# Calculating binding energies of ground state electrons

## 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, $$E_{n}$$ = - $$E_{o}$$ / $$n^{2}$$

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

gabbagabbahey
Homework Helper
Gold Member
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?

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

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

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