Question

In a hydrogen atom, if the radius of the orbit of the electron is doubled, then its energy will ...

a)Decrease by a factor of 2
b)Remain the same
c)Increase by a factor of 2
d)Actually, it is not possible for the radius of the orbit to double
e)Increase or decrease by a factor not listed

Is it going to increase by a factor of 2?

Answer 1

U = - (1/4 pi eo)e^2 / r
Negative because 0 at infinity and drops as it gets closer

F = (1/4 pi eo) e^2/r^2 = m a = m v^2 /r

m v^2 = (1/4 pi eo) e^2/r
so
KE = (1/2) m v^2 = (1/2) (-U)
the KE is half the magnitude of the potential energy at every radius.

Total energy = (1/2)(1/4 pi eo)e^2/r - (1/4pi eo) e^2/r

= -(1/2) (1/4 pi eo) e^2/r

Energy at R =-(1/2) (1/4 pi eo) e^2/R
Energy at 2R = -(1/2)(1/4 pi eo) e^2/2R
Because of - sign, the total energy is higher at the bigger radius by a factor of 2

Answer 2

The potential energy goes up by more than the kinetic energy goes down.