2. Photo-production. Suppose a photon collides with a particle (call it particle

Question

2. Photo-production. Suppose a photon collides with a particle (call it particle 0) with mass m0, that is initially at rest. After the collision, particle 0 disappears and two new particles emerge, one with mass m1 and one with mass m2. Suppose further that after the reaction, particle 1 is left at rest.

(a) (15 points) Calculate the momentum of the initial photon in terms of the masses m0, m1 and m2.

(b) (15 points) There is a condition involving the values of m0, m1, and m2 that determines whether the reaction as described can take place. What is it? Possibly useful hint: express your answer by replacing the quantity (m0-m1) by the symbol m.

Explanation / Answer

iniitally there is a photon
photon collides with a aprticlew to produce two particels of masses m1 and m2, m1 is at rest m2 is moving

a. momentum of initial photon = h/lambda ( where lambda is wavelength of this photon)
   now, from conservation of energy
   hc/lambda = 0.5m2*v^2 + (mo - m1 - m2)c^2
   so velocity of the m2 mass is

   v2 = sqroot{2[hc/lambda - (mo - (m1 + m2))c^2]/m2}

   now from conservation of momentum
   h/lambda = m2v2 = m2*sqroot{2[hc/lambda - (mo - (m1 + m2))c^2]/m2}
   h^2/lambda^2 = 2hc*m2/lambda - 2(mo*m2 - m2(m1 + m2))c^2

   let 1/lambda = t then
   h^2*t^2 - 2hc*m2*t + 2(mo*m2 - m2(m1 + m2))c^2 = 0
   solving for t
   t = [2c*m2 +- sqroot(- 2mo*m2c^2 + 2m2*m1c^2 + 3*m2^2*c^2)]/h
   where h is planks constant and c is speed of light
   so initial momentum of photon = h*t = [2c*m2 +- sqroot(- 2mo*m2c^2 + 2m2*m1c^2 + 3*m2^2*c^2)]

b. for this to be true, ht > 0 and ht is real
       so - 2mo*c^2 + 2*m1c^2 + 3*m2*c^2 > 0
       3*m2 > 2(mo - m1)
       3m2 > 2 dm

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