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Kinetic Energy of a hydrogen electron

  1. Oct 29, 2007 #1
    1. The problem statement, all variables and given/known data

    In the Bohr model of the atom, the ground-state electron in hydrogen has an orbital speed of 2190 km/s. What is its kinetic energy?

    2. Relevant equations

    KE = 1/2mv^2

    3. The attempt at a solution

    I'm not sure what I am doing wrong but the program we use to enter homework is saying I'm wrong. So anyway kinetic energy = 1/2*m*v^2. Mass of 1 hydrogen atom = 1.008g/mol*(1mol/6.022*10^23 )*(1kg/1000g) = 1.674*10^-27 g/atom.
    The speed is 2190km/s*1000m/km = 2190000m/s therefore
    KE = 1/2(1.674*10^-27)(2190000)^2 = 4*10^-15 Joules
  2. jcsd
  3. Oct 29, 2007 #2


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    It asks for the kinetic energy of the electron. What is the mass of an electron?
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