Bohr Radius and Ground State Energy

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SUMMARY

The discussion focuses on deriving the Bohr radius (r_B) and ground state energy (E_0) for a hydrogen atom using the uncertainty principle and the virial theorem. The potential energy is defined as V(r) = -\frac{1}{4\pi \epsilon_0}\frac{e^2}{r}. Participants clarify that by substituting the Bohr radius into the kinetic energy equation derived from the uncertainty principle, one can simplify the calculations. The virial theorem is also highlighted as a crucial tool, allowing for a more straightforward solution to find r_B.

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  • Understanding of the uncertainty principle in quantum mechanics
  • Familiarity with the virial theorem and its applications
  • Knowledge of the time-independent Schrödinger equation (TISE)
  • Basic concepts of potential energy in quantum systems
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  • Study the derivation of the Bohr radius using the uncertainty principle
  • Learn about the virial theorem and its implications in quantum mechanics
  • Explore the time-independent Schrödinger equation (TISE) in detail
  • Investigate the relationship between kinetic and potential energy in quantum systems
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Students and educators in quantum mechanics, physicists focusing on atomic structure, and anyone interested in the mathematical foundations of the Bohr model of the hydrogen atom.

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Homework Statement


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

Hint: write down the kinetic energy in terms of r_B using the uncertainty principle.

Homework Equations


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

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

EDIT: As an added note, I tried reverse engineering the solution: r_B=\frac{4\pi \epsilon_0\hbar^2}{m_e e^2}=\frac{-\hbar^2}{m_e r V(r)}, but that didn't make the problem any clearer to me.
 
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Hmm, I'm not sure how you can find the Bohr radius without an additional piece of information. If you use the virial theorem, which says <T> = -<V>/2, you can then solve for a0.
 
Our book only mentions the special cases of the virial theorem for stationary states ,2&lt;T&gt;=&lt;x\frac{dV}{dx}&gt;, and stationary states of the harmonic oscillator, &lt;T&gt;=&lt;V&gt;.

Even if I use your version of it, though, I don't see how that helps.
 
For a potential of the form rn, the virial theorem reduces to what I said.

http://en.wikipedia.org/wiki/Virial_theorem

In your expression for T after you substituted with the uncertainty principle, set x to the Bohr radius. Similarly, in the potential, set r to the Bohr radius. Then solve for the Bohr radius.
 
Wow, that theorem made it a much easier problem to solve. Thanks very much.
 

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