Thread: Gamma rays - need refreshing for exam

  1. #1
    Hail to the king, baby. Akkernight's Avatar
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    Gamma rays - need refreshing for exam

    hello!

    I'm having physics exam tomorrow and need refreshing on the topic.
    I've tried looking through the wikipedia ( Gamma ray - Wikipedia, the free encyclopedia ) and my school papers, yet it isn't fully clear.

    What is the reason of a gamma ray's existance? Like, how does a gamma ray get unleashed? Where does it come from? As for now I understand it to be the cracking of an atom, but I ain't fully sure...

    So please refresh my mind on this topic and thanks in advance.
    Currently research OpenGL

  2. #2
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    Gamma rays are higher energy light, basically the next level in the list of infrared, visible, ultraviolet, xray.

    They are not, directly, created by fission or fusion ("cracking/combining atoms"), although they often occur in combination with that because the energy released by the atom causes such high excitement in the electrons that they jump like crazy, "spewing photons" everywhere. But it's really like any other light-generation: atoms jumping between their levels in the atom. Just that they jump MORE.

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  3. #3
    Hail to the king, baby. Akkernight's Avatar
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    So am I gettin' this right now... When an e- moves between e- 'borders' it changes its energy consumption and thus when moving back to the correct e- 'border' it has too much energy or e- energy which it unleashes, this is called a photon.
    This photons are then waves with different wave lengths that are based of the unleashed e- energy. If the energy is powerful/high enough, it's an gamma ray?
    Currently research OpenGL

  4. #4
    Banned ಠ_ಠ's Avatar
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    Quote Originally Posted by Akkernight View Post
    So am I gettin' this right now... When an e- moves between e- 'borders' it changes its energy consumption and thus when moving back to the correct e- 'border' it has too much energy or e- energy which it unleashes, this is called a photon.
    This photons are then waves with different wave lengths that are based of the unleashed e- energy. If the energy is powerful/high enough, it's an gamma ray?
    close enough
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  5. #5
    Guest Sebastiani's Avatar
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    Basically, yes. There are other way that gamma rays could be generated, such as rapidly rotating electromagnetic fields, etc. Also, as the special theory of relativity points out, although light in vaccuo travels a constant speed regardless of the reference frame, light originating from an accelerated reference frame would be measured as either a higher or lower frequency (depending on whether the other reference frame was travelling toward or away from you, respectively) than it would be from the originating reference frame. In other words, a microwave originating from a reference frame accelerated towards you could possible be shifted into a gamma wave!
    Code:
    #include <cmath>
    #include <complex>
    bool euler_flip(bool value)
    {
        return std::pow
        (
            std::complex<float>(std::exp(1.0)), 
            std::complex<float>(0, 1) 
            * std::complex<float>(std::atan(1.0)
            *(1 << (value + 2)))
        ).real() < 0;
    }

  6. #6
    spurious conceit MK27's Avatar
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    Quote Originally Posted by Sebastiani View Post
    In other words, a microwave originating from a reference frame accelerated towards you could possible be shifted into a gamma wave!
    This is why you shouldn't stare into the sun while sprinting.

    Good luck Akkernight. If you find a question you can't answer, just put "It is because of superstrings."
    C programming resources:
    GNU C Function and Macro Index -- glibc reference manual
    The C Book -- nice online learner guide
    Current ISO draft standard
    CCAN -- new CPAN like open source library repository
    3 (different) GNU debugger tutorials: #1 -- #2 -- #3
    cpwiki -- our wiki on sourceforge

  7. #7
    Hail to the king, baby. Akkernight's Avatar
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    ok, I think I'm feeling a bit confident in this now... Also, when something casts a alpha-ray, it's bound to have a 5% chance of casting a gamma-ray? there are no exceptions?
    Currently research OpenGL

  8. #8
    Guest Sebastiani's Avatar
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    I'm not sure what the overall emission probablility is, but an "alpha-ray", as you call it, is simply a helium nuclei and is released in many different processes, and whether or not it emits a gamma wave just depends on how energetic the event is.
    Code:
    #include <cmath>
    #include <complex>
    bool euler_flip(bool value)
    {
        return std::pow
        (
            std::complex<float>(std::exp(1.0)), 
            std::complex<float>(0, 1) 
            * std::complex<float>(std::atan(1.0)
            *(1 << (value + 2)))
        ).real() < 0;
    }

  9. #9
    Officially An Architect brewbuck's Avatar
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    On this planet, the only source of gamma rays is radioactive nuclei. You can divide the EM spectrum crudely into three regimes:

    Low energy: Radio waves, infrared, visible, ultraviolet
    Medium energy: X-rays
    High energy: Gamma rays

    Low energy waves are produced by oscillating EM fields, and by electron quantum state changes.

    Medium energy waves are produced by sudden acceleration or deceleration of electrons, for instance, by crashing high speed electrons into a chunk of metal.

    High energy waves cannot easily be produced. They are naturally generated by relaxations of the quantum state of atomic nuclei. These energies are beyond anything we can easily generate in a controlled manner (you won't see a "gamma ray laser" any time soon).

    Out in the universe, gamma rays can also be generated by extreme acceleration of electrons, for instance near a black hole, or during the explosion of a supernova.

    EDIT: Another important point is that the higher the energy, the less "wave-like" and the more "particle-like" the radiation becomes. All EM radiation is composed of photons, but the particle nature of these emissions is only apparent at higher energies or when using extremely sensitive equipment. A gamma particle is so high energy that its wave-like character is hard to measure, and it acts more like a bullet than a wave.
    Last edited by brewbuck; 06-03-2009 at 10:00 AM.
    Code:
    //try
    //{
    	if (a) do { f( b); } while(1);
    	else   do { f(!b); } while(1);
    //}

  10. #10
    Guest Sebastiani's Avatar
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    >> Another important point is that the higher the energy, the less "wave-like" and the more "particle-like" the radiation becomes.

    I disagree. I would argue that the distinction is simply a matter of convention (ie: what effect is being studied) and also applies equally well to matter (the orbits of electrons around an atom, for instance, can be explained by simply interpreting the electron as a wave). Another example is the photo-electric effect, which can be stimulated by frequencies much lower than gamma rays, which can be explained by adopting a particle-centric viewpoint.

    >> On this planet, the only source of gamma rays is radioactive nuclei.

    The difference between a gamma wave and an x-ray is really just a matter of origin. Often, the frequencies are equivalent.
    Last edited by Sebastiani; 06-03-2009 at 10:26 AM. Reason: clarity
    Code:
    #include <cmath>
    #include <complex>
    bool euler_flip(bool value)
    {
        return std::pow
        (
            std::complex<float>(std::exp(1.0)), 
            std::complex<float>(0, 1) 
            * std::complex<float>(std::atan(1.0)
            *(1 << (value + 2)))
        ).real() < 0;
    }

  11. #11
    Officially An Architect brewbuck's Avatar
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    Quote Originally Posted by Sebastiani View Post
    I disagree. I would argue that the distinction is simply a matter of convention (ie: what effect is being studied) and also applies equally well to matter (the orbits of electrons around an atom, for instance, can be explained by simply interpreting the electron as a wave). Another example is the photo-electric effect, which can be stimulated by frequencies much lower than gamma rays, which can be explained by adopting a particle-centric viewpoint.
    The photo-electric effect is what I was referring to when I said "sensitive equipment." Without deliberately looking for it, you will not perceive the particle nature of low-energy light.

    I side with Feynman. Light is made of particles, not waves. The wave-like behavior is explained by the phasors of quantum field theory. But it isn't some strange thing that's both a wave and a particle -- it's a particle, period.

    (If you doubt that Feynman took such a strong stance on the issue, read through his book "The strange theory of light and matter")

    EDIT: Also, the original explanation of the energy levels of hydrogen in terms of an integral number of electron wavelengths "fitting" around the orbit came up with the right values, but only by coincidence. That explanation is not considered correct under modern quantum physics. It hinges on the concept of an "orbital radius" which is nonsense. It does say something about the beauty of physics, though, that you can come up with the right answer even if you use the wrong method
    Last edited by brewbuck; 06-03-2009 at 11:37 AM.
    Code:
    //try
    //{
    	if (a) do { f( b); } while(1);
    	else   do { f(!b); } while(1);
    //}

  12. #12
    Guest Sebastiani's Avatar
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    >> I side with Feynman. Light is made of particles, not waves.

    Interesting. I would lean toward the exact opposite - that both photons and matter are essentially waves.

    >> Also, the original explanation of the energy levels of hydrogen in terms of an integral number of electron wavelengths "fitting" around the orbit came up with the right values, but only by coincidence.

    I think it applies to all atoms (DeBroglie's equations).

    Louis de Broglie tried to expand on Bohr's ideas, and he pushed for their application beyond hydrogen. In fact he looked for an equation which could explain the wavelength characteristics of all matter. His equation was experimentally confirmed in 1927 when physicists Lester Germer and Clinton Davisson fired electrons at a crystalline nickel target and the resulting diffraction pattern was found to match the predicted values.
    >> That explanation is not considered correct under modern quantum physics.

    Assuming that modern quantum physics is correct, maybe (although I don't believe that it is).
    Last edited by Sebastiani; 06-03-2009 at 12:27 PM.
    Code:
    #include <cmath>
    #include <complex>
    bool euler_flip(bool value)
    {
        return std::pow
        (
            std::complex<float>(std::exp(1.0)), 
            std::complex<float>(0, 1) 
            * std::complex<float>(std::atan(1.0)
            *(1 << (value + 2)))
        ).real() < 0;
    }

  13. #13
    Officially An Architect brewbuck's Avatar
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    Quote Originally Posted by Sebastiani View Post
    I think it applies to all atoms (DeBroglie's equations).
    DeBroglie's equations are certainly valid, but applying them to electron orbitals to derive their energy levels isn't. It happens to work for hydrogen because the potential field is so simple. If you tried the same thing with a multi-electron atom, you'd get answers which are slightly different from reality (or maybe not so slightly, depending on the atom).

    Assuming that modern quantum physics is correct, maybe (although I don't believe that it is).
    Me neither, but I don't believe a theory can be "correct," it can only make correct predictions in all known circumstances. A theory can definitely be incorrect if it makes a wrong prediction.
    Code:
    //try
    //{
    	if (a) do { f( b); } while(1);
    	else   do { f(!b); } while(1);
    //}

  14. #14
    Guest Sebastiani's Avatar
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    >> If you tried the same thing with a multi-electron atom, you'd get answers which are slightly different from reality (or maybe not so slightly, depending on the atom)

    Hmm. Well, ok. I stand corrected then.

    >> Me neither, but I don't believe a theory can be "correct," it can only make correct predictions in all known circumstances. A theory can definitely be incorrect if it makes a wrong prediction.

    Good point. And as far as giving correct predictions, modern quantum theory succeeds. But the insistance of some to attribute some sort of reality to probability and randomness, instead of seeing that it is just a necessary mathematical generalization in order to compensate for the fact that we simply can't factor in all of the variables involved is the biggest mistake made by many proponents of quantum theory.
    Code:
    #include <cmath>
    #include <complex>
    bool euler_flip(bool value)
    {
        return std::pow
        (
            std::complex<float>(std::exp(1.0)), 
            std::complex<float>(0, 1) 
            * std::complex<float>(std::atan(1.0)
            *(1 << (value + 2)))
        ).real() < 0;
    }

  15. #15
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    One question I asked my high school science teacher once was whether radio waves can kill someone. My theory was that since Gamma & X-rays & microwaves can kill you, as well as highly focused Ultra Violet, Visible, & Infrared light (lasers) can kill you, then why not also highly concentrated radio waves? He thought I was nuts.
    "I am probably the laziest programmer on the planet, a fact with which anyone who has ever seen my code will agree." - esbo, 11/15/2008

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