Sample Problem
The electron in a hydrogen atom is in the n = 2 state. When it drops to the ground state a photon is emitted. What is the wavelength of the photon? Is this in the visible spectrum?
One way to do this is to first calculate the energy of the electron in the initial and final states using the equation:
| En = |
| -13.6 eV
|  |
| n2
|
|
| E2 = |
| -13.6 eV
| |
| 4
|
|
= -3.4 eV |
| E1 = |
| -13.6 eV
| |
| 1
|
|
= -13.6 eV |
In dropping from the n = 2 state to the ground state the electron loses 10.2 eV worth of energy. This is the energy carried away by the photon.
Converting this to joules gives:
E = 10.2 eV * 1.60 x 10-19 J/eV = 1.632 x 10-18 J
| For a photon E = hf = |
| hc
| |
| l
|
|
| l = |
| hc
| |
| E
|
|
= |
| 6.63 x 10-34 * 3 x 108
| |
| 1.632 x 10-18
|
|
= 1.22 x 10-7 m = 122 nm |
This is in the ultraviolet part of the spectrum, so it would not be visible to us.