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  8. MÅNE LANDING-> skalkeskjul for 3 Verdenskrig ?

    i Naturvitenskap

    Skrevet · Endret av bebes

    [...]atom bombe[...]

     

    Fint det, kanskje du snart skal øse av din uendelige visdom, og vise sammenhengen mellom atombombe, ICBM og månelandinger? :roll:

    The Compton shift formula

    A photon of wavelength \lambda \, comes in from the left, collides with a target at rest, and a new photon of wavelength \lambda ' \, emerges at an angle \theta \,.

    See also: Klein-Nishina formula

     

    Compton used a combination of three fundamental formulas representing the various aspects of classical and modern physics, combining them to describe the quantum behavior of light.

     

    * Light as a particle, as noted previously in the photoelectric effect.

    * Relativistic dynamics: special theory of relativity

    * Trigonometry: law of cosines

     

    The final result gives us the Compton scattering equation:

     

    \lambda' - \lambda = \frac{h}{m_e c}(1-\cos{\theta})

     

    where

     

    \lambda\, is the wavelength of the photon before scattering,

    \lambda'\, is the wavelength of the photon after scattering,

    me is the mass of the electron,

    \theta\, is the angle by which the photon's heading changes (between 0° and 180°),

    h is Planck's constant, and

    c is the speed of light.

     

    \frac{h}{m_e c} = 2.43 \times 10^{-12}\,\text{m} is known as the Compton wavelength. λ' − λ can be between 0 (for θ = 0°) and two times the Compton wavelength (for θ = 180°).

     

    [edit] Derivation

     

    Begin with conservation of energy and conservation of momentum:

     

    E_\gamma + E_e = E_{\gamma^\prime} + E_{e^\prime} \quad \quad (1) \,

    \vec p_\gamma = \vec{p}_{\gamma^\prime} + \vec{p}_{e^\prime} \quad \quad \quad \quad \quad (2) \,

     

    where

     

    E_\gamma \, and p_\gamma \, are the energy and momentum of the photon and

    E_e \, and p_e \, are the energy and momentum of the electron.

     

    [edit] Solving (Part 1)

     

    Now we fill in for the energy part:

     

    E_{\gamma} + E_{e} = E_{\gamma'} + E_{e'}\,

    hf + mc^2 = hf' + \sqrt{(p_{e'}c)^2 + (mc^2)^2}\,

     

    The square of the second equation gives an equation for pe':

     

    p_{e'}^2c^2 = (hf + mc^2-hf')^2-m^2c^4 \quad \quad \quad \quad \quad (3) \,

     

    [edit] Solving (Part 2)

     

    Rearrange equation (2)

     

    \vec{p}_{e'} = \vec{p}_\gamma - \vec{p}_{\gamma'} \,

     

    and square it to see

     

    p_{e'}^2 = (\vec{p}_\gamma - \vec{p}_{\gamma'}) \cdot (\vec{p}_\gamma - \vec{p}_{\gamma'})

    p_{e'}^2 = p_{\gamma}^2 + p_{\gamma'}^2 - 2\vec{p_{\gamma}} \cdot \vec{p_{\gamma'}}

    p_{e'}^2 = p_\gamma^2 + p_{\gamma'}^2 - 2|p_{\gamma}||p_{\gamma'}|\cos(\theta) \,

     

    Energy and momentum of photons are connected by the relativistic equation p_{\gamma}=\frac{E_{\gamma}}{c}, so E_{\gamma}^2=p_{\gamma}^2c^2=(h f)^2.

     

    Therefore, multiplying by c2, we have also

     

    p_{e'}^2c^2 = (h f)^2 + (h f')^2 - 2(hf)(h f')\cos{\theta} \quad \quad \quad (4)

     

    [edit] Putting it together

     

    Now we have the two equations (3 & 4) for p_{e'}^2c^2, which we equate:

     

    \left(h f\right)^2 + \left(h f'\right)^2 - 2h^2 ff'\cos{\theta} = (hf + mc^2-hf')^2 -m^2c^4 \,

     

    Energies of a photon at 500 keV and an electron after Compton scattering.

     

    Next we multiply out the right-hand term (hf + mc2 − hf')2 and cancel square terms on both sides and get:

     

    -2h^2ff'\cos{\theta} = -2h^2ff'+2hfmc^2-2hf'mc^2. \,

     

    Combining two terms, this becomes:

     

    2h^2ff'(1-\cos(\theta)) = 2hfmc^2 - 2hf'mc^2. \,

     

    After dividing both sides by \left(2hff'mc^2\right) , we get:

     

    \frac{h}{mc^2}\left(1-\cos \theta \right) = \frac{1}{f^\prime} - \frac{1}{f} \,

     

    This is equivalent to the Compton scattering equation, but it is usually written in terms of wavelength rather than frequency. To make that switch use

     

    f=\frac{c}{\lambda} \,

     

    so that finally,

     

    \lambda'-\lambda = \frac{h}{mc}(1-\cos{\theta}) \,

    ] Applications

     

     

     

     

    her har du sammenhengen :ohmy:

  19. Viftestyring & strømstyring på bærbar

    i Datamaskiner

    Skrevet

    Har en acer ferrari som vifta går all the time, ser det er noen programmer i 1. post om viftestyring så kunne jo prøvd dem - men formaterer pc'n nå så får ikke prøvd på en stund.. Det jeg lurer på er om det er en bedre\mer lydløs vifte det går ann å putte i, evt hvilket program det er noe vits i å prøve. Forslag?

    det er fordi du trenger turbogear.exe for å velge hastighet og lydnivå på viften!! men Acer er et dårlig merke og Asus er et bra merke selvsagt!!! :roll:

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