Module 5 : MODERN PHYSICS
Lecture 25 : Derivation of Compton's Formula
   
$\displaystyle h\nu_0 + m_0c^2 = h\nu + \sqrt{m_0^2c^4 + p_e^2c^2}$ (4)
  From Eqn. (4), we get, on squaring,
 
$\displaystyle m_0^2c^4 + p_e^2c^2$ $\displaystyle =$ $\displaystyle (h\nu_0-h\nu + m_0c^2)^2$
$\displaystyle =$ $\displaystyle (h\nu_0-h\nu)^2 + m_0^2c^4 + 2m_0c^2(h\nu_0-h\nu)$
  Thus,
 
$\displaystyle p_e^2c^2 = (h\nu_0-h\nu )^2+2m_0c^2(h\nu_0-h\nu)$
  On substituting expression (3) for $ p_e$ in the above equation, we get
 
$\displaystyle p_i^2c^2+p_f^2c^2 -2p_ip_f\cos\theta c^2 = (h\nu_0-h\nu )^2+2m_0c^2(h\nu_0-h\nu)$
  Recalling $ p_i = h\nu_0/c$ and $ p_f= h\nu/c$ and on simplification, we get
 
$\displaystyle h\nu\nu_0(1-\cos\theta) = m_0c^2 (\nu_0-\nu)$
  Using $ \lambda_s = c/\nu$, we get Compton's formula
 
$\displaystyle \boxed{\lambda_s-\lambda_0 = \frac{h}{m_0c}(1-\cos\theta)}$
  $ h/m_0c\equiv \lambda_c $ is known as the Compton Wavelength of an electron.
   
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