Is There a Calculation Error in Wikipedia's Helium Atom Energy Formula?

Click For Summary
SUMMARY

The forum discussion centers on the calculation of the Helium Atom energy using the Thomas-Fermi approximation, specifically the formula &langle H &rangle = [-2Z^2 + \frac{27}{4}Z]E_1, where E_1 is defined as the Hartree unit of -13.6 eV. The minimum energy value occurs at Z=1.69, yielding an energy of -77.45 eV. However, the Wikipedia article incorrectly concludes that \frac{1}{2} \Bigg(\frac{3}{2}\Bigg)^6 = -77.5 eV without including E_1. The correct approach involves multiplying 729/128 by 13.6 eV, revealing that the optimal Z value is actually 27/16.

PREREQUISITES
  • Understanding of the Thomas-Fermi approximation
  • Familiarity with quantum mechanics terminology
  • Knowledge of energy units, specifically the Hartree unit
  • Basic algebra for manipulating equations
NEXT STEPS
  • Research the Thomas-Fermi approximation in quantum mechanics
  • Explore the derivation of the Hartree unit and its applications
  • Study the implications of screening parameters in atomic energy calculations
  • Investigate common errors in scientific articles and how to verify calculations
USEFUL FOR

Physicists, students of quantum mechanics, and researchers interested in atomic energy calculations will benefit from this discussion.

conway
Messages
397
Reaction score
0
The Wikipedia article on the Helium Atom goes through the Thomas-Fermi approximation and shows that when you use a screening parameter Z you can optimize the energy of the atom according to the following formula:

\langle H \rangle = [-2Z^2 + \frac{27}{4}Z]E_1

(E_1 is elsewhere defined as the Rydberg unit -13.6 eV)

The minimum value for energy occurs for a choice of Z=1.69, and you can plug in the numbers (I did) to confirm that this gives a value of -77.45 eV for the helium atom.

QUESTION: The article then concludes by saying that using this value "we obtain the most accurate result yet:"

\frac{1}{2} \Bigg(\frac{3}{2}\Bigg)^6 = -77.5 eV

What is this formula? Of course it is a mistake because it is missing E_1; but that's not the problem. We can correct the mistake and just multiply 729/128 by 13.6 eV to get the quoted result. But how does this come from the original calculation? Where is the value Z=1.69 which came from the optimization?
 
Last edited:
Physics news on Phys.org
conway said:
The Wikipedia article on the Helium Atom goes through the Thomas-Fermi approximation and shows that when you use a screening parameter Z you can optimize the energy of the atom according to the following formula:

\langle H \rangle = [-2Z^2 + \frac{27}{4}Z]E_1

(E_1 is elsewhere defined as the Hartree unit -13.6 eV]

The minimum value for energy occurs for a choice of Z=1.69, and you can plug in the numbers (I did) to confirm that this gives a value of -77.45 eV for the helium atom.

QUESTION: The article then concludes by saying that using this value "we obtain the most accurate result yet:"

\frac{1}{2} \Bigg(\frac{3}{2}\Bigg)^6 = -77.5 eV

What is this formula? Of course it is a mistake because it is missing E_1; but that's not the problem. We can correct the mistake and just multiply 729/128 by 13.6 eV to get the quoted result. But how does this come from the original calculation? Where is the value Z=1.69 which came from the optimization?

It's a Wiki article... GOK. :wink:
 
Oops. Found the answer. The Z value that comes from the minimization is actually an exact fraction, 27/16.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • Sticky
  • · Replies 2 ·
Replies
2
Views
10K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
10K
  • · Replies 2 ·
Replies
2
Views
6K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
2
Views
2K