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Homework Help: Modern Physics Problem

  1. Feb 3, 2005 #1
    The neutral pion is an elementary particle of the meson family that has a rest mass energy of 135 MeV. This particle is unstable and decays into two photons ("light particles" of no rest mass and energy E=pc). Consider now the following situation: a neutral pion has a kinetic energy of 270 MeV as measured in a given frame. Find the momentum, the direction of propagation and the energy of each of the two emitted photons, given that one of the photons is emitted in a direction perpendicular to the initial velocity of the neutral pion.

    My solution: by conservation of energy Kpion+Mpion*c^2=(p1+p2)c^2
    p1+p2=(270+135)/c=405/c

    by conservation of momentum:

    x-direction: Ppion=p1cos(theta1) + p2cos(theta2)=p2cos(theta2)
    y-direction: 0=p1-p2sin(theta2)

    For the pion we have Ppion*c^2=(E^2-mc^2)^0.5=4.498E-15*c^2

    My problem now is solving for theta2. theta2=arctan(P2y/P2x)
    I already know P2x=Ppion, but I can't get P2y.

    Thanks.
     
  2. jcsd
  3. Feb 3, 2005 #2

    dextercioby

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    You have a system of 3 equations with 3 unknowns
    [tex] p_{2}\cos \vartheta=P [/tex] (1)

    [tex]p_{2}\sin\theta=p_{1} [/tex] (2)

    [tex] p_{1}+p_{2}=\frac{K}{c}+Mc [/tex] (3)

    ,where M is the rest mass of the neutral pion,K is the KE of the pion & P is the pion's (relativistic) momentum...

    Solve the system & find the 3 unknowns...

    Daniel.
     
  4. Feb 4, 2005 #3
    Hi Daniel I am getting the following,

    from (1): p2=P/cos(theta2)
    from (2): p1=p2sin(theta2)

    substitution in (3) yields p2sin(theta2)+P/cos(theta2)=K/c+Mc
    and after a few manipulations I get tan(theta2)+
    sec(theta2)=(K/c+Mc)/P I am not sure how to solve
    for theta2.
     
  5. Feb 4, 2005 #4

    Curious3141

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    Hint : take equation (1) squared plus equation (2) squared.
     
    Last edited: Feb 4, 2005
  6. Feb 4, 2005 #5

    dextercioby

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    That doesn't work...The "theta's" are different... :tongue2: Cf. [itex] \theta [/itex] to [itex] \vartheta [/itex].

    It was a joke...They're the same.

    Good advice... :smile:

    Daniel.
     
  7. Feb 4, 2005 #6
     
  8. Feb 4, 2005 #7

    Curious3141

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    Were the "different" thetas a typo ? :wink:

    Anyway, let me just work this out as I would do it.

    Using Daniel's notation,

    Square (1) and (2) and add :

    [tex]p^2_2 = p^2_1 + P^2[/tex]


    [tex]p^2_2 - p^2_1 = P^2[/tex]


    [tex](p_2 - p_1)(p_2 + p_1) = P^2[/tex] ---(4)

    Put equation (3) into (4) and rearrange,

    [tex]p_2 - p_1 = \frac{P^2}{\frac{K}{c} + Mc}[/tex] ---eqn(5)

    Take (3) + (5) :

    [tex]2p_2 = \frac{P^2}{\frac{K}{c} + Mc} + \frac{K}{c} + Mc[/tex]

    and you can find [itex]p_2[/itex] and then [itex]p_1[/itex]

    How to get [itex]P^2[/itex] in terms of what's given in the question ? For that, I would use [tex]E^2 = P^2c^2 + m^2c^4[/tex]

    where [tex]E = K + mc^2[/tex]. You're given [itex]K[/itex] and the rest energy.

    I'll leave the orig. poster to do the final simplifications.
     
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