Tag: option c: imaging

By compton formula
$\Delta \lambda = \lambda _{f} -\lambda _{i}= \dfrac{h}{m _{e}C}(1-cos\theta )$

$0.22\ A^{\circ}-\lambda \ _{i}= 0.024A^{\circ}(1-\dfrac{1}{2}) \ \ \ \ (\because  cos60^{\circ}= 1/2)$

$\lambda   _{i}= 0.22-0.012$
$= 0.208  A^{\circ}$

$\lambda -\lambda ^{1}  =  \dfrac{h}{m _e c}(1-cos  \theta )$

where
$\lambda $ is initial wavelength
$\lambda^{1} $ is the wavelength after & scattering
$h$ is the plank constant
$m _e$ is the electron rest mass.
$c$ is the speed of light
$\theta $ is the scattering angle.

Compton effect happens with free electrons.
Photoelectric effect happens to bounded electrons.

The quantity $(\dfrac {h}{m _0c})$ is known as the  Compton wavelength of the electron; it is equal to $2.43\times 10^{-12}$ $m$ or $0.0243 A$. 

 From Scattering Formula
$\lambda'-\lambda=\dfrac{G}{M _{e}c}(1-cos \theta )$
We can see, $\lambda '> \lambda $
So $v'< v$

Compton scattering is an inelastic scattering of a photon by a quasi-free charged particle, usually an electron. It results in a decrease in energy (increase in wavelength) of the photon (which may be an X-ray or gamma ray photon), called the Compton effect. Part of the energy of the photon is transferred to the recoiling electron. Binding energy is equal to work function of metal. In most of metals, the threshold frequency is equal to that of ultravoilet light. So Compton effect is associated with X-rays.

Compton shift refers to proton.

The wave nature of light is exhibited by interference, diffraction and polarisation of light.
Compton effect is the strongest evidence to show the particle nature of light, and hence the quantum theory of light.