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B Photoelectric effect and atomic excitation

  1. Oct 10, 2016 #1
    A few quick questions I'd like cleared up:

    1) Alkali metals are said to have a really low threshold energy, enough for visible light to cause the photoelectric effect. Does this mean if I aim a flashlight (turned on) at a piece of sodium, I could ionise it? Simply flashing a light over a piece of alkali metal could give it a positive charge? Surely they'd always be positively charged then?

    2) It doesn't have to be IR-UV radiation does it? Not an 'actual colour' all the time, right? X-rays and gamma rays are more capable of the photoelectric effect, correct?

    3) An electron dropping back to its ground state after excitation releases a photon of equal energy to the energy level difference (energy it gained from another electron or photon). Why is it I commonly read when waves are absorbed, they are re-emitted at a longer wavelength (therefore lower frequency)? E.g: Greenhouse effect.

    4) When we see an absorption spectrum, can the photons only be absorbed by gases or do the black lines also account for photons absorbed by liquids/solids also in the way?

    5) How can an electron emit more than one photon when relaxing?

    6) In fluorescent tubes, do the electrons release UV photons because of how high the energy from the incident electrons is?

    Much appreciated.
  2. jcsd
  3. Oct 10, 2016 #2

    Charles Link

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    For an answer to your first question, I think the experiment with visible light is best performed by having a positive voltage difference between a collecting plate and the metal that is being illuminated with visible light. Otherwise the electron as it starts to leave the surface will induce a positive charge around it and get pulled back to the surface. And also, once the surface gets slightly positively charged, additional electrons would find it more difficult to leave the surface. The photons of visible light have an energy of only 2-3 eV, so that they would never be able to escape from the surface when illuminated with visible light if the voltage got as high as +3 volts. I'm no expert at the photoelectric effect, but I think I successfully answered your first question. ## \\ ## For question 6, I think you need to get the electrons accelerated across a sufficient voltage in the fluorescent tube to ionize and also excite the atoms of the gas in the tube. (The arc would quickly become extinguished if the electrons didn't have sufficient energy to ionize additional atoms. Oftentimes in a fluorescent tube you have a start-up potential of a couple thousand volts, but once the arc gets established you can run it on 15 volts or thereabouts.) Upon recombination and/or dropping down to the ground or lower states, many of the transitions are in the uv range. If you look up ionization potentials for some of the noble gases, their ionization potentials are in the neighborhood of 12-25 volts which result in uv photons for any subsequent recombinations of an electron with the ion. The purpose of the fluorescent coating is to convert uv photons to visible light.
    Last edited: Oct 10, 2016
  4. Oct 20, 2016 #3
    I can settle for 1 and 6, though 2 - 5 still remain. These are relatively quick to answer, I just need a brief explanation and/or confirmation on whether I am correct or not.

    Many thanks.
  5. Oct 20, 2016 #4

    Charles Link

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    I can give a simple answer to #3. When visible light from the sun is absorbed by the earth's surface, it serves to raise the temperature of the earth so that more infrared (blackbody type radiation) is emitted. Much of this is in the 5-20 micron wavelength region of the spectrum. (Peak wavelength of radiated blackbody spectrum is given by Wien's law ## \lambda_{peak} T_{kelvin}=2898 \, microns \,deg \,K ##. Temperature of earth's surface is approximately T=300 Kelvin). ## \\ ## The greenhouse effect is caused by greenhouse gases ( ## CO_2 ## and chlorofluorocarbons, etc., you can google this), that have absorption bands in this region of the spectrum. Without the greenhouse gases, a higher percentage of IR would get radiated into space, thereby allowing for additional radiative cooling. ## \\ ## Editing...In addition, I think you have #2 and # 4 correct, and #5 can certainly occur by an electron making a couple of successive transitions to the ground state. For example, for a hydrogen atom, going from 3s to 2p and then from 2p to 1s. Some transitions are forbidden by the selection rules=you can google the details. (e.g. A quick google shows ## \Delta L=\pm 1 ## is required for a transition.)
    Last edited: Oct 20, 2016
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