physics
posted by jude on .
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?

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