1.

The observation of the tracks of secondary electrons showed that a neutral pion decayed into two identical photons. The angle of separation of the photons is 90^(@) Find the kinetic energy of the pion and the energy of each photon.

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Solution :Let us denote the TOTAL energy of the neutral pion by `epsi_(pi)` and its rest energy by `epsi_(0)=135MeV`, By the law of conservation of energy `epsi_(pi)=2epsi_(gamma)`, where `epsi_(gamma)` is the PHOTON energy. From the law of conservation of momentum. We obtain `p_(pi)=2p_(gamma)COS45^(@)`. Noting that the photon momentum is `p_(gamma)=epsi_(gamma)//c` and the pion momentum is `p_(pi)=1/csqrt(epsi_(pi)^(2)-epsi_(0)^(2))`, we obtain
`epsi_(pi)^(2)-epsi_(0)^(2)=4epsi_(gamma)^(2)cos^(2)45^(@),or,epsi_(pi)^(2)-epsi_(0)^(2)=epsi_(pi)^(2)cos^(2)45^(@)`
It then follows that `epsi_(pi)=epsi_(0)sqrt2`.
The KINETIC energy of the pion is `K_(pi)=epsi_(0)(sqrt2-1)` and the photon energy is `epsi_(gamma)=epsi_(0)//sqrt2`.


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