The ground state of hydrogen has total electronic angular momentum J = 1/2 and nuclear spin I = 1/2. How many distinct hyperfine energy levels does this state split into?
AOne level with F = 1, because the spins always align in the lowest energy state
BTwo levels: F = 0 and F = 1
CThree levels: F = −1, F = 0, and F = 1 (one per projection of F)
DFour levels corresponding to all combinations of the two spin-1/2 particles
F ranges from |I − J| to I + J in integer steps: from |1/2 − 1/2| = 0 to 1/2 + 1/2 = 1, giving F = 0 and F = 1 — two levels. Option C confuses the quantum number F with its magnetic projection m_F. Option D counts the four uncoupled spin states (which are reorganized by coupling into F = 0 with 1 state and F = 1 with 3 states). The F = 1 → F = 0 transition is the famous 21 cm hydrogen line.
Question 2 Multiple Choice
Why do electrons in s-orbitals experience larger hyperfine splitting than electrons in p-orbitals?
As-orbitals have higher energy, so the interaction Hamiltonian is stronger
Bs-orbitals have nonzero probability density at the nucleus, enabling the Fermi contact interaction; p-orbitals have a node at the nucleus so the contact term vanishes
Cs-orbitals have no orbital angular momentum, leaving the nuclear magnetic moment completely unshielded
Dp-orbitals interact through a stronger magnetic quadrupole term that actually suppresses splitting
The dominant mechanism for s-orbital hyperfine splitting is the Fermi contact interaction, which depends on |ψ(0)|² — the electron probability density at the nucleus. Only s-orbitals have nonzero density at the origin; p, d, and f orbitals all have ψ(0) = 0. Without the contact term, only the weaker magnetic dipole interaction from the orbital current contributes. This is why the hydrogen 1s ground state has the largest hyperfine splitting of its levels.
Question 3 True / False
Hyperfine splittings are much smaller than fine structure splittings because the nuclear magnetic moment is roughly 1,836 times smaller than the Bohr magneton.
TTrue
FFalse
Answer: True
The nuclear magnetic moment is μ_I = g_I(e/2m_p)I, where the proton mass m_p appears in the denominator rather than the electron mass m_e. Since m_p/m_e ≈ 1836, the nuclear magnetic moment is ~1836 times smaller than the Bohr magneton that governs fine structure. The hyperfine interaction energy is correspondingly ~1836 times smaller — which is why hyperfine transitions like the 21 cm line fall in the radio band rather than optical frequencies.
Question 4 True / False
Most orbital types (s, p, d, f) contribute equally to hyperfine splitting through the Fermi contact interaction.
TTrue
FFalse
Answer: False
Only s-orbitals contribute through the Fermi contact interaction, because only s-orbitals have nonzero electron density at the nucleus (|ψ(0)|² ≠ 0). For all other orbital types (l ≠ 0), the wavefunction vanishes at the nucleus, so the contact term is zero. p, d, and f orbitals experience hyperfine splitting through the magnetic dipole interaction, which is weaker. This is a direct consequence of the angular node structure of non-s wavefunctions.
Question 5 Short Answer
Explain why the 21 cm hydrogen line is important in radio astronomy, and what feature of s-orbitals makes the ground-state hyperfine splitting measurable despite its extremely long radiative lifetime.
Think about your answer, then reveal below.
Model answer: The 21 cm line (F = 1 → F = 0 in the hydrogen ground state) is important because hydrogen is the most abundant element in the universe, and 21 cm photons pass through interstellar dust that blocks visible light — allowing radio telescopes to map galactic structure. The ground-state hyperfine splitting is large enough to be measurable precisely because the 1s wavefunction has maximum amplitude at the nucleus (Fermi contact interaction), giving the strongest possible coupling. Despite the ~10 million year radiative lifetime, the enormous quantity of interstellar hydrogen ensures a detectable signal.
The Fermi contact interaction maximizes for the 1s state because |ψ(0)|² is largest there. The resulting energy gap — though tiny (5.9 × 10⁻⁶ eV) — is precisely defined and falls at 1420 MHz, a frequency radio telescopes detect easily. The long lifetime is not a barrier because the interstellar medium contains so much hydrogen that even one-in-ten-million-year decays produce a constant, bright signal. Atomic clocks exploit similarly precise hyperfine transitions for the same reason: the precision of the transition frequency, not its rate, is what matters.