A particle with rest energy 8680 GeV decays at rest into two

A particle with rest energy 8.680 GeV decays at rest into two daughter particles, one with rest energy 7.450 GeV and the other with rest energy 290.000 MeV.

(a) What is the mass of the parent particle? (in kg)

(b) How much energy is released in the decay? (in MeV)

(c) What is the momentum of the lighter daughter particle? (in MeV/c)

(d) What is the energy of the lighter daughter particle? (in MeV)

Solution

a) m*c² =9*10⁹*1.6*10^-19

m=9*10⁹*1.6*10^-19 / (3*10⁸)² =1.6*10^-26kg

b) energy released =9-6.86-0.28= 1.86GeV =1860MeV

c) mass of lighter particle = 280*10⁶ *1.6*10^-19/ (3*10⁸)² = 4.98*10^-28kg

from momentum energy relation

p² / (2m) = energy =280*10⁶

p² =280*10⁶ *2*4.98*10^-28 = 2.78*10^-19

momentum = 5.28*10^-9 *3*10⁸= 1.584MeV/C

d)energy of lighter daughter particle=280MeV