Hi need help with energy, momentum and Compton effect, quantum mechanics? please

What was so surprising about the Compton effect in terms of the classical theory of how electromagnetic waves are scattered by atoms? (Hint: think about the wavelengths of the incident and scattered radiation).
(ii) Using the relationship between energy E, mass m and momentum p of a particle, namely, E^2 = m^2 + c^4 + p^2 c^2 , together with E = hf, where h is a constant and f the frequency of light, derive an expression for the momentum of a photon in terms of the wavelength λ.
(iii) Draw a schematic diagram of a Compton collision. Defining the appropriate variables, write down the equations for the conservation of momentum in directions parallel and perpendicular to the direction of the incident photon.

1) The surprise was that the only logical way to explain the change in wavelength was to assume that the X-ray photons that he used were acting like particles. When they collided with the graphite particles, they lost momentum, in the same way that a small particle would have done. The loss of momentum caused a change in energy and thus the change in wavelength.

2) Your equation for the total energy is not quite correct. It should read

E^2 = m^2c^4 + p^2c^2.

For a photon, the rest mass is zero, so

E^2 = p^2c^2

E = pc

E = h f = h c / λ

So

pc = h c / λ

p = h / λ

3) Lets assume that the incoming photon is travelling along the positive x-direction and that it collides with a stationary electron. After the collision, the photon recoils into the first quadrant making an angle Ѳ with the positive x-direction, and the electon recoils into the fourth quadrant making an angle α with the positive x-direction.
Before the collision, the electron's momentum is zero and the photon's momentum is ( h/λ).

After the collision, the electron has a recoil momentum of m(e)v. The photon has a reduced momentum and thus has a reduced frequency of (f - ∆f) and therefore has an increased wavelength of (λ + ∆λ ).