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Speed and Momentum of a Beta Particle

  1. Apr 28, 2010 #1
    1. The problem statement, all variables and given/known data

    A 14C nucleus, initially at rest, emits a beta particle. The beta particle is an electron with 127 keV of kinetic energy.

    A) What is the speed of the beta particle?
    B) What is the momentum of the beta particle?
    C) What is the momentum of the nucleus after it emits the beta particle?
    D) What is the speed of the nucleus after it emits the beta particle?

    2. Relevant equations

    v = √[ 2K / m]

    3. The attempt at a solution

    Okay, so this is my work, but it gives the wrong answer.

    Kinetic energy of beta particle K = 127 keV = 127 * 10 ^ 3 eV
    = 127 * 10 ^ 3 * 1.6 * 10 ^ -19 J
    = 203.2* 10^ -16 J

    K = ( 1/ 2) mv ^ 2
    v = √[ 2K / m]
    m=9.11* 10 ^ -31 kg
    v = 2.31* 10 ^ 8 m / s
    THIS IS NOT RIGHT.

    Can someone help me out where my thinking was flawed please?
     
  2. jcsd
  3. Apr 28, 2010 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi ihearyourecho! :smile:

    (try using the X2 tag just above the Reply box :wink:)
    The speed of light is c = 3*108 m/s,

    so your v is about 0.77 c, which is far too large to be able to use non-relativistic formulas. :redface:

    You need the relativistic formula for energy. :wink:
     
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