What is the kinetic energy of the electron?

[jamespetrovitch]

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) = $$\sqrt{}GM/r$$
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...

 

[Hootenanny]

The Attempt at a Solution

a)KE = ke^2/2r

Correct

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

Look up the equation for electic potentional energy

 

[jamespetrovitch]

ah, so for ...
b) U=kqq/r

and

c) KE = ke^2/2r
(1/2)(kqq/r) = KE = kee/2r

do I need to relate
KE = GMm/2r = kqq/2r any further or do the constants pretty much switch out since it is dealing with electric potential and not gravitation?

 

[Hootenanny]

ah, so for ...
b) U=kqq/r

and

c) KE = ke^2/2r
(1/2)(kqq/r) = KE = kee/2r

do I need to relate
KE = GMm/2r = kqq/2r any further or do the constants pretty much switch out since it is dealing with electric potential and not gravitation?

The gravitational field is negligable on the atomic scale, you only need to consider the electromagnetic field.

 

[jamespetrovitch]

Awesome! Thank you so much for the help! Any ideas with the other post I made about electrical fields?

https://www.physicsforums.com/showthread.php?t=186613