What is the linear momentum of the electron if the energy of incident photon was 1.024 MeV? What is the angle between the directions of the scattered photon and recoiled electron?

Answer 1

Assuming the electron is at rest (and hence starts with zero momentum), we can use conservation of momentum here. The final momentum of the electron can be derived using the following definition:

p = E / c, where E is the energy and c is the speed of light

Momentum is being transferred from the photon to the electron. So, we can write this as

p[electron] = sqrt( E[photon]^2 - E[electron]^2 ) / c

All we did was expand the energy term from the definition to account for the momentum transfer. We can also write out the energy of the electron using mass-energy equivalence E=mc^2

p[electron] = sqrt( E[photon]^2 - (m[electron]*c^2)^2 ) / c

You can calculate that out, as you now have all the needed quantities.

For the second part, we know that momentum must be conserved.

p[photon, initial] = p[photon, final] + p[electron]

We just calculated p[electron]. We can find p[photon, initial] simply by using the definition of momentum as stated above.

p[photon, initial] = E[photon, initial] / c

Now we know p[photon, final].

We can then use the path length difference equation (modified algebraically for momentum) for the photon to find the angle.

wavelength[final] - wavelength [initial] = h/mc*(1 - cos(theta))

1/p[final] - 1/p[initial] = h/mc*(1-cos(theta)).

Note that this theta is the difference in the angle of the incident photon and the scattered photon. To find the angle of the recoiled electron, you can apply conservation of momentum (in the direction perpendicular to the incident photon trajectory) and some basic trig.