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

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In the Bohr Model of the hydrogen atom, the kinetic energy (KE) of the electron is expressed as KE = ke^2/2r, while the electrical potential energy (U) is related to the interaction between the electron and proton. The relationship KE = -1/2U is a key point of confusion, as participants struggle to derive U correctly, with some suggesting U should be expressed in terms of gravitational constants rather than electric charge. The centripetal force required for the electron's circular motion must be linked to the electric force, which is essential for deriving the correct expressions for KE and U. The discussion highlights the need for clarity in applying the correct formulas for electric potential energy in the context of the hydrogen atom.
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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...

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

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
 
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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...
 

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