The following six particles (with the rest energy E, and kinetic energy KE matte
ID: 3162292 • Letter: T
Question
The following six particles (with the rest energy E, and kinetic energy KE matter partners traveling in the opposite direction with the same kioetie antimatter annihilation produces a pair of photons in opposite directions) indicated) collided with their are The resulting matter E. (Met) KE Mev) a. Rank these particles on the basis of their momentum. Largest 1. 2. 3. 4. 5. 6 Smallest The ranking cannot be determined based on the information provided. Explain the reason for your ranking: b. Rank these particles on the basis of energy of the photons they create. Largest I 2. 3. 4. 5. 6. smallest The ranking cannot be determined based on the information provided. Explain the reason for your ranking: c. Rank these particles on the basis of speed of the photons they create. Largest 1 Smallest The ranking cannot be determined based on the information provided. Explain the reason for your ranking:Explanation / Answer
For reletevisit particles
E2 = P2c2 - mc2
reletivisitc momentum P = E2 - (mc2 )2 , E is total energy (rest mass energy(mc2) + KE
particle A = P2 = (3002 -1002 )/c2 = 8.0e+4/c2
B = P2 = (3002 -2002 )/c2 = 5.0e+4/c2
C = P2 = (5002 -1002 )/c2 = 24.0e+4/c2
D = P2 = (6002 - 4002 )/c2 = 20e+4/c2
E = P2 = (6002 -2002 )/c2 = 32.0e+4/c2
F = P2 = (9002 -8002 )/c2 = 17.0e+4/c2
ranking on momentum : E,C,D,F,A,B
b) The total energy of the matter and the anti-matter is converted into two photon emitted in oppiste direction and the total energy is shared by the two photon equally, hence each phton energy is equal to the total energy of the particle
phton energy A = 100 +200 = 300 Mev
B = 200 +100 = 300 Mev
C = 100 +400 = 500 Mev
D = 400 +200 = 600 Mev
E = 200 +400 = 600 Mev
F = 800 +100 = 900 Mev
ranking on photon energy largst to smallest
F,E,D,C,B,A
c) all photons have the same speed c and ranking on the speed does not make senese.