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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?

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# Question about the energy of an electron orbiting the nucleaus

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