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Maximum photon energy in the photoelectric effect

  1. Jul 1, 2007 #1
    Hi I'm a materials engineer and we have a subject about the optical and electrical properties of materials. In this subject I was asked a challenging question. I hope this is the right section for it.

    We were asked why gamma rays do not exhibit true absorbance in metals whereas ultraviolet rays do. I answered that gamma rays had too much energy and were above the photoelectric energy band in metals whereas ultraviolet rays were within the photoelectric band. If I remember correctly this got me full marks but it did not seem right because surely the electrons would just be ejected faster and faster as the photon energy gets higher and higher. I continued to think about this problem until the other day I realised what if the electrons would need to be ejected faster then the speed of light. This is the calculation I came up with (sorry if it's a bit messy I'm not sure what all the right characters are);

    1) E(kinetic) = 0.5*m(electron)*V(electron)^2
    2) but there is a certain amount of energy from the photon associated with destabilising the original electron bond as opposed to new kinetic energy.
    3) E(kinetic) = 0.5*m(electron)*V(electron)^2 + E(bond)
    4) E(photon)=h(constant)F(photon)
    5) max v ~ 3x10^8 m/s
    6) m(electron) ~ 9.11*10^-31 kg
    7) h(constant) = 6.626x10^-34 JS
    8) therefore f(max, photon)= [0.5*m(electron)*V(max, electron)^2 + E(bond)] / h(constant)
    9) By considering equation 8 and by looking at the ionisation energy of certain element I decided to assume that E(bond) was negligible in comparison to E(photon, max)
    10) I got a result of E(photon, max) of 6.2*10^19 Hz, which I think is mid X-rays. This is in line with the result I was expecting.

    I was exited by this result so I went to talk to one of my buddies who was a physicist. He said he didn't really know but he expected there would be no maximum due to relativistic effects. I argued that it's not like the electron is accelerating to the speed of light the photon disappears and the electron must immediately embody all the energy of the original photon. So he suggested I used a forum.

    The questions I have for you guys are;
    Q1) Have I completely misinterpreted the question and in fact there is no maximum energy for the photoelectric effect.
    Q2) Is my calculation above close to correct, if not why not and why is there this maximum?
    Q3) Feel free to make any other comments about my thinking.

    P.S. Sorry this is a bit wordy isn't it.
  2. jcsd
  3. Jul 2, 2007 #2


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    Electrons is never emitted with a speed faster than the speed of light.
    When their kinetic energy is approaching and exceeding the rest mass energy of an electron, we must take account for the relativistic effects. Even tough the electron is not accelerated, the relativistic effects is the right thing. Nowere in the theory of special relativity is there a theorem that states that a body must accelerate for these effects to occur.
  4. Jul 2, 2007 #3


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    You might also consider beta decay of atomic nuclei.
  5. Jul 8, 2007 #4
    The problem goes away if you use relativistic (as opposed to classical) kinetic energy:

    [ (1 - (v/c)^2)^(-1/2) - 1] m c^2.

    You will see that if you try to solve for the velocity from this, you will always have something less than c, so arbitrarily large energies are permissible.

    This does of course reduce to the classical energy for small v. If (v/c)^2 is small, you can use the binomial approximation on the first part to get:

    [1 + 1/2(v/c)^2 - 1] mc^2 = 1/2 (v/c)^2 m c^2 = 1/2mv^2
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