Regarding PD of xray tube

Discussion in 'Physics' started by mentaaal, May 14, 2008.

  1. mentaaal

    Thread Starter Senior Member

    Oct 17, 2005
    451
    0
    Hey guys, a friend and I are having our physics exam tomorrow and a friend asked my help in the following question which i am unsure of.

    In the question, the three lowest energy levels of an atom of the target material in the xray tube are shown and the question is: "What is the minimum Potential Difference at which the tube can operate if the transition from n=3 to n=1 is possible? Straight away i figured you just subtract n = 1 from n=3 and hey presto but my friend said he recalls the teacher saying something about you have to use the PD relating to the maximum transition, i.e. to allow the atom to be ionised entirely: 0 - (n=3). My thoughts on this is that the PD needed here is only enough to give the elctron enough energy to allow the electron to jump from n = 1 to n = 3.
    The atom's model is:

    n = 3 ----------------------- -11 x 10^3
    n=2 ------------------------- -26 x 10^3
    n=1 -------------------------- -98 X 10^3

    Could anyone tell me if i am wrong and why?

    Thanks loads.

    Not sure if this question should be here or in the homework section so Mr Admin, feel free to move it if you like.
     
    Last edited: May 14, 2008
  2. mentaaal

    Thread Starter Senior Member

    Oct 17, 2005
    451
    0
    I am guessing here that if an electron contained exactly the right energy pertaining to an energy difference, excitation would occur, but this is extremely unlikely and photon production is far more likely if the incident electron has enough energy to ionise the atom??
     
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