Special Relativity - Relativistic Energy

A photon strikes an electron of mass m that is initially at rest, creating an electron-positron pair. The photon is destroyed and the positron and two electrons move off at equal speed along the initial direction of the photon. The energy of the photon was

A. mc²
B. 2mc²
C. 3mc²
D. 4mc²
E. 5mc²

(GR0177 #99)

Solution

Relativistic Energy: 

E = (p²c² + m₀²c⁴)^1/2

Photon has no mass, m₀ = 0, therefore,

E_(photon) = (p²c² )^1/2 = pc

Initially, electron is at rest, thus momentum is zero, therefore,

 E_(electron) = (m₀²c⁴)^1/2 = m_ec²

 
Total initial energy of the system,

E_i = E_(photon) + E_(electron) = pc + m_ec²

After collision, 3 particles (2 electrons and a positron) move off at equal speed, thus each has p/3 momentum.

Electron and positron has the same mass.

Thus, each particle has energy,

E_{f (electron)} = E_{f (positron)} = [(p/3)²c² + m_e²c⁴]^1/2

Total final energy of the 3 particles, E_f  = 3 [(p/3)²c² + m_e²c⁴]^1/2

Conservation of energy, E_i = E_f 

pc + m_ec²  = 3 [(p/3)²c² + m_e²c⁴]^1/2

(pc + m_ec²)² = 9 [(p/3)²c² + m_e²c⁴]

p²c² + m_e²c⁴ + 2pm_ec³ = 9(p/3)²c²  + 9m_e²c⁴

p²c² + m_e²c⁴ + 2pm_ec³ = p²c²  + 9m_e²c⁴

m_e²c⁴ + 2pm_ec³ = 9m_e²c⁴

2pm_ec³ = 8m_e²c⁴

p = 4m_ec

Energy, E = pc = 4m_ec²

Answer: D