Calculating binding energies of ground state electrons

[beecher]

Homework Statement

What is the calculated binding energy of the electron in the ground state of (a) deuterium, (b) He^{+} and (c) Be^{+++}?

Homework Equations

For the hydrogen atom, E_{n} = - E_{o} / n^{2}

E_{o} = me^{4} / 2hbar^2(4\piEo)^2

The Attempt at a Solution

Not sure how to do this. Can I apply the above equations which are meant for a hydrogen atom to these other atoms? I don't believe so but I'm unsure how to change them.

I know how to correct the Rydberg constant for the different atoms, but it doesn't even appear in any of these formulas so I don't think that would change anything.

Not sure where to start. Help?

 

[gabbagabbahey]

Since you aren't dealing with Hydrogen, shouldn't you be using the equation for the approximate energy levels for any Hydrogenic atom with Z protons instead of the equation for Hydrogen's energy levels?

 

[beecher]

Yes, that would make sense, although I can't find that equation anywhere.

 

[beecher]

Ok, I finally found it online. E_{n} = (-Z^{2}R_{e}) / n^{2}
Where R_{e} = 0.5(m_{e}c^{2})\alpha^{2}
Where \alpha = 1/137
Therefore R_{e} = 0.5 [(9.11x10^-31 kg)(3.00x10^8 m/s)^2](1/137)^2
=2.184x10^-18 J = 13.65 ev

So for deuterium,
E_{n} = [(-1)^{2}(13.65ev)] / 1^{2}
E_{n} = 13.65ev

For He, the difference is that Z = 2, and thus E = 4(-13.65ev) = -54.6ev
and for Be Z = 4, and thus E = 16 (-13.65ev) = -218.4 ev