• Tags: reduced-mass, correction, isotope
• Difficulty: Advanced

In the Bohr model, the electron mass m should be replaced by the reduced mass μ = mM/(m+M), where M is the nuclear mass. For hydrogen: μ = m × 1836m/(m + 1836m) = m × 1836/1837 ≈ 0.99946m. The correction is tiny (~0.05%) but detectable. For positronium (e^{+}$$e^{-}): μ = m/2, so all energies are halved and radii are doubled compared to hydrogen. For muonic hydrogen (muon replaces electron, m_μ = 207m_e): μ ≈ 186m_e, so the radii are ~200× smaller and energies ~200× larger. JEE occasionally tests the reduced mass concept for exotic atoms.