Question

An electron and a positron collide head on, annihilate, and create two 0.804 MeV photons traveling in opposite directions. What was the initial kinetic energy of an electron? What was the initial kinetic energy of a positron?

Answer 1

Answer:

Ke- = Ke+ = 0.294MeV

Explanation:

To fins the kinetic energy of both electron and positron you use the following formula, for the case of annihilation of one electron an positron:

2$E_p=2E_o+K_{e^-}+K_{e^+}$   (1)

Ep: photon energy = 0.804MeV

Eo: rest energy of one electron (and positron) = 0.51MeV

Ke-: kinetic energy of electron

Ke+: kinetic energy of positron

You replace the values of Ep and Eo in the equation (1):

$K_{e^-}+K_{e^+}=2E_p-2E_o=2(0.804MeV-0.51MeV)=0.588MeV$

Iy you assume both positron and electron have the same speed, then, the kinetic energy of them are equal, and the kinetic energy of each one is:

$K_{e^-}=K_{e^+}=\frac{0.588MeV}{2}=0.294MeV$