- #1
UrbanXrisis
- 1,196
- 1
as the radius of an electron orbit increases, the total energy of the electron decreases because of the relationship: [tex]E=\frac{-Z^2E_o}{n^2}[/tex] 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:
[tex]\frac{1}{\lambda}=R[\frac{1}{n_1^2}-\frac{1}{n_2^2}][/tex]
[tex]\frac{1}{\lambda}=1.0973731x10^7[\frac{1}{1}-\frac{1}{infinity}][/tex]
[tex]\lambda=91.12667nm[/tex]
so E=hc/91.12667nm=13.605698eV
is this correct?
if I was asked to find the energy of the shortest wavelength photon that can be emitted by a hydrogen atom, I would use:
[tex]\frac{1}{\lambda}=R[\frac{1}{n_1^2}-\frac{1}{n_2^2}][/tex]
[tex]\frac{1}{\lambda}=1.0973731x10^7[\frac{1}{1}-\frac{1}{infinity}][/tex]
[tex]\lambda=91.12667nm[/tex]
so E=hc/91.12667nm=13.605698eV
is this correct?