What Are the Energies in the Bohr Model of a Hydrogen Atom?

  • Thread starter Thread starter jamespetrovitch
  • Start date Start date
  • Tags Tags
    Electric
Click For Summary
SUMMARY

The discussion focuses on calculating the kinetic energy (KE) and electrical potential energy (U) of an electron in the Bohr Model of a hydrogen atom. The kinetic energy is derived as KE = ke²/2r, while the potential energy is expressed as U = -ke²/r. The relationship between these energies is established, confirming that KE equals half of U, specifically KE = -1/2U. Additionally, a secondary problem addresses the electric charge required for a person to counteract the Earth's electric field.

PREREQUISITES
  • Understanding of the Bohr Model of the hydrogen atom
  • Familiarity with kinetic energy and potential energy equations
  • Knowledge of electric fields and forces
  • Basic grasp of centripetal acceleration concepts
NEXT STEPS
  • Study the derivation of kinetic energy in circular motion
  • Learn about electric potential energy between point charges
  • Explore the implications of the Bohr Model on atomic structure
  • Investigate the effects of electric fields on charged objects
USEFUL FOR

Students studying physics, particularly those focusing on atomic models, electric forces, and energy calculations. This discussion is beneficial for anyone preparing for exams in classical mechanics or electromagnetism.

jamespetrovitch
Messages
4
Reaction score
0

Homework Statement


In the Bohr Model of a hydrogen atom, a single electron revolves around a single proton in a circle of radius r. Assume that the proton remains at rest.
(a) what is the kinetic energy of the electron?
(b) what is the electrical potential energy?
(c) show that the electron's kinetic energy is equal to half of the electric potential energy.
(give answers in terms of e, Me, Mp, and r)

Homework Equations


KE = 1/2mv^2
F = Ma(centripetal accel.)
a(centripetal accel.) = v^2/r
F = (mv)^2/r = (kqq)/r^2

KE = -1/2U
U = GMm/r^2 = (kqq)/r
v(orbit) = [tex]\sqrt{}GM/r[/tex]
1/2mv^2 = GMm/2r
F = q|E|
|E| = F/q = kq/r^2


The Attempt at a Solution


a)KE = ke^2/2r

b) I am having trouble finding a way to say that KE = -1/2U because I keep getting that...
KE = ke^2/2r
and that
U = GMm/2r even though U should be equal to something like...
U = -GMm/4r

c)cannot find a way to relate them...

----------------------------------------------------------------

Homework Statement


1. The Earth has a net electric charge that causes an electric field at its surface equal to 150N/C and directed inward to the center of the Earth.
(a) What magnitude (and sign) of charge would a 60kg person have to acquire to overcome the weight of the force exerted by the Earth's electric field?
(b) What would be the force of repulsion between two people, each with the charge calculated above and separated by 100m?

r = 6.37 x 10^6 m
Me = 5.98 x 10^24 kg
E = 150 N/C
Mp = 60 kg
d = 100m

Homework Equations



i) E = f/q = (ma)/q

ii) E = (kq)/r^2

iii) F = q|E| = (kqq)/d^2

The Attempt at a Solution



1a)
using eq. i,
150 = 60(-9.81)/q
q = -3.924 C

1b)
using eq. iii,
F = [(9 x 10^9)(-3.924)^2]/(100)^2
F = 1.39 x 10^7 N
 
Physics news on Phys.org
I think your life is going to be easier here if you look at the first problem first (the second ought to have been submitted in a separate thread.

jamespetrovitch said:

The Attempt at a Solution


a)KE = ke^2/2r

You're going to find the units for this result don't check. You have the equation for KE, which involves v^2 . The equation for centripetal acceleration *also* involves v^2 . What physical force provides the centripetal force that keeps the electron on its circular orbit? What would be the expression for the centripetal force then? Solve that for v^2 and substitute it into the equation for KE. What do you find? (Remember which quantities we want the results expressed in.)

b) I am having trouble finding a way to say that KE = -1/2U because I keep getting that...
KE = ke^2/2r
and that
U = GMm/2r even though U should be equal to something like...
U = -GMm/4r

Since we are finding the *electric* potential energy between the electron and proton, we probably don't want G in there. What is U for a pair of charges?

We'll come back to the second problem later...
 

Similar threads

Replies
8
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 0 ·
Replies
0
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
6
Views
4K
  • · Replies 4 ·
Replies
4
Views
5K