem: Calculate the energy, orbital radius, and angular momentum for an electron in the 2p orbital (n=2, l=1, m=0) of hydrogen (Z=1).
Solution: Energy = -13.6 * 1^2 / 2^2 = -13.6 / 4 = -3.4 eV\nOrbital radius = a0 * n^2 / Z = 52.9 pm * 4 / 1 = 211.6 pm\nAngular momentum = hbar * sqrt(l*(l+1)) = hbar * sqrt(2) = 1.414 * hbar\nz-component = m * hbar = 0\nRadial nodes = n - l - 1 = 2 - 1 - 1 = 0\nAngular nodes = l = 1\nDegeneracy of 2p = 2*(2*1+1) = 6
Result: E = -3.4 eV | r = 211.6 pm | L = 1.414 hbar | 0 radial nodes, 1 angular node
Example 2: He+ Ion Ground State
Problem: Calculate the properties of the ground state (1s) of a He+ ion (Z=2). Compare with hydrogen.
Solution: Energy = -13.6 * 2^2 / 1^2 = -13.6 * 4 = -54.4 eV (4x hydrogen)\nOrbital radius = a0 * 1^2 / 2 = 52.9 / 2 = 26.45 pm (half of hydrogen)\nIonization energy = 54.4 eV\nVelocity = 2 * 2188 km/s = 4376 km/s\nWavelength to n=2: 1240 / (13.6 * 4 * (1 - 1/4)) = 1240 / 40.8 = 30.4 nm (EUV)
Result: E = -54.4 eV | r = 26.45 pm | Ionization = 54.4 eV | v = 4376 km/s
Frequently Asked Questions
What are hydrogen orbitals and why are they important?
Hydrogen orbitals are the quantum mechanical wavefunctions that describe the probability distribution of the electron in a hydrogen atom or hydrogen-like ion. They are the exact analytical solutions to the Schrodinger equation for the Coulomb potential and form the foundation of all atomic physics and chemistry. Each orbital is characterized by three quantum numbers: principal (n), angular momentum (l), and magnetic (m), which determine the energy, shape, and orientation of the electron distribution. Hydrogen orbitals are important because they serve as the basis for understanding multi-electron atoms, chemical bonding, molecular structure, and spectroscopy across all of chemistry and materials science.
Why are hydrogen atom energy levels degenerate?
In the hydrogen atom, all orbitals with the same principal quantum number n have the same energy regardless of the l and m values. This means the 2s and 2p orbitals have identical energy, and the 3s, 3p, and 3d orbitals are all degenerate. This accidental degeneracy is unique to the pure Coulomb potential and arises from a hidden symmetry related to the Laplace-Runge-Lenz vector. The total degeneracy of level n is 2n squared (including spin). In multi-electron atoms, electron-electron repulsion breaks this degeneracy, causing subshells with different l values to have different energies. This lifting of degeneracy explains the structure of the periodic table and chemical properties.
What are radial and angular nodes in hydrogen wavefunctions?
Nodes are regions where the wavefunction (and therefore the probability of finding the electron) is exactly zero. Radial nodes are spherical surfaces where the radial wavefunction crosses zero, and their number equals n minus l minus 1. Angular nodes are planes or cones where the angular wavefunction is zero, and their number equals l. The total number of nodes (excluding the node at infinity) is always n minus 1. For example, the 3d orbital has zero radial nodes and two angular nodes (total of 2), while the 3s orbital has two radial nodes and zero angular nodes (also total 2). Node structure determines the orbital shape and influences chemical bonding properties.
How do hydrogen orbital energies relate to atomic spectra?
When an electron transitions between energy levels in hydrogen, it emits or absorbs a photon with energy equal to the difference between the levels. The energy formula E equals negative 13.6 eV times Z squared divided by n squared predicts discrete spectral lines organized into series. The Lyman series (transitions to n=1) produces ultraviolet light, the Balmer series (to n=2) produces visible light, and the Paschen series (to n=3) produces infrared radiation. These spectral lines were observed experimentally before quantum mechanics and provided crucial evidence for the quantized nature of atomic energy levels. The hydrogen spectrum remains the most precisely measured and theoretically understood atomic spectrum.
What are hydrogen-like ions and how do their properties scale with Z?
Hydrogen-like ions are atoms or ions with only one electron, such as He+ (Z=2), Li2+ (Z=3), and Be3+ (Z=4). Their energy levels scale as Z squared times the hydrogen values, so He+ has four times the binding energy of hydrogen. Orbital radii scale as 1/Z, making the orbitals more compact. The velocity scales as Z times the hydrogen value. These scaling relationships are exact because the single-electron Coulomb problem has an analytical solution for any Z. Hydrogen-like ions are important in plasma physics, astrophysics, and spectroscopy. The study of highly charged hydrogen-like ions (large Z) tests quantum electrodynamics corrections to high precision.
What is the significance of orbital angular momentum in hydrogen?
Orbital angular momentum in hydrogen is quantized with magnitude equal to hbar times the square root of l times (l plus 1). The z-component is quantized as m times hbar, where m ranges from minus l to plus l. This quantization means the angular momentum vector can only take specific orientations relative to a chosen axis, a phenomenon called space quantization. The angular momentum determines the orbital shape, the number of angular nodes, the magnetic properties of the atom, and selection rules for transitions. In a magnetic field, different m values acquire different energies (Zeeman effect), splitting spectral lines and demonstrating the reality of space quantization experimentally.