Kinetic energy (Question 1 (20 Marks)) — UPSC Physics Optional CSE 2026 Paper I

The energy of a photon is expressed as E = h\nu, where h is the Planck's constant and \nu is the frequency of the photon. The momentum of the photon is \frac{h\nu}{c}, where c is the speed of light. Show that if a photon scatters from a free electron (of mass m_e), the scattered photon has energy E' = E \left[ 1 + \frac{E}{m_e c^2} (1 - \cos\theta) \right]^{-1} where \theta is the angle through which the photon scatters. Also, show that the electron acquires a kinetic energy T = \frac{E^2}{m_e c^2} \left[ \frac{1 - \cos\theta}{1 + \frac{E}{m_e c^2}(1 - \cos\theta)} \right]