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To find the wavelength of an electron in its ground state in a hydrogen atom, would I or could I do the following?

  1. Use the ground state energy (-13.6eV) in $E^2 = m^2c^4 + p^2c^2$
  2. Solve for $p$
  3. Use $p$ to find $λ$ in $λ = h/p$

Also, if I wanted to find the total energy of an electron in another state given the wavelength, could I just do the above process backwards (use $λ$ to get $p$, then find $E$)?

Let me know if I'm simply completely incorrect, missing a step, or if there's anything wrong here please.

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Seems good to me. Make sure your units are correct. eV can be pesky, so convert everything into Joules. A little trick I learned in undergrad was that if I have everything in MKS (meters kilograms and seconds,and my constants in those units) then you wont have unit problems EVER! Ya its annoying to use $\hbar$ in $10^{-34}$ J$\cdot$s but $10^{-15}$ eV isn't much better. –  John M Apr 24 '13 at 2:21
    
The electron in an H atom isn't a plane wave or any superposition of plane waves, so it isn't clear what wavelength means in this context. I suppose if you approximate the potential by a harmonic potential you could calculate the wavelength for the corresponding harmonic oscillator ground state. –  John Rennie Apr 24 '13 at 9:38

1 Answer 1

I believe you will find that the expectation value of the momentum is zero, which will sort of mess up your calculation of the wave length. Calculating the wavelength of the ground state of any bound system is folly anyway, since it will not have any nodes. The characteristic you may be looking for is the average radius, or the uncertainty in the position.

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Given an average radius of the ground state ($5.29*10^-11m$) could I use the aforementioned method? Moreover, you mentioned finding the uncertainty in position - would $ΔxΔP = h/4π$ be of use here somehow? –  ThroatOfWinter57 Apr 24 '13 at 3:30

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