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## Homework Statement

Given the potential energy [tex]V(r)=-\frac{1}{4\pi \epsilon_0}\frac{e^2}{r}[/tex] (where e is the unit charge), use the uncertainty principle [tex]\Delta x \Delta p \geq \hbar[/tex] to find the Bohr radius [tex]r_B[/tex] for a hydrogen atom and the ground state energy [tex]E_0[/tex].

Hint: write down the kinetic energy in terms of [tex]r_B[/tex] using the uncertainty principle.

## Homework Equations

-TISE: [tex]-\frac{\hbar^2}{2m}\frac{d^2u}{dr^2}+[V+\frac{\hbar^2}{2m}\frac{l(l+1)}{r^2}]u=Eu[/tex], where [tex]u(r)=r*R(r)[/tex] and R(r) is the radial component of the separable wave function [tex]\Psi[/tex].

-Alternatively, [tex]{V}_{eff} =V(r)+\frac{\hbar^2}{2m}\frac{l(l+1)}{r^2}[/tex]

## The Attempt at a Solution

The hint is really what throws me off. Finding the ground state energy is done out entirely in the textbook, but there is nothing about the modified uncertainty relation or writing the kinetic energy in terms of the Bohr radius before finding it.

I tried doing that, so [tex]T=\frac{p^2}{2m}=\frac{\hbar^2}{2mx^2}[/tex] if we assume equality in the uncertainty relation. Then I plugged that into an equation for the defined quantity [tex]\kappa=\frac{\sqrt{-2mE}}{\hbar}=\frac{i}{x}[/tex]. Do I just plug in [tex]r_B[/tex] for x now? I just don't see at all how doing that, if it's even correct, is leading closer to finding [tex]r_B[/tex].

EDIT: As an added note, I tried reverse engineering the solution: [tex]r_B=\frac{4\pi \epsilon_0\hbar^2}{m_e e^2}=\frac{-\hbar^2}{m_e r V(r)}[/tex], but that didn't make the problem any clearer to me.

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