Question about the energy of an electron orbiting the nucleaus

  • Thread starter MohdAziz
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  • #1
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Hello,
I have a difficulty calculating the energy of an electron orbiting a nucleus, let's say we have a a hydrogen atom, there is only one electron orbiting around the proton. And since the radius never changes.
r = 5.29 x 10^-11 m
charge (qp) of proton= 1.60 x 10^-19 Coulumbs
charge (qe) of electron= -1.60 x 10^-19 Coulumbs
mass proton= 1.67 x 10^-27 kg
mass electron= 9.11 x 10^-31 kg

F = k (charge of proton) (charge of electron)/ (r^2)

If you solve for the Coulombic force, you get that it is approximately 8.221 * 10^-8 N.

This is equal to the centripetal force of any object in uniform circular motion: m(v^2)/r
Therefore:

8.221 * 10^-8 = (mass of the electron) v^2/ (r)
v^2 = 4.773 * 10^12
v is approximately 2.184 * 10^6 m/s

And after that to find the total enegry which is equal to total energy = U + KE

KE = 1/2 * m v^2 = 1/2 * 9.901 * 10^-31 * (2.183 * 10^6)^2 = 2.36 * 10^-18.

U = qE, E ( produced by the proton) = kcq/r^2 thus E = 5.141 * 10^11
thus U = qe = 8.226 * 10-8 J

Is my answer correct?
 

Answers and Replies

  • #2
1,006
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First, note that the electron in a hydrogen atom can only be treated properly with quantum mechanics, and it's quantum mechanics that lets you derive the atomic radius. If we're given the radius, your calculation will give an answer of the right order of magnitude. But you can't hope for more accuracy than that, because you are doing classical mechanics, not quantum mechanics.

Your kinetic energy is indeed the right order of magnitude, assuming your answer is in Joules (units are important!).

You are using the wrong formula for potential energy. qE is the force on a charge, not its potential energy. The potential energy is given by qV, where the potential V of a point charge is kq/r. You should end up with a potential energy equal to -2 times the kinetic energy.
 

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