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Determine the energy of an alpha particle

  1. Oct 5, 2016 #1
    I have been asked to determine the energy in MeV of an alpha particle and an Rn nucleus in the decay of Ra-226 using the conservation of energy and momentum. (assume Ra is at rest)
    Here is what i have done, im just wondering if i did it properly. Thanks for your help :)

    Ek=kinetic energy
    m=mass
    V=velocity
    p=momentum
    α=alpha particle
    Rn= radon nucleus
    Q= Energy

    226Ra→222Rn + 4He + Q
    88 86 2

    Ekα= 1/2mV^2
    = 1/2 (4.002606amu/6.02x10^26kg/amu) (1.52x10^7m/s)^2
    = 7.67x10^-13J x 1MeV/1.6022x10^-13J
    = 4.78 MeV for the kinetic energy of the alpha particle
    (I read that the velocity of an alpha particle is 5% the speed of light, so thats where I got 1.52x10^7 but this could be wrong)

    Next since momentum is conserved

    mα(Vα)=mRn(VRn)
    Vα/VRn=mRn/mα
    4.78MeV/VRn=222.017578amu/4.002603amu
    VRn= 0.09MeV

    So the speeds in MeV are α= 4.78MeV and Rn=0.09MeV

    Did I miss any steps or make any mistakes?
     
  2. jcsd
  3. Oct 5, 2016 #2

    mfb

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    There is no fixed speed. If they have more energy they are faster, if they have less energy they are slower.

    To solve this problem you'll need some information about the decay, like the decay energy.

    A speed in MeV does not make sense. What is your height in Volts?
     
  4. Oct 5, 2016 #3
    Here is the exact question
    using conservation of momentum and energy, determine the energies in MeV of the alpha particle and the Rn nucleus in the decay Ra-226. You may assume that the Ra nucleus is initially at rest.
     
  5. Oct 5, 2016 #4

    vanhees71

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    I don't understand what this "+Q" in your question refers to. If this is a usual ##\alpha## decay, you just need to do the usual decay kinematics of a two-body decay. Usually one works in the reference frame, where the decay particle is at rest. Then you have (in natural units with ##\hbar=c=1##)
    $$p=(M,0,0,0)$$
    for the radium nucleus in the initial state, and energy-momentum conservation tells you that
    $$p=q_1+q_2,$$
    where ##q_1## and ##q_2## are the four-momenta of the ##\alpha particle## and the Radon nucleus. Additionally you have the on-shell conditions ##q_1^2=m_{\alpha}^2## and ##q_2^2=m_{\text{Rn}}^2##.

    Now to get the momentum ##\vec{q}_1=-\vec{q}_2## of the ##\alpha## particle just take the square of the energy-momentum conservation equation (in the sense of the Minkowski product)!
     
  6. Oct 5, 2016 #5
    Q represents the total energy released in the process of radium decay. I found this value to be 4.87 MeV in my textbook
     
  7. Oct 5, 2016 #6

    vanhees71

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    I see, but that's implied in my formalism too. It's implied by the masses of the involved particles. So let's see, whether I can reproduce your results. The simplest way is to use the energy-momentum conservation equation in the form
    $$q_2=p-q_1 \; \Rightarrow \; q_2^2=m_{\text{Rn}}^2=p^2+q_1^2-2 p \cdot q_1=m_{\text{Ra}}^2+m_{\alpha}^2-2 m_{\text{Ra}} E_{\alpha}$$
    Solving for ##E_{\alpha}## gives
    $$E_{\alpha}=\frac{m_{\text{Ra}}^2+m_{\alpha}^2-m_{\text{Rn}}^2}{2m_{\text{Ra}}}=3733 \;\text{MeV}$$
    From this you get
    $$q_1=p_{\alpha}=\sqrt{E_{\alpha}^2-m_{\alpha}^2}=189 \;\text{MeV}.$$
    The speed of the ##\alpha## particle thus is
    $$v=\frac{p_{\alpha}}{E_{\alpha}}=0.051,$$
    i.e., it goes with 5.1% of the speed of light. So that's correct.
     
  8. Oct 5, 2016 #7

    mfb

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    I moved the thread to our homework section.

    @vanhees71: Your numbers are inconsistent, see WolframAlpha.

    Good, that is the sum of the kinetic energies of the two nuclei. You can use this plus momentum conservation to get the energies and speeds of both products. Newtonian mechanics should be fine as approximation.
     
  9. Oct 6, 2016 #8

    vanhees71

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    Hm, I took the masses of the particles from some online nuclide data base. I attatch the pdf of the Mathematica Notebook used to do the calculation.
     

    Attached Files:

  10. Oct 6, 2016 #9

    mfb

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    Ah, your energy includes the rest energy of the alpha particle. I think the problem asks for the kinetic energy.
    Anyway, we shouldn't do OPs homework.
     
  11. Oct 6, 2016 #10

    vanhees71

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    Of course, I was too lazy to think about the non-relativistic approximation. For the kinetic energy you get ##E_{\text{kin},\alpha}=E_{\alpha}-m_{\alpha}=4.79 \, \text{MeV}##.
     
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