- #1
jamespetrovitch
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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) = [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...
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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