# Calculating binding energies of ground state electrons

1. Oct 10, 2009

### beecher

1. The problem statement, all variables and given/known data

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

2. Relevant equations

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

$$E_{o}$$ = $$me^{4}$$ / 2$$hbar$$^2(4$$\pi$$Eo)^2

3. 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 dont believe so but I'm unsure how to change them.

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

Not sure where to start. Help?

2. Oct 12, 2009

### 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?

3. Oct 12, 2009

### beecher

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

4. Oct 12, 2009

### 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

Last edited: Oct 12, 2009