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    Relativistic Electric Potential Difference

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    For this question you need to know that an electron accelerated through an electric potential difference of V volts acquires a kinetic energy of V eV. (This question is best done by not converting to joules, but by exploiting the fact that mec2 is given as 0.511 MeV).

    i)
    What is v/c for an electron?
    a) Striking the screen of a TV tube with accelerating voltage 30 kV.
    b) In an electron microscope accelerated to an energy of 511 keV?

    ii)
    When the ratio v/c is appreciable, the classical expression for kinetic energy is no longer accurate. In both cases in part i), give the factor by which the classical expression for kinetic energy must be multiplied in order to get the relativisitically correct kinetic energy.

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    Solution Preview

    Please see the attached file.

    =======
    Here is the plain TEX source

    centerline{bf Relativistic kinetic energy }

    It appears to be an omission that in questions i)a and i)b it does not say explicitly that the

    energy cited is kinetic.
    We shall assume that it is kinetic.

    The total energy of a particle is
    $$
    E = gamma mc^2,
    eqno(1)
    $$
    where
    $$
    gamma = {1oversqrt{1-v^2/c^2}}
    eqno(2)
    $$
    and its ...

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