# Energy of proton

1. Feb 14, 2006

### UrbanXrisis

as the radius of an electron orbit increases, the total energy of the electron decreases because of the relationship: $$E=\frac{-Z^2E_o}{n^2}$$ right? does this means that the electron's kinetic energy would also decreas too?

if I was asked to find the energy of the shortest wavelength photon that can be emitted by a hydrogen atom, I would use:

$$\frac{1}{\lambda}=R[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$$
$$\frac{1}{\lambda}=1.0973731x10^7[\frac{1}{1}-\frac{1}{infinity}]$$
$$\lambda=91.12667nm$$

so E=hc/91.12667nm=13.605698eV

is this correct?

2. Feb 14, 2006

### SpaceTiger

Staff Emeritus
Are you using the Bohr model? If so, keep in mind that it's a classical approximation, not a correct model of the atom. Within the context of the model, however, you first statement isn't quite right. As you move to larger radius (larger n), the total energy becomes less negative -- that is, increases. The kinetic energy does, however, decrease.

That's correct. This is also known as the "ground state energy" of hydrogen.

Last edited: Feb 14, 2006