Posted by **jude** on Sunday, April 4, 2010 at 5:59pm.

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?

- physics -
**Damon**, Sunday, April 4, 2010 at 6:22pm
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

- physics -
**Damon**, Sunday, April 4, 2010 at 6:24pm
The potential energy goes up by more than the kinetic energy goes down.

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