Questions: Hydrogen Atom: Quantum Energy Levels and Orbitals

Question 1

Two hydrogen orbitals have quantum numbers (n=2, l=0, m=0) and (n=2, l=1, m=0) respectively. What can you say about their energies?

• A. The 2p orbital (l=1) has lower energy because angular momentum stabilizes the electron
• B. The 2s orbital (l=0) has lower energy because s electrons penetrate closer to the nucleus
• C. Both orbitals have the same energy — in hydrogen, energy depends only on the principal quantum number n
• D. Their energies cannot be compared without knowing the electron's spin quantum number

Answer: (C)

In hydrogen, the energy eigenvalues E_n = −13.6 eV/n² depend only on n, not on l or m_l. Both the 2s and 2p orbitals have n=2, so both have energy E₂ = −3.4 eV. Options A and B describe real effects — in multi-electron atoms, l does affect energy through shielding and penetration — but these effects arise from electron-electron repulsion, which doesn't exist in hydrogen. This confusion between hydrogen and many-electron atoms is one of the most common errors in atomic physics.

Question 2

The 1s orbital of hydrogen has its highest probability density at the nucleus (r=0). What does this mean about where the electron is most likely to be found?

• A. The electron is most likely found at r=0, inside the nucleus itself
• B. The electron follows a circular orbit closest to the nucleus, as Bohr predicted
• C. The most probable radius to find the electron is the Bohr radius a₀, because the radial probability distribution 4πr²|ψ|² peaks there despite |ψ|² being highest at r=0
• D. The electron is uniformly distributed throughout a sphere of radius a₀

Answer: (C)

This question targets the distinction between probability density |ψ(r)|² and the radial probability distribution P(r) = 4πr²|ψ|². The probability density |ψ|² is indeed highest at r=0 for the 1s orbital. But the probability of finding the electron in a thin shell at radius r involves both |ψ|² and the volume of that shell (4πr²dr). Near r=0, the shell volume goes to zero, so even though |ψ|² is large there, the electron is rarely found at the nucleus. The radial probability distribution P(r) = 4πr²|ψ|² peaks at exactly r = a₀, the Bohr radius — which is why Bohr got the right scale but for the wrong reason (he had definite orbits; QM has probability distributions).

Question 3

According to quantum mechanics, the electron in a hydrogen atom follows a definite circular orbit at a distance determined by the quantum number n, just as Bohr described.

• T. True
• F. False

Answer: False

This is the central conceptual error the topic addresses. Quantum mechanics replaces Bohr's definite orbits with probability densities: |ψ(r)|² gives the probability per unit volume of finding the electron near position r. There is no trajectory; between measurements the electron doesn't 'travel' anywhere in a classically meaningful sense. The Bohr model predicts correct energy levels by lucky cancellation of errors, but its physical picture — electrons in circular orbits at fixed radii — is wrong. Nodes in the wavefunction, where |ψ|² = 0, have no classical analogue (an orbiting particle can't simply skip a region), and confirm that the orbital picture is fundamentally different from Bohr's.

Question 4

In a hydrogen atom, the 2s and 2p orbitals have the same energy because the energy formula E_n = −13.6 eV/n² depends only on the principal quantum number n.

• T. True
• F. False

Answer: True

This is correct for hydrogen specifically. Both 2s (n=2, l=0) and 2p (n=2, l=1) have E₂ = −13.6/4 = −3.4 eV. This degeneracy is a special feature of the Coulomb potential — it breaks down in multi-electron atoms where electron repulsion makes energy depend on both n and l. For hydrogen, the three quantum numbers (n, l, m_l) describe the shape and orientation of the orbital but only n determines the energy.

Question 5

What is the fundamental conceptual difference between a Bohr orbit and a quantum mechanical orbital, and why does this difference matter physically?

Model answer: A Bohr orbit is a definite circular path: the electron travels around the nucleus at a fixed radius with a specific speed, like a planet orbiting a star. A quantum mechanical orbital is a probability density distribution |ψ(r)|² — it describes the probability per unit volume of finding the electron near each point in space if you make a measurement, not a trajectory the electron follows. Physically this matters because orbitals can have nodes (surfaces where |ψ|² = 0) that a trajectory could never pass through, because orbitals have complex spatial shapes (p, d, f) with lobes and angular nodes that have no classical counterpart, and because the electron's position is genuinely indeterminate between measurements — not merely unknown but undefined.

The key phrase is 'probability density, not path.' The Bohr model is conceptually wrong even when numerically correct: the electron doesn't 'travel' in a circle, its position has no determinate value between observations, and the orbital shapes (including nodes) are intrinsically quantum phenomena with no classical analogue. Understanding this shift is essential for everything from chemical bonding to spectroscopy to quantum computing.