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Kinetic energy of two particles

  1. Feb 20, 2016 #1
    If light of certain wavelength falls on two particles say electron and a neutron(isolated) then since they are absorbing the same amount of energy their kinetic energy must be same.

    But using the formula:
    $$ \lambda = \frac{h}{\sqrt{2mE}} $$
    we get
    $$ E = \frac{h^2}{2m \lambda^2}$$
    which states us that for same wavelength kinetic energy varies inversly with mass thus neutron would have less kinetic energy than electron mathematically but logical approach states that for same wavelength of light falling on both, they absorb same amount of energy thus kinetic energy must be same.

    Why mathematical and logical aspects are not giving the same result?
     
  2. jcsd
  3. Feb 20, 2016 #2

    mfb

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    Staff: Mentor

    The (de-Broglie) wavelength of particles (=your formula) has nothing to do with the wavelength of incoming light.

    Isolated particles cannot absorb photons without any further interaction, this would violate energy or momentum conservation.
     
  4. Feb 20, 2016 #3
    Your question is very interesting, let's analise it.

    First, we have the De Broglie's equation:
    $$λ=\frac{h}{p}=\frac{h}{mv}=\frac{h}{\sqrt{2mE}}$$
    but this equation tells you the wavelength associated to a particle (in this particular case they are an electron and a neutron). So if you impress these particles with photons of frecuency ##f## the particles will have the same energy ##E=hf=h\frac{c}{λ}##, but they will have different wavelength associated, (their masses are different). Furthermore, the wavelength of the photons is not the wavelegth of the particles.
    Hence, the energy absorbed is the same, but the wavelegth acquired by each particle is different:
    $$E_e=E_n,\frac{h^2}{2m_e \lambda_e^2}=\frac{h^2}{2m_n \lambda_n^2},m_e \lambda_e^2=m_n \lambda_n^2$$

    A similar situation in Newtonian mechanics to this situation would be as follows:

    If you gives the same energy to two bodies of different masses ##M>>m## the body with mass ##m## will get a higher velocity than the mass of ##M## and it seems it has more energy. However, both have the same.
     
  5. Feb 20, 2016 #4
    i think the above position is correct vis-a-vis the problem posed - its a unphysical situation and absorption of light is not possible through de broglie wave picture of the particles.
    further one can apply scattering process to the two cases and can calculate the energy shared by electron and neutron using coservation of energy and momentum and in that case my guess is that neutron may not share the photon energy much
    should be much less than the electron case.

    well i can not figure out the above process.
     
  6. Feb 20, 2016 #5

    mfb

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    Staff: Mentor

    Elastic scattering of photons and neutrons is incredibly rare - it is called neutron for a good reason. Inelastic scattering has a reasonable cross-section, but then we get new particles.
    For elastic scattering, the electron will get more energy than a neutron, sure.
     
  7. Feb 20, 2016 #6
    agreed

    some additional info -may not be very relevant to the question
    Photo neutron reactions
    • Photon brings enough energy to drive reaction.
    • Photo neutron sources emit mono energetic neutrons ( if a single energy photon comes in)
    . • Requires photons of > several MeV.

    ref;http://ocw.mit.edu/courses/nuclear-engineering/22-01-introduction-to-ionizing-radiation-fall-2006/lecture-notes/energy_dep_neutr.pdf [Broken]
     
    Last edited by a moderator: May 7, 2017
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