Which of the radiation will release greater number of electrons?

Discussion in 'Physics & Math' started by chikis, Jul 11, 2014.

  1. chikis Registered Senior Member

    Messages:
    328
    In a photoelectric emission, a
    faint blue light of wavelenght
    \(5.0\times10^{-7}m\) and a bright
    red light of wavelenght
    \(7.0\times10^{-7}m\) are incident on
    a metal plate. Which of the
    radiation will release: (i)
    greater number of electrons;
    (ii)electrons of higher
    maximum kinetic energy?
    Give reasons for your
    answers.

    Though am not sure about the answer am about to give, I would say that since the wavelenghths of the radiations have different magnititude, and if the electrons emitted is proportional to the magnititude of the wavelenght of radiation, the bright red light having a much greater wavelenght of \(7.0\times10{-7}m\) will release greater number of electrons.

    The same applies to (ii), wavelenght of \(7.0\times10{-7}m\) will release electrons of higher maximum kinetic energy. Am not sure of my answers, please help my poor understing of this concept.
     
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  3. origin Heading towards oblivion Valued Senior Member

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    11,888
    According to the photoelectric effect light is made up of photons and a single photon will cause a single electron to gain enough energy to leave the material.

    So for the electrons to have a higher KE the individual photons must have more energy. So on ii your thinking was right but you made a mistake in which light has more energy. A shorter wavelength has more energy not a longer. So the blue light makes the electrons have a higher KE.

    For the first part of the question the brighter red light means there are MORE photons so there will be more electrons leaving the material but they will have a lower KE than the blue light photons.

    I am pretty sure that is right if not feel free to correct me.
     
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  5. James R Just this guy, you know? Staff Member

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    39,397
    Maximum kinetic energy is given by Einstein's photoelectric equation. Higher frequency of incident light means higher maximum kinetic energy of emitted electrons.

    Number of emitted electrons depends on the intensity of the light. Brighter (more intense) light will release more electrons than dimmer light, provided that electrons are actually emitted (which will only happen if the photon energy is greater than the work function).
     
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  7. chikis Registered Senior Member

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    328
    But looking at this

    Please Register or Log in to view the hidden image!

    THER FUNDAMENTAL LAWS OF PHOTO ELECTRIC EMISSION
    from http://www.citycollegiate.com/physicsXII_17b.htm It looks as if the the bright red light having high intensity and high wavelenght will do both one and (i) and (ii)
     
  8. origin Heading towards oblivion Valued Senior Member

    Messages:
    11,888
    ,

    No it doesn't. The longer wavelength means lower energy so the red light will not give the highest KE to the electrons, as I said in my first post.
     
  9. chikis Registered Senior Member

    Messages:
    328
    You are confusing me by saying yes and no at the same time.


    From the my post, you saw that K.E. Increases with the increase in frequency.
    Based on this, the bright red light having a higher wavelenght should have more energy.
     
  10. origin Heading towards oblivion Valued Senior Member

    Messages:
    11,888
    Frequency and wavelength are inversely proportional to each other. So as frequency increases the wavelength becomes shorter.

    \(\nu = \frac{c}{\lambda}\)
     
  11. chikis Registered Senior Member

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    328
    I was even thinking that frequency and wavelenght are the same thing. Thank you for the clearance.
     
  12. origin Heading towards oblivion Valued Senior Member

    Messages:
    11,888
    No problem sorry I wasn't clearer.
     
  13. dumbest man on earth Real Eyes Realize Real Lies Valued Senior Member

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    3,523
  14. chikis Registered Senior Member

    Messages:
    328
    One more question:
    Is there anything like threshhold wavelenght? If there is, does threshhold wavelenght and threshhold frequency mean the same thing?
     
  15. origin Heading towards oblivion Valued Senior Member

    Messages:
    11,888
    Yes and yes. If the incident photon is below a certain energy the electron that absorbs the photon will not have enough energy to leave the atom. The energy of the photon is expressed in the following equations:

    \(E = h\nu\)

    \(E = {\frac{hc}{\lambda}\)
     
  16. OnlyMe Valued Senior Member

    Messages:
    3,914
    It would seem that as far as the OP is concerned, it is implied that the photocell/metal plate will absorb both wavelengths (equally). That suggests that for the OP there is no threshold involved. The only real questions you answered earlier.

    Comparing, blue and red photons one to one; The blue photons are higher frequency, shorter wavelength and carry more energy. The red photons are lower frequency, longer wavelength and carry less energy. But this only reflects the inherent energy of the photons. How those photons interact with the atoms of the metal plate/photo cell is far more complex.., and may not result in displaced/emitted electrons of any significantly different energies.

    Comparing blue and red beams of light, is different. The red beam being brighter than the blue beam, and composed of more photons, assuming the metal plate/photo cell interacts equally with both wavelengths, the red beam would result in more electrons being displaced/emitted... But each individual electron may or may not have a higher kinetic energy.

    If these were homework questions, I would assume that the correct answer is not based on any measurable difference in electron kinetic energies, which would be indistinguishable and has more to do with the atoms than the particular wavelength (of visible light). Instead it is likely a hypothetical or theoretical question, where all of the photon energy is transferred to an electron in the process. Thus from that perspective the blue photon results in electrons with more kinetic energy than the red photons.., and the bright red beam of photons results in more electrons than the faint blue beam of photons.

    If this was a homework question, there is classroom or course background context that is missing.
     
  17. origin Heading towards oblivion Valued Senior Member

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    11,888
    There is a threshold according to the photoelectric effect and I think that is precisely what the questions were getting at.

    Again the photoelectric effect indicates that an electron cannot be ejected from the orbital until a certain energy per photon is reached, the excess energy will be reflected as an increased KE of the electron.

    The photoelectric effect says that the different energy photons will NOT interact equally.

    I am afraid you would get low marks on your homework!

    The photoelectric effect.
     
  18. brucep Valued Senior Member

    Messages:
    4,098
    He would need to read the chapter on the photoelectric effect to be able to begin to understand this physics. I doubt only me has read much experimental literature. Good way to learn about the natural phenomena being evaluated.
     
  19. origin Heading towards oblivion Valued Senior Member

    Messages:
    11,888
    I remeber that I went over this in my college physics courses, so this is pretty basic stuff. I agree that the beauty of the photoelectric effect is that it is a very easy way to demonstrate quantum phenomena.
     
  20. OnlyMe Valued Senior Member

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    3,914
    First, the only reason I posted anything at all, is that the way I read it, the OP itself eliminated the threshold issue and it seemed as though it was being assumed that the red and blue photons both interact with the same electron energy state. Something that could only be assumed, if the questions were limited to an introductory explanation of the photoelectric effect.

    .... I attempted to qualify my post and comment as in reference to the way the OP was worded.

    The photoelectric threshold is the upper and lower frequency limits that a photo sensitive material will interact with. The OP began with an assumption that the blue and red beams of light were both within that threshold. Thus as far as the OP was concerned threshold was not an issue.

    Beyond that the energy of any ejected photon includes a component associated with its initial energy state and the added energy of the photon. Not all electrons that may interact photoelectrically are the same... The kinetic energy of even A single electron associated with an atom, varies. If you are assuming that within the context of the question, all light interacts with the same electron energy state, then only the energy associated with the photon is important. But is that the case in practice? Without information about the composition of the metal plate and envionment, you cannot know that only electrons of a fixed initial energy state are ejected.

    Is the temperature of the photo emissive plate constant? Or does it change over time.., or even when exposed to a bright beam of light? How does the temperature affect the starting energy level of the - to be ejected electrons? Do both wavelengths interact with electrons occupying the same energy states? In the OP that can be assumed, but the OP leaves a great deal of detail unstated.

    And yes different energy photons will not interact equally. They may not even interact with the same electron energy states. My initial point was that the OP assumes that both beams of light are within the threshold limits, thus threshold is not an issue. The second point was that the kinetic energy of an ejected electron is the product of more than just the wavelength of the photon, that initiates the electron ejection. Gamma rays can even interact with any electron associated with the atom, or even directly with the nucleus.

    As for grades, when I was being graded, which was many decades ago, I never failed anything. I have never even had to take a professional licensing exam twice. And I have taken I think 8 different licensing exams, in four unrelated fields. And I have likely forgotten more than most... And yes I almost always post off the cuff, on the spur of the moment, and frequently fail to convey my original intent efficiently.

    Bruce, it has been a long time since I dealt with anything approaching the specifics of the photoelectric effect.., but 90% of what I do read is research and or theoretical papers... And the kind of stuff I am working through now is not even discussed on forms like this. The text books I read were back in the late 60s, so it is likely I have not read much if any of most of the texts often referenced around here. The choice for me is do I study SED and a bit about quantum gravity or read a text book? It is harder going but I choose the research papers.
     
  21. origin Heading towards oblivion Valued Senior Member

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    Post deleted
     
  22. Walter L. Wagner Cosmic Truth Seeker Valued Senior Member

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    2,559
    Interestingly, Einstein obtained a Nobel Prize for his work on quantum theory, not relativity, and his explanation of the photoelectric effect. http://www.einsteinyear.org/facts/photoelectric_effect/

    He should have gotten a second one for relativity, but it apparently was too controversial at the time.
     
  23. chikis Registered Senior Member

    Messages:
    328
    Since threshhold wavelenght and frequency are not the same.
    If am asked to compute the threshhold wavelenght, I use
    \(E=hƒ=\frac{hc}{\lambda}\)
    If am asked to compute the threshold frequency,
    I use either
    \(E=hƒ-hƒ\omicron \)or
    \(\omega=hƒ\omicron\)

    where
    [/tex]ƒ\omicron[/tex] is the threshold frequency.
    But are these Greek letters called/pronounced \(ƒ, h, c\)?
     

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