1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Basic quantum physics, Balmer series

  1. Sep 14, 2011 #1
    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. jcsd
  3. Sep 14, 2011 #2

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    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[tex]E = \frac{hc}{\lambda} = \frac{1240~\mathrm{nm~eV}}{\lambda}[/tex]

    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?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Basic quantum physics, Balmer series
  1. Balmer series (Replies: 3)

  2. Balmer series (Replies: 12)

  3. Balmer Series (Replies: 1)

Loading...