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A negative pion with negative energy T= 50MeV disintergrated during its flight into a moun and a neutrino. Find the energy of the neutrino outgoing at right angles to the pion's motion derection. |
Answer» Solution : See the DIAGRAM. By CONSERVATION of ENERGY `sqrt(m_(pi)^(2)c^(4)+c^(2)p_(pi)^(2))=cp_(V)+sqrt(m_(mu)^(2)+p_(pi)^(2)c^(2)+c^(2)P_(v)^(2))` or `(sqrt(m_(pi)^(2)c^(4)+c^(2)p_(pi)^(2))-cp_(v))^(2)=m mu^(2)c^(4)+c^(2)p_(pi)^(2)+c^(2)P_(v)^(2)` or `m_(pi)^(2)c^(4)-2cp_(v)sqrt(m_(pi)^(2)c^(4)+c^(2)p_(pi)^(2))=m_(mu)^(2)c^(4)` Hence the energy of the neutrino is `E_(v)=cPv=(m_(pi)^(2)c^(4)-m_(mu)^(2)c^(4))/(2(m_(pi)c^(2)+T))` on writing `sqrt(m_(pi)^(2)c^(4)+c^(2)p_(pi)^(2))=m_(pi)c^(2)T` Substitution gives `E_(v)= 21.93MeV` |
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