Problem Find the minimum energy of the hydrogen atom by using uncertainty principle a. Take the uncertainty of the position Δr of the electron to be approximately equal to r b. Approximate the momentum p of the electron as Δp c. Treat the atom as a 1-D system My step 1. Δr Δp ≥ h/4(pi) Δp ≥ h/4(pi)r 2. Total energy E = (p^2 /2m) - ke^2 /r ≥ (h^2/8(pi)^2 m r^2) - (ke^2)/r 3. rearranging the term (Emin)r^2 + (ke^2)r - (h^2 /8(pi)^2 m ) = 0 Require Δ = 0 for the quadratic equation I obtain E = -54.7 eV ≠ -13.6 eV If I replace Δr by (1/2)Δr, I can obtain the correct result. But I don't know why.