A student claims: 'The electron in a hydrogen 1s orbital is circling the nucleus at a fixed radius of a₀, like a planet orbiting a star.' What is the most fundamental error in this picture?
AThe radius is wrong — electrons in 1s orbitals are much closer to the nucleus than a₀
BElectrons do not circle — they oscillate back and forth through the nucleus
CThere is no electron trajectory at all — |ψ|² is a probability density, and between measurements the electron has no defined position or path
DThe picture is approximately correct for 1s but fails for higher orbitals
The orbital picture is a probability distribution, not a trajectory. The 1s wavefunction gives |ψ₁s|² peaked near a₀, meaning the electron is most likely to be *found* near a₀ when measured. Between measurements, there is no classical path — no orbit, no oscillation, no trajectory. The Bohr model assumed circular orbits, which quantum mechanics replaces with probability clouds. Options A and B retain the classical-trajectory picture and are therefore wrong at the level of foundations.
Question 2 Multiple Choice
An electron has quantum numbers n=3, ℓ=2, m=−1. What do these quantum numbers tell you?
An=3 gives the third energy level; ℓ=2 means it is a d orbital with angular momentum magnitude √6ℏ; m=−1 gives the z-component of angular momentum as −ℏ
Bn=3 gives the energy; ℓ=2 means the electron is in the second excited state; m=−1 gives the spin orientation
CAll three quantum numbers together give the energy, with degeneracy determined by spin
Dn=3 determines shape; ℓ=2 determines energy level; m=−1 determines the radial distribution
Each quantum number encodes distinct physics. n (here 3) determines energy: E = −13.6 eV/n² = −1.51 eV. ℓ (here 2) determines the orbital angular momentum magnitude: |L| = √(ℓ(ℓ+1))ℏ = √6ℏ, and the orbital shape (ℓ=2 → d orbital). m (here −1) determines the z-component: Lz = mℏ = −ℏ. The m quantum number has nothing to do with spin (that requires a fourth quantum number, mₛ). Options B, C, and D misassign the physical meaning of one or more quantum numbers.
Question 3 True / False
Two hydrogen electrons with quantum numbers (n=2, ℓ=1, m=0) and (n=2, ℓ=1, m=+1) have the same energy.
TTrue
FFalse
Answer: True
In the hydrogen atom, the energy depends only on the principal quantum number n: E_n = −13.6 eV/n². All states with the same n are degenerate in energy — they have different shapes and orientations (determined by ℓ and m) but identical energies. For n=2, there are four degenerate states: (ℓ=0, m=0), (ℓ=1, m=−1), (ℓ=1, m=0), (ℓ=1, m=+1). This degeneracy is broken by spin-orbit coupling and other relativistic effects, but at the leading-order hydrogen solution all n=2 states are equal in energy.
Question 4 True / False
An atomic orbital directly represents the path that an electron travels around the nucleus — the denser regions of the orbital diagram show where the electron spends most of its time as it moves.
TTrue
FFalse
Answer: False
Orbitals are probability *densities* — |ψ(r)|² gives the probability per unit volume of finding the electron near position r upon measurement. Between measurements, quantum mechanics assigns no definite position or trajectory. Describing the electron as 'moving through' the orbital is a classical intuition that quantum mechanics discards. The denser regions do indicate higher probability of detection, but this is a statement about measurement outcomes, not a classical trajectory. The distinction is not semantic — it is the core conceptual break from Bohr-model thinking.
Question 5 Short Answer
Why is it more accurate to describe the 1s orbital as a 'probability cloud' than as an 'orbit,' and what physical quantity does |ψ|² actually represent?
Think about your answer, then reveal below.
Model answer: |ψ(r,θ,φ)|² is the probability density: the probability per unit volume of finding the electron near position (r,θ,φ) if a position measurement is made. The total probability of finding the electron somewhere integrates to 1 (normalization). An 'orbit' implies a definite trajectory — a path the electron follows in time. Quantum mechanics replaces trajectories with probability amplitudes: the electron has no defined position between measurements, and asking 'where is it right now?' has no answer within the theory. The cloud picture visualizes the probability distribution, not a blurred trajectory.
This conceptual shift — from deterministic trajectories to probability distributions — is the foundational break of quantum mechanics from classical mechanics. The orbital is a stationary state: the probability cloud does not change over time for an energy eigenstate. It is not that the electron is moving too fast to track; it literally has no trajectory in the quantum description.