Calculating binding energies of ground state electrons

  • Thread starter beecher
  • Start date
  • #1
15
0

Homework Statement



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

Homework Equations



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

[tex]E_{o}[/tex] = [tex]me^{4}[/tex] / 2[tex]hbar[/tex]^2(4[tex]\pi[/tex]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?
 

Answers and Replies

  • #2
gabbagabbahey
Homework Helper
Gold Member
5,002
7
Since you aren't dealing with Hydrogen, shouldn't you be using the equation for the approximate energy levels for any Hydrogenic atom with [itex]Z[/itex] protons instead of the equation for Hydrogen's energy levels?
 
  • #3
15
0
Yes, that would make sense, although I can't find that equation anywhere.
 
  • #4
15
0
Ok, I finally found it online. [tex]E_{n}[/tex] = ([tex]-Z^{2}[/tex][tex]R_{e}[/tex]) / [tex]n^{2}[/tex]
Where [tex]R_{e}[/tex] = 0.5([tex]m_{e}[/tex][tex]c^{2}[/tex])[tex]\alpha^{2}[/tex]
Where [tex]\alpha[/tex] = 1/137
Therefore [tex]R_{e}[/tex] = 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,
[tex]E_{n}[/tex] = [[tex](-1)^{2}[/tex](13.65ev)] / [tex]1^{2}[/tex]
[tex]E_{n}[/tex] = 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:

Related Threads on Calculating binding energies of ground state electrons

  • Last Post
Replies
5
Views
4K
Replies
2
Views
9K
Replies
13
Views
6K
Replies
0
Views
1K
Replies
11
Views
423
Replies
3
Views
10K
Replies
1
Views
2K
Replies
2
Views
7K
  • Last Post
Replies
2
Views
5K
Top