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A hydrogen atom transitions from n= 5 state to the ground state

  1. May 17, 2017 #1
    1. The problem statement, all variables and given/known data
    A hydrogen atom transitions from ni= 5 state down to the ground state.
    a) What is the energy of photon emitted from the transition of the hydrogen atom?
    b) What is the ratio of the momentum of the emitted photon to the momentum of an electron which possesses the same kinetic energy as that of the photon. That is calculate pe/pγ.

    2. Relevant equations
    ΔEphoton = Ef - Ei

    En= -13.6 eV / (n)2

    3. The attempt at a solution

    a)
    ΔEphoton = E1 - E5
    ΔEphoton = -13.6 eV [1/(1)2 - 1/(5)2] = - 13.0 eV = - 2.08 x 10 -18 J (shouldn't the answer here be positive?)

    b)
    Momentum of electron
    Pelectron = √(2 ⋅ m ⋅ Ephoton)
    Pelectron = √(2 ⋅ 9.11 x 10-31 ⋅ 2.08 x 10 -18)
    Pelectron = 1.95 x 10-24

    Momentum of photon
    Pphoton = h /λ = Ephoton / c = (2.08 x 10-18) / (3.00 x 108)= 6.97 x 10-27

    Ratio

    Pelectron / Pphoton = 1.95 x 10-24 / 6.97 x 10-27 = 280
     
    Last edited: May 17, 2017
  2. jcsd
  3. May 17, 2017 #2

    mfb

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    2016 Award

    Staff: Mentor

    The answer should be positive. Your initial formula gives the correct sign if you absorb a photon.

    Working with eV everywhere would make calculations easier, but working with SI units is possible as well, of course.
     
  4. May 17, 2017 #3
    So i'm using the wrong formula for photon emission?
     
  5. May 17, 2017 #4

    mfb

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    2016 Award

    Staff: Mentor

    You can just change the sign at the end. Or use the absolute value of the energy difference, then it works in both cases.
    Signs should come from understanding the situation. Good if the formulas also have the correct sign, but you should understand which sign the result has to have.
     
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