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Homework Help: Photon energy, frequency, wavelength

  1. Apr 27, 2007 #1
    1. The problem statement, all variables and given/known data

    I'm quite stuck on several parts of this problem and I've been trying it for a while... I've looked up a lot of equations and attempted a lot of work but I am still very unsure. I would really just like someone to give any hint (but not answer) as to where my difficulties are so I can try to understand my homework problem better.

    A free electron with negligible kinetic energy is captured by a stationary proton to form an excited state of the hydrogen atom. During this process a photon of energy Ea is emitted, followed shortly by another photon of energy 10.2 eV. No further photons are emitted. The ionization energy of hydrogen is 13.6 eV.

    a. Determine the wavelength of the 10.2 eV photon
    b. Determine the following for the first proton emitted.
    bi. The energy Ea of the photon
    bii. The frequency that corresponds to this energy
    c. The following diagram shows some of the energy levels of the hydrogen atom, including those that are involved in the processes described above. Draw arrows on the diagram showing only the transitions involved in these processes.
    The diagram looks exactly like this http://spiff.rit.edu/classes/phys301/lectures/comp/hyd_levels.gif
    d. The atom is in its ground state when a 15 eV photon interacts with it. All the photon's energy is transferred to the electron, freeing it from the atom. Determine the following:
    di. The kinetic energy of the ejected electron
    dii. The de Broglie wavelength of the electron

    2. Relevant equations
    E = hf
    Hf = KE + W
    p = h / wavelength
    (Compton effect) Wavelength' - wavelength = (h/mc)(1-cos(theta))
    Ei-Ef = hf
    1/wavelength = 1.097 x 10^7 m^-1(Z^2)(1/n_f^2 - 1/n_i^2)

    3. The attempt at a solution
    a. I first put the ev into joules
    (10.2 eV)(1.6 x 10^-19 J/ev) = 1.632 x 10^-18 J
    Then used Hf = KE (0) + W
    f = W/h = 1.632 x 10^-18 J / 6.63 x 10^-34 J-s = 2.462 x 10^15 s^-1
    wavelength = c/f = 3 x 10^8 m/s / 2.462 x 10^15 s^-1 = 1.219 x 10^-7 m = 121.875 nm

    bi. I thought that because there are only 2 photons emitted, I could use Ei-Ef = hf where Ei is the first photon and Ef is the second. Therefore:
    Ea - 1.632 x 10^-18 J = (6.63 x 10^-34 J-s)(2.462 x 10^15 s^-1)
    Ea = 3.26 x 10^-18 J = 20.4 eV
    But then I realized this can't be possible... and I'm confused as to why. The only other thing I can think of is that Ei is the ionization energy and that the difference between it and Ef is the Ea energy I'm looking for, in other words 13.6 eV - 10.2 eV = 3.4 eV ... which would correspond on the energy diagram in problem c.

    bii. I can do this part easily once I know the answer to part bi

    c. I just draw arrows describing the transition processes above... which would mean to the -3.4 mark from the -13.6 mark .. I don't believe any other processes were involved... though I'm not entirely sure based on my work above

    di. If a 15 Ev photon interacts with a 13.6 eV ground state atom, then the difference is 1.4 eV which is close to n = 3 level. In that case, am I allowed to do hf = KE + W where W = 1.4 eV (converted to J of course), and f is derived from the equal 1/wavelength = 1.097 x 10^7 m^-1)(1^2)(1/1^2 - 1/3^2)... where you get the wavelength and then just do c/f = wavelength to solve for f?
    dii. I can do this if I know the frequency from di using wavelength = h/p and p = hf/c

    I know this is lengthy.. but I appreciate any help. Even just for taking the time to read this means a lot :redface:
  2. jcsd
  3. Apr 28, 2007 #2


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    Homework Helper

    this part is ok

    I believe the picture is the following:

    during the capture: energy is released (because energy is needed to break them apart again.. to overcome the binding energy)...
    you will find that:
    [tex]\text{rest Energy of electron} + \text{rest Energy of Proton} > \text{rest energy of hydrogen}[/tex]

    the energy difference is the binding energy which is released during the capture

    after the electron was captured, it underwent de-excitation, going from n=2 to n=1 level hence the emitted 2nd photon has energy of 10.2ev (exactly the band gap you mentioned)

    13.6eV is the energy needed to ionised the hydrogen atom, ie setting the electron free from its orbit... so the remaining energy is the KE of the electron.
  4. Apr 28, 2007 #3
    For part b:
    Okay.. I can see how the photon went from n=1 to n=2. In that case, I plugged it into the last equation I mentioned and got the wavelength to be 1.823 x 10^-7 m or 182 nm. Converting this to frequency, I got 1.6455 x 10^15 Hz and therefore the energy is E = hf or 1.091 x 10^-18 J or 6.82 eV... which actually makes some sense to me. I notice that 6.82 eV is almost double 3.4 eV.. is there any connection?

    For part d:
    The kinetic energy is the remaining energy after the 13.6 eV of energy is needed to ionize the H atom... so it's simply 1.4 eV (15 - 13.6)?
    If so... how do you find the work involved? Or is there no work at all?

    (Thanks a lot for your help ^^. I feel like I'm understanding this a little more)
  5. Apr 28, 2007 #4


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    Homework Helper

    what work are you talking about?
  6. Apr 28, 2007 #5
    In the equation hf = KE + W where w is work. I believe it's the work involved in moving the photon from one state to the next.. oo;;
  7. Apr 28, 2007 #6


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    Homework Helper

    it seems to me that the work is the energy needed to liberate the e- from the hydrogen atom, in this case, it should be equivalent to the ionization energy
  8. Apr 28, 2007 #7
    That makes sense.. thank you very much for your help ^_^;
  9. May 4, 2009 #8
    thanks for this im doing it right now!
  10. May 5, 2009 #9
    Just a quick note to say that the W in hf= Ek + W, is the work function.
    This equations is usually associated with photo-electric effect =]
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