Undergrad Riddle: Where is Epsilon in Bohr Model of Atom?

  • Thread starter Thread starter AhmedHesham
  • Start date Start date
  • Tags Tags
    Epsilon
Click For Summary
SUMMARY

The discussion centers on the Bohr model of the atom, specifically addressing the role of the permittivity constant (epsilon) in the energy formula for electrons in orbit. The original poster questions the absence of epsilon in the formula for emitted photons, which leads to clarification that the formula is expressed in units where epsilon is set to 1. The Bohr radius formula, r_n = n^2 \hbar^2 / (Z e^2 m_e), is also referenced, highlighting the importance of understanding the constants involved in atomic physics.

PREREQUISITES
  • Understanding of the Bohr model of the atom
  • Familiarity with atomic physics terminology
  • Knowledge of SI units and their application in physics
  • Basic grasp of quantum mechanics concepts
NEXT STEPS
  • Research the implications of using natural units in physics
  • Study the derivation and applications of the Bohr radius formula
  • Explore the relationship between energy levels and emitted photons in quantum mechanics
  • Learn about the role of permittivity in electromagnetic theory
USEFUL FOR

Students of physics, educators teaching atomic theory, and anyone interested in the mathematical foundations of the Bohr model and its applications in quantum mechanics.

AhmedHesham
Messages
96
Reaction score
11
Hi curious people
Today i had a lecture about bohr model of the atom .we know that the total energy of electron in orbit is minus e squared divided by 8 pi times epsilon times R of the orbit . this is written in my book and i agree with this .But when the author writes a formula for an emitted photon he writes the difference of the law mentioned above but without epsilon . why ?
Thankes
 
Physics news on Phys.org
AhmedHesham said:
this is written in my book

Which book?
 
PeterDonis said:
Which book?

His book! He just said so! :wink:

OP, we're looking for the book's title, and author, and if you can do a good job, maybe a scan of that page. If not, please write the entire formula.
 
  • Like
Likes vanhees71
This is a picture from the book
 

Attachments

  • Screenshot_٢٠٢١٠٤١١-٢٠٣٦٢١.png
    Screenshot_٢٠٢١٠٤١١-٢٠٣٦٢١.png
    40.5 KB · Views: 184
haushofer said:
We are curious, not psychics.
There is no such thing as psychic.
- Patrick Jane
 
  • Like
Likes dextercioby
I have no idea what units the equation you posted is supposed to be in. There's a missing electron mass, for one thing.
 
Vanadium 50 said:
I have no idea what units the equation you posted is supposed to be in. There's a missing electron mass, for one thing.

No, it's not missing. This formula is in terms of the ##n^{th}## Bohr radius ##r_n##, which is given by:

##r_n = \dfrac{n^2 \hbar^2}{Z e^2 m_e}##

(where ##Z## is the number of protons in the nucleus, and ##m_e## is the mass of the electron)

The answer to the original poster's question is that they are using units where ##\epsilon_0 = 1##.
 
  • Like
Likes Vanadium 50
Ah...it's buried in the r's. I see,
 
  • #10
stevendaryl said:
he answer to the original poster's question is that they are using units where .

Are you sure? Isn't it buried in the r's as well?
 
  • #11
No ...i think it is missing
We always us SI in my institute
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 36 ·
2
Replies
36
Views
8K
Replies
14
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
10
Views
2K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 15 ·
Replies
15
Views
4K
  • · Replies 9 ·
Replies
9
Views
3K