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Homework Help: Calculating the wavelength of an electron

  1. Jan 18, 2012 #1
    1. The problem statement, all variables and given/known data

    Calculate the wavelength of a beta particle (electron) that has an energy of 4.35 × 104 eV

    2. Relevant equations

    E = hf
    λ = h/mv
    V = W/q
    eV = .5mv^2

    3. The attempt at a solution

    I can't figure out how to get electron volts into a form of energy that I'm familiar with in an equation. I tried

    eV = .5mv^2 , but that gave me a speed above the speed of light.

    I also tried using E = hf, but that also gave me a ridiculous number.
  2. jcsd
  3. Jan 18, 2012 #2


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    The conversion factor for swapping between eV and Joules is 1.6 x 10-19.

    You either divide or multiply depending which way you are changing.

    You should know which figure is bigger: either a few joules = lots of eV or a few eV = lots of joules. You thus multiply or divide to get the size of answer you need.

    Think: if you multiply by 1.6 x 10-19 will that give a bigger number or a smaller number? Dividing will give the opposite. What size of answer do you want?
  4. Jan 18, 2012 #3


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    1 eV = elementary charge * 1 V=1.602 x 10-19 J.

    Apply relativity theory. The given energy is the kinetic energy of the particle (the energy it gains when accelerated by 4.35 x 10 4 V. Find the relativistic momentum, and you get the wavelength as λ=h/p.

  5. Jan 18, 2012 #4
    I can't get it. I'm pretty sure I have to divide 4.35 * 10^4 by 1.6 * 10^-19, but that gives me 2.7 *10^23. I then plug that into the equation

    E/h = f

    which gives me 56 orders of magnitude. I then divide 1.6 * 10^-19 by 4.35 * 10^4 and that also gives me the wrong answer. I have a feeling that my second step of using the

    E/h = f equation is wrong.
  6. Jan 18, 2012 #5


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    In what units? You have to get joules.

    It is valid for photons. The electron is not photon.

    What do you know about Relativity Theory?

  7. Jan 18, 2012 #6
    ok, I got the joules to be 7.53 * 10^-18, using this website


    I don't know how they got that number and would like to know. In any case, I then plugged that number into

    .5mv^2 and E = mc^2, using 9.11*10^-31 for the mass. After I got v, I plugged that into

    lambda = h/mv

    and I was off by an order of magnitude, so I still am doing something wrong.
  8. Jan 18, 2012 #7


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  9. Jan 18, 2012 #8


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    See above
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