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Clarification on electric potential and kinetic energy

  1. May 1, 2008 #1
    One form of nuclear radiation, beta decay, occurs when a neutron changes into a proton, an electron, and a chargeless particle called a neutrino: n -> (p+) + (e-) + tex]\upsilon[/tex] where nu is the symbol for a neutrino. When this change happens to a neutron within the nucleus of an atom, the proton remains behind in the nucleus while the electron and neutrino are ejected from the nucleus. The ejected electron is called a beta particle. One nucleus that exhibits beta decay is the isotope of hydrogen ^{3}H, called tritium, whose nucleus consists of one proton (making it hydrogen) and two neutrons (giving tritium an atomic mass m = 3u Tritium is radioactive, and it decays to a helium ion: 3H -> 3He + (e-) + tex]\upsilon[/tex].

    The nuclei of both 3H and 3He have radii of 1.5 *10^-15 With what velocity must the electron be ejected if it is to escape from the nucleus and not fall back?

    I don't quite understand what this question is asking, but given the information (and given that neutrons have no charge) I treated this problem as if there were only two particles: the proton (nucleus) and the electron being ejected. I am unsure if this is the right setup, because of the '3' in front of the 3H and 3He.

    I am also unsure of how to calculate this, because the equations we have learned imply that the force acting on a charge is only zero at a "very large distance", ie, the distance is approaching infinity. In theory this means that there is no point where the electron would not be influenced by the proton's force. So how does it escape?

    I would really appreciate it if someone could give me a nudge in the right direction.
  2. jcsd
  3. May 2, 2008 #2
    "Escape" means go to infinity.

    The problem is asking this: An electron starts at the surface of a helium nucleus and escapes to infinity. What is its minimum initial velocity?
  4. May 2, 2008 #3
    Got it!
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