Question

An electron and positron are moving toward each other and each has speed 0.500c in the lab frame. a) What is the kinetic energy of each particle? b) The e+ and e- head -on and annihilate. What is the energy of each photon that is produced? What is the wavelength of each photon? How does the wavelength compare to the photon wavelength when the initial kinetic energy of the e+ and e- is negligibly small?

Answer 1

a) There are a couple of ways you could work out kinetic energy.
Try subtracting rest energy from total energy.
\(E_k = E - E_0 = \sqrt{p^2c^2+m^2c^4}-mc^2 = \gamma m c^2 - m c^2\)
remember
\(\gamma = \sqrt{\frac1{1-\beta^2}}\)
b) Conservation of energy and conservation of momentum
In standard scattering numbering, the incident particles are 1 and 2, the outgoing ones are 3 and 4.
Momentum conservation:
\(\bf{p}_1 + \bf{p}_2 = \bf{p}_3 + \bf{p}_4 = 0\)
\(|\bf{p}_1| = |\bf{p}_2|\) and \(|\bf{p}_3| = |\bf{p}_4|\)
Energy conservation:
\(E_1 + E_2 = E_3 + E_4\)
since
\(E = \sqrt{p^2c^2+m^2c^4}\)
and the values of p and m is the same for 1 and 2, as well as 3 and 4 (but not the same for all 4):
\(E^2_e = E^2_\gamma\)
\(p_e^2c^2+m_e^2c^4 = p^2_\gamma c^2\)
Alternatively
\(\gamma m_ec^2 = p_\gamma c\)

Wavelength is found from
\[p=\frac{h}\lambda\]

The last part wants you to work out the wavelength when the total energy of each photon is
\(E_\gamma = p_\gamma c = m_e c^2 \)