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Question regarding radius of circular paths (chapter the nucleus)

  1. May 19, 2005 #1
    I have my doubt,solution and question in the attachment that followed.Thanks for anybody that spend some time on this question.

    Attached Files:

  2. jcsd
  3. May 19, 2005 #2
    OK the problem with your solution is that You have made it a bit too complicated.
    First of all ,
    After accelerating through a potential , the electrons gain certain amount of K.E , which can be calculate din the following way:

    [itex]eV= \frac{1}{2}mv^2[/itex]

    From The KE , calculate velocity gained by the electron beam.

    Now if you have studied how charged particles behave in magnetic fields, you should know that electrons when enter prependicular to a magnetic field, start moving in a circle.

    Force due to a Magnetic field on an electron provides the centripedal force necessary for moving in circle. therefore,

    F= -e(VxB) = m \frac{v^2}{R}

    From here calculate the radius of the beam...easy ..isnt it?
  4. May 20, 2005 #3
    Yes,Dr Brain.Thanks for your help and I really appreciate it.Thanks again for your explaination in detail.
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