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How does the line spectra work?

  1. Nov 16, 2005 #1
    i'm about to tear off all the hair on my head, learning about line spectra...

    i got the lyman,balmer and paschen series in my text book and the formula for it. but how do i apply it in a physics question?

    like for example if i have an electron that is excited to the n = 6 state, which formula do i apply? or when the question ask me wat is the longest wavelength that can be used to excite the ground state of hydrogen atom?

    better yet, can anyone explain plainly wat is it all about? my textbook does a poor job of that...or maybe my brains aint working too well
    Last edited: Nov 16, 2005
  2. jcsd
  3. Nov 16, 2005 #2


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    um um um um um. All the answer to these questions you asked are found not in any of the Lyman, Balmer or Orpaschen series, but simply in the Bohr postulates for the hydrogen atom.

    An electron in orbit can only have certain values of energy. Those are noted by [itex]E_n[/itex] and I'm sure you can find the formula in your book. When the electron's energy is [itex]E_1[/itex], we say that the electron is in ground state. The electron can jump from state to state by absorbing or emitting a photon (light).

    When an electron is in a state of energy [itex]E_m[/itex], and it absorbs a photon of energy corresponding to the difference btw its state and a state of higher energy, say [itex]hf = E_n - E_m[/itex], it will jump to state [itex]E_n[/itex].

    On the contrary, if an electron is currently in state [itex]E_m[/itex], it stay there for a little while, and then it will emit a photon of energy corresponding to the difference btw its state and a state of lower energy, say [itex]hf = E_m - E_i[/itex], then jump to state [itex]E_i[/itex].
    Last edited: Nov 16, 2005
  4. Nov 16, 2005 #3
    When an electron from an axcited state 'falls' to another state with a lower energy this difference in energy is emitted in the form of a photon. The more energy E is released the higher the frequency [itex]\nu[/tex]

    [tex] E=h \nu[/tex]

    With h, Plancks constant.
    Now, this energy has to do with the interaction between the electron and the nucleus and in case of hydrogen can be written

    [tex] E_n = -\frac{E_0}{n^2} [/tex]

    for the nth excited state ([itex]E_0[/tex] is the ground state (lowest) energy). Notable is the minus sign, the energy of the electron is negative, you have to add energy ([itex]E_0[/tex] for the ground state e.g.) by firing a photon at the atom for example to release the electron from it's 'bound state'.

    Another notable thing is the dependence on n, the energy is inversely related to the square of n. The relation can be found both using quantum mechanics or Bohr's model.

    The energy released from an electron 'falling' from [itex]n=n_i[/tex] to[itex]n=n_f[/tex] is ofcourse

    [tex] E = E_f-E_i = E_0(\frac{1}{n_i}-\frac{1}{n_f}) [/tex]

    Using the relation between the energy of a photon and its frequency, you can calculate the frequency to be

    [tex] \nu = E/h = \frac{E_0}{h}(\frac{1}{n_i}-\frac{1}{n_f}) [/tex]

    Using the relation between the frequency, wavelength and velocity (c, the speed of light) of a wave (a photon is a wave in some way...)

    [tex]c=\nu \lambda[/tex]

    we can rewrite the expression in terms of the wavelength of the emitted photon

    [tex] \frac{1}{\lambda} = \nu / c = \frac{E_0}{hc}(\frac{1}{n_i}-\frac{1}{n_f}) = R_H }(\frac{1}{n_i}-\frac{1}{n_f}) [/tex]

    Where the new constant R_H is called the Rydberg constant but is just equal to [itex]\frac{E_0}{hc}[/tex]. Hope this helps...
  5. Nov 18, 2005 #4
    thanks everyone.
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