The energy of an electron in a hydrogen atom is: [itex]E = p^2/2m_e - \alpha e^2/r[/itex]; where [itex]p[/itex] is the momentum, [itex]m_e[/itex] is the electron charge magnitude, and [itex]\alpha[/itex] the coulomb constant. Use the uncertainty principle to estimate the minimum momentum in terms of [itex]m_e, a, e, \hbar[/itex].
[itex]\Delta p \Delta r = \hbar/2[/itex]
The Attempt at a Solution
The answer sheet set [itex]dE/dp = 0[/itex] to find [itex]r[/itex], and solved for [itex]p[/itex] using the uncertainty principle, but I'm confused why [itex]dE/dP[/itex] would give you minimum momentum, and whether the minimum momentum corresponds to minimum energy. All the questions I've seen that are related to this confines the electron to a certain radius. If [itex]r \to \infty[/itex], shouldn't both potential and kinetic energy (thus momentum) go to 0?