# Basic quantum physics, Balmer series

1. Sep 14, 2011

### Quaint

1. The problem statement, all variables and given/known data
The first five ionization energies of hydrogen are 1312, 328, 146, 82 and 52 kJ/mol. Calculate the wawelength of the first three in Balmer series.

2. Relevant equations
Avokadro constant, NA = 6,022 x 1023 /mol
Planck constant, h = 6,626 x 10-34Js
speed of light, c = 2,998 x 108ms-1
energy, E in J
wavelength, lambda in nm = 10-9m
frequency of radiation, v in s-1

E = hv
c = lambda x v

3. The attempt at a solution
So, in my knowledge you only need the first three energies given in problem and the energies represents the value of one mole of photons. So dividing the 1312 with Avokadro you will get the energy of one photon which is 2.17868 x 10-21J

Using the equation E = hv you will get v = 3,28807 x 1012s-1
and using c = lambda x v you will get lambda = 9,1178 x 10-5nm

So the exponent is definitely wrong but so are numbers because this wavelength should be visible to human eye (400-780nm). You can solve this other way when you are not using the given energies and the rydberg value is present in the equation but this should be another way to do this.
Thx beforehand

2. Sep 14, 2011

### vela

Staff Emeritus
One suggestion I have is that you get comfortable working with the electron-volt as a unit of energy. The conversion is 1 eV = 1.6x10-19 J. It makes the numbers quite a bit easier to deal with.

Here's another trick that can save you some time. The combination hc is approximately 1240 nm eV, so the energy and wavelength of a photon are related by$$E = \frac{hc}{\lambda} = \frac{1240~\mathrm{nm~eV}}{\lambda}$$

Regarding the actual problem, you want to look what the definition of the Balmer series is. Also, how are the ionization energies related to the energy of the electron states?