Question

An electron-positron pair (positron is electron's antiparticle, it has the same mass as electron, but opposite charge) can be produced what two photon are collided. Two photons of frequency w are collided head-on. What will be the electron's momentum? Electron's rest mass is m(e).

Answer 1

Answer:

$p_e = \sqrt{ \frac{(\ h \ w \ )^2}{{c^2}} - m_o^2c^2}$

Explanation:

If the photons got frequency w, the energy of each photon must be

$E \ = \ h \ w$,

so the total energy of the system must be

$E_{total} \ = \ 2 \ h \ w$.

The momentum for each photon will be:

$p \ = \ \frac{h \ w}{c}$.

But, as they are colliding head on, the total momentum of the system must be zero.

Now, for the particles, the energy must be

$E \ = \ \sqrt{p^2c^2 + m_o^2c^4}$.

Momentum conservation implies that the total momentum must be zero, so:

$| \ \bar{p}_{electron} \ | = | \ \bar{p}_{positron} \ |$,

so the squares of the momentum will be the same.

Now, this implies that the energies for the electron and the positron must be the same, so we can write:

$E_{total} \ = \ 2 \ \sqrt{p^2c^2 + m_o^2c^4}$.

Taking conservation of energy in consideration:

$E_{total} \ = \ 2 \ \sqrt{p^2c^2 + m_o^2c^4} = \ 2 \ h \ w$.

$\sqrt{p^2c^2 + m_o^2c^4} = \ h \ w$.

$p^2c^2 + m_o^2c^4 = (\ h \ w \ )^2$.

$p^2c^2 = (\ h \ w \ )^2 - m_o^2c^4$.

$p^2 = \frac{(\ h \ w \ )^2 - m_o^2c^4}{c^2}$.

$p^2 = \frac{(\ h \ w \ )^2}{{c^2}} - m_o^2c^2$.

$p = \sqrt{ \frac{(\ h \ w \ )^2}{{c^2}} - m_o^2c^2}$.

And this its the electron's momentum