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Starting velocity of a electron

  1. Aug 26, 2010 #1
    Hello, I am conducting some cathode ray tube experiments and I need to know how to calculate the starting velocity of an electron when it is emitted from a cathode. Does anyone know how to do this? Will the electron starting travelling velocity vary depending on the heat of the cathode? I know as you heat the cathode up more electrons are liberated and hence emitted.

    I thank you in advance

  2. jcsd
  3. Aug 26, 2010 #2

    The final velocity of the electron can be controlled by controlling the accelerating potential. The cathode temperature will determine the no: of electron/intensity of the beam.

    I am not sure about calculating the starting velocity of electron but if you equate the energy required to heat up the cathode to the Kinetic Energy of the electron emitted due to thermionic emission, may be you can get the value.

    Everybody: Please correct me if required.
  4. Aug 26, 2010 #3
    The energy (and velocity) of the electron is reduced by the work function W (~4 volts) of the filament (probably tungsten):


    So the electron velocity is given by

    v = [2(V-W)/511,000]½·c

    where V is the voltage (potential) relative to the filament at any given location, and c is the velocity of light. There is a very small energy spread, due in part to the thermal spread in the thermionic emission energy, and Coulomb forces within the electron beam. See also


    Bob S
  5. Aug 27, 2010 #4
    Hi Bob where is this equation derived from? I cant find what equation your re-arranged to get that?

    511000 eV is the electrons energy at rest ?

    Kind Regards

    Last edited: Aug 27, 2010
  6. Aug 27, 2010 #5
    Hi Jon-

    The kinetic energy of the electron is the applied voltage potential V (equivalent to electric field E times distance x) minus the filament work function W. So, using energy in electron-volt (eV) units,

    V-W = ½mv2 (non-relativistic form)

    where m is electron mass and v = electron velocity. Multiplying and dividing by c, the velocity of light, we get

    V - W = ½·mc2·v2/c2

    using the electron mass in eV units as mc2 = 511,000 eV

    we get V - W = ½·511,000 eV·v2/c2

    et cetera

    Bob S
  7. Sep 2, 2010 #6
    Hi Bob thanks for the clarification on that. Just to cure my curiosity. How do you arrive at the fact that V-W = ½mv2 ???
  8. Sep 3, 2010 #7
    Read about work function at


    The electrons have to be pulled out of a potential well to leave the hot cathode. The depth of this potential well varies from one element to another. See picture and table in above URL. This reduces the effective voltage (kinetic energy) of the accelerated electrons by a voltage W (in eV).

    Bob S
  9. Sep 10, 2010 #8
    Just one last question, if I was to exceed the work function voltage for tungsten, say I took it upto 10V. Would I get more emitted electrons but all with the same energy (all be it with a little spread)?
  10. Sep 10, 2010 #9


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    I don't know if people are simply missing pertinent information here, or this has been simplified way too much.

    There is no one single velocity emitted via thermionic emission. It is a distribution of energy and thus, velocity. See, for example, Fig. 3 in R.D. Young, Phys. Rev. v.113, p.110 (1959). The energy distribution has a form of Eq. 28, with a spread of roughly 2.45 kT, with the peak at kT.

    So your starting velocity has a spread in values!

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