A subatomic particle X spontaneously decaysinto two particles, A and B , eachof

Question

A subatomic particle X spontaneously decaysinto two particles, A and B, eachof rest energy 1.40 × 102 MeV. The particles flyoff in opposite directions, each with speed 0.827crelative to an inertial reference frame S. Reference: Ref 28-1
Use energy conservation to determine the rest energy ofparticle X.

a. 280 MeV
b. 392 MeV
c. 498 MeV
d. 206 MeV
e. 249 MeV A subatomic particle X spontaneously decaysinto two particles, A and B, eachof rest energy 1.40 × 102 MeV. The particles flyoff in opposite directions, each with speed 0.827crelative to an inertial reference frame S. Reference: Ref 28-1 Use energy conservation to determine the rest energy ofparticle X. 280 MeV 392 MeV 498 MeV 206 MeV 249 MeV

Explanation / Answer

   The rest energy of the particle x is              E = 1/[1-(v/c)2]mc2 + 1/[1-(v/c)2]mc2                 =1/[1-(0.827c/c)2]140 MeV + 1/[1-(0.827/c)2]140 MeV               = 498 MeV                 =1/[1-(0.827c/c)2]140 MeV + 1/[1-(0.827/c)2]140 MeV               = 498 MeV

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---