• $summary_{type}$: concept
• $word_{count}$: 160

The Bohr quantization condition L = nhbar is equivalent to the de Broglie standing wave condition 2pir = nlambda. Proof: substituting lambda = $\frac{h}{mv}$ gives mvr = $\frac{nh}{2pi}$ = nhbar. This physical insight shows stable orbits are those where the electron wave constructively interferes with itself around the orbit. The de Broglie wavelength in the nth orbit: $lambda_n$ = 2pi*$r_n$/n = 2pin*$a_0$/Z. For reduced mass correction, replace electron mass m with mu = $\frac{mM}{m+M}$. For hydrogen: mu approximately equals 0.99946m (tiny correction). For positronium (electron-positron): mu = m/2, so energies are halved and radii doubled. For muonic hydrogen (muon replaces electron, mass = 207*$m_e$): mu approximately equals 186*$m_e$, radii are approximately 200 times smaller, energies approximately 200 times larger. JEE occasionally tests exotic atom calculations using reduced mass.