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Photon, Wave/Particle Nature of light questions

  1. May 6, 2007 #1
    I have a lot of quesitons relating to Photons and the Wave/Particle Nature of light. Answers to any one of my questions would be appreciated! =] Thanks in advance.

    -Can you define photons as particles of EM radiation? or "wave bundles of EM radiation?" Are photons "matter waves," wave particles, or particles that travel in waves? According to the equation: [tex]\lambda=\frac{h}{mv}[/tex]
    would all particles (like gas particles) have a wave nature and therefore a wavelength related to its momentum? Can you elaborate on this? Do all solids have a wave nautre also? Are there any everyday phenomena that can exemplifly this?

    -When producing a photoelectron, do the electrons actually "absorv" the photon and its energy or does the photon collide with the electron and transfer some of its energy to the electorn upon collision? (an example would be the compton shift) Is the colision perfectly ellastic?

    -What is the significance of the Compton wave length?

    -How are [tex]p=E\c \; and \; p=mv[/tex] related? Can someone derive or explain the concepts behind the first equation of momentum?

    Thanks!
     
  2. jcsd
  3. May 7, 2007 #2

    Astronuc

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    Please show some work and try to answer the questions oneself.

    Yes. Obviously photons are EM radiation, and photons have both wave and particle attributes. The particle nature is reflected in the quantized nature, i.e. we can ascribe a particular energy to a photon.

    Particles have a wave attribute - wavelength - given by [tex]\lambda=\frac{h}{mv}[/tex] - which is sometimes called the deBroglie wavelength.

    Please review these pages

    http://hyperphysics.phy-astr.gsu.edu/hbase/debrog.html

    http://hyperphysics.phy-astr.gsu.edu/hbase/davger.html
     
  4. May 7, 2007 #3
    Thanks a lot. I'll look into the pages you've provided. :smile:
     
  5. May 7, 2007 #4
    All particles have wavelike properties, but usually the mass and momentum are so large that the DeBrogile wavelength is too small to be detectable by any means (smaller than Plank length). The DeBrogile wavelength of certain electrons have been measured and confirmed experimentally
     
  6. May 8, 2007 #5
    So does it mean that all objects, for example my eraser, have wave natures and a "de Broglie" wavelength too? I don't get why they have these wavelengths. Do the particles that compose the piece of matter oscillate or something?? I know that the wavelengths are small but why do they even have them?? It seems very obscure to connect wave like attributes to soldis, like my eraser.

    According the hyperphysics: If you explore the wavelength values for ordinary macroscopic objects like baseballs, you will find that their DeBroglie wavelengths are ridiculously small. Comparison of the power of ten for the wavelength will show what the wavelengths of ordinary objects are much smaller than a nucleus. The implication is that for ordinary objects, you will never see any evidence of their wave nature, and they can be considered to be particles for all practical purposes.
     
  7. May 8, 2007 #6
    Everybody knows the equation E=mc^2, but les people know this actually only holds for particles at rest, i.e. with momentum p zero. The general relation between mass, energy and momentum from special relativity is

    [tex]E^2=m^2 c^4 + p^2 c^2[/tex]

    which can easily seen to be yielding E=mc^2 for p=0.

    Now for particles without mass, the first term on the right hand side vanishes and yields the relation E=pc, for e.g. photons, which are massless.

    Finally from this relation

    [tex]E=\sqrt{m^2 c^4 + p^2 c^2}=mc^2 \sqrt{1+\frac{p^2 c^2}{m^2 c^4}}[/tex]

    Now for particles with small velocities we can Taylor expand this to yield

    [tex]E \approx mc^2(1+\frac{p^2}{2m^2c^2})=mc^2 + \frac{p^2}{2m}[/tex]

    which is probably fimiliar to you. It states that the energy of a particle with small velocity compared to c is the sum of a constant 'rest' energy mc^2 and a 'kinetic' energy p^2/2m which yields with p=mv just a kinetic energy of (1/2)mv^2! So the definition of momentum p=mv for small velocities yields the correct (observed) kinetic energy.
     
    Last edited: May 8, 2007
  8. May 8, 2007 #7
    Note also that the correct relativistic definition of energy and momentum are

    [tex]E=\frac{mc^2}{\sqrt{1+\frac{v^2}{c^2}}}[/tex]
    [tex]p=\frac{mv}{\sqrt{1+\frac{v^2}{c^2}}}[/tex]

    which can easily be seen to yield the relation between energy momentum and mass I stated earlier. Moreover for small velocities the square root factor is approximately unity and the familiar relations for energy and momentum remain.
     
    Last edited: May 8, 2007
  9. May 8, 2007 #8

    Hootenanny

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    With one caveat, these definitions are only valid for massive particles. The general relationship [itex]E^2=m^2 c^4 + p^2 c^2[/itex] is valid for all particles. It is therefore, usually more preferable to define energy and momentum using this general expression.
     
  10. May 8, 2007 #9
    Oh okay, this is making more sense to me. So photons don't have a "rest mass" because they are never at rest?
     
  11. May 8, 2007 #10

    Hootenanny

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    You've got the right idea. I'm not brushing you off with a hand wave, but we have an article about this; FAQ: Do Photons Have Mass? with an more extended version available at the bottom of the article.
     
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