Ground State Energy of an Electron

Click For Summary
SUMMARY

The discussion centers on the calculation of the ground state energy of an electron in a hydrogen atom using the Bohr model. The equation referenced is E_n = \frac{2 \pi m_e e^4}{n^2 (4 \pi \epsilon_{0}) h^2}, where constants such as electron mass (m_e = 9.11 x 10-31 kg), elementary charge (e = 1.60 x 10-19 C), and Planck's constant (h = 6.63 x 10-34 J·s) are utilized. The user encountered discrepancies in their calculations, specifically a factor of 3.14, suggesting a misunderstanding of the equation's structure, particularly regarding the inclusion of π in the denominator. Clarifications indicate that a factor of 2π is indeed necessary in the numerator.

PREREQUISITES
  • Understanding of the Bohr model of the atom
  • Familiarity with fundamental constants: electron mass, elementary charge, permittivity of free space, and Planck's constant
  • Basic knowledge of quantum mechanics
  • Ability to perform unit conversions between joules and electron volts
NEXT STEPS
  • Study the derivation of the Bohr model energy levels in detail
  • Learn about the significance of fundamental constants in quantum mechanics
  • Explore the concept of angular momentum quantization in atomic systems
  • Investigate the implications of the Rydberg formula for spectral lines
USEFUL FOR

Students and professionals in physics, particularly those focusing on quantum mechanics and atomic theory, as well as educators seeking to clarify the Bohr model's applications and calculations.

jbowers9
Messages
85
Reaction score
1
This is what I've tried to work out and I'm not getting -13.7 eV. What am I doing wrong?


E 2 Π m e^4 / (4 Π ε0 )^2 h^2 6.90E-19 J=4.31eV

m 9.11 x 10-31 kg 9.11E-31
e 1.60 x 10-19 C 1.60E-19
ε0 8.85 x 10-12 C2/Nm2 8.85E-12
h 6.63 x 10-34 J S 6.63E-34

1 joule = 6.24150974 × 10^18 electron volts
 
Physics news on Phys.org
Sorry but I have absolutely no idea what that equation is supposed to be, or where you got it from. Try TeXing it perhaps? How did you arrive at it?
 
Need more information, have no idea what you're saying.
 
The equation is based on the Bohr atom for energy levels. The version I wrote above is for n=1. The development in the text I'm using:

AUTHOR Mortimer, Robert G.
TITLE Physical chemistry / Robert G. Mortimer.
PUB INFO San Diego, Calif. : Academic Press, c2000.
pgs. 511-520 roughly

uses En = 2 Π m e4 / n2 (4 Π ε0 )2 h2

When I plug in the constants, n=1, the value is off from 13.7 eV, after conversion from Joules, by a factor of 3.14, as if Pi doesn't belong in the denominator. I'm thinking that it is already included in the permitivity constant ε0.
 
jbowers9 said:
The equation is based on the Bohr atom for energy levels. The version I wrote above is for n=1. The development in the text I'm using:

AUTHOR Mortimer, Robert G.
TITLE Physical chemistry / Robert G. Mortimer.
PUB INFO San Diego, Calif. : Academic Press, c2000.
pgs. 511-520 roughly

uses En = 2 Π m e4 / n2 (4 Π ε0 )2 h2

When I plug in the constants, n=1, the value is off from 13.7 eV, after conversion from Joules, by a factor of 3.14, as if Pi doesn't belong in the denominator. I'm thinking that it is already included in the permitivity constant ε0.

I presume you mean E_n = \frac{2 \pi m_e e^4}{n^2 (4 \pi \epsilon_{0}) h^2}

Incidentally, my quantum mech book gives the equation for the energy according to the Bohr model as \frac{m_e Z^2 e^4}{(4 \pi \epsilon_0)^2 2 \hbar^2}\frac{1}{n^2}

so you're missing a factor of 2 \pi up top and you're missing a 2 from down below... in other words, you're missing a factor of pi. Which is what you say you're missing. =)
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 12 ·
Replies
12
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
7K
  • · Replies 113 ·
4
Replies
113
Views
13K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K