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

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

$cf   \theta = 0^{0}$

then  $\Delta \lambda = 0$

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

where $\dfrac{h}{m _e c}$ is called compton wavelength.

$\lambda -\lambda ^{1}=\dfrac{h}{m _e c}\ (1+cos  \theta )$
where $\theta$ is scattering angle.

Compton wavelength $= \ \dfrac{h}{m _{e}c}$

$= \ \dfrac{6.62\times 10^{-34}}{9.1\times 10^{-31}\times 3\times 10^{8}}$

$= \ 0.242\times 10^{-11}m$

$= \ 0.0242\times 10^{-10}m$

$= \ 0.0242\ A^{\circ}$

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.

Compton shift is given as $\Delta \lambda =\dfrac{h}{mc}(1-cos \phi )=\dfrac{6.62\times 10^{-34}Js}{9.1\times 10^{-31} Kg\times 3\times 10^8m/s}(1-cos 85^0)$ 

According to Compton's scattering 

In compton scattering it is necessary that the energy of the photon should be very much greater than binding energy of electron. Binding energy is equal to work function of metal. In most of metals, the threshold frequency is equal to that of ultravoilet light that is why we do not observe comption effect with visible light.

After scattering the wavelength of scattered photon increases due to the loss of energy and hence, the frequency decreases.
So, $f'<f$
So, the answer is option (C).

de Broglie wavelength,

If atomic number of an atom is high- suggests that more energy is required to eject the electron or electron is more tightly bound . If energy of x rays is low- suggests that x-ray photon possesses insufficient energy to cause any measurable effect on electron. Thus, considering both these factors, A is the correct option

According to the Compton's Equation

If electrons are accelerated to a velocity v by a potential difference V and then allowed to collide with a metal target, the minimum wavelength is given by:

The modified line occurs due to collision of photon with single electron.
Compton scattering usually refers to the interactive involving only the electrons of atoms,if  photon does not collide with any of electron of an atom, then it shows unmodified lines.

$\theta = \ 90^{\circ}$
so, $cos \theta =  0$
$\Delta \lambda  =  \dfrac{h}{m _{e}C}(1-cos \theta )  =  \dfrac{h}{m _{e}C}(1-0)  =  \dfrac{h}{m _{e}C}$

Compton wavalength $=\dfrac{h}{m _{e}c}(1-cos\theta )$

Compton formula
$\Delta \lambda = \lambda _f-\lambda _i= \dfrac{h}{m _ec}  (1- cos  \theta )$

${\lambda }'-\lambda =\dfrac{h}{m _{e}c}(1-cos\theta)$

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

For maximum increase $ cos\theta =   -1$

so $\Delta \lambda =  \dfrac{h}{m _{e}C} (1-(-1))$

$=  \dfrac{2h}{m _{e}C}$

$=  \dfrac{2\times 6.63\times 10^{-34}}{9.1\times 10^{-31}\times 3\times 10^{8}}$

$=  0.484 \times 10^{-11}m$
$=  4.84 \times 10^{-12}m$

Compton effect formula

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

$\lambda _{f}-1.0 A^{\circ} = \dfrac{6.63\times 10^{-34}}{9.1\times 10^{-31}\times 3\times 10^{8}}\ (1-cos  90^{\circ})$

$\lambda _{f} = 1  A^{\circ}+ \dfrac{6.63\times 10^{-34}}{9.1\times 10^{-31}\times 3\times 10^{8}}  (1-0)$             $(\because  cos  90^{\circ}=0)$

$= 1  A^{\circ}+  0.24 \times 10^{-11}m$
$= 1  A^{\circ}+  0.024  A^{\circ}   (\because  10^{-10}m= 1  A^{\circ})$
$= 1.024  A^{\circ}$

$\lambda -\lambda \ '=\dfrac{h}{m _{e}c}(1-cos\theta )$

Compton effect

When a light wave (Photon) is incident on an electron, there is a decrease in energy of a photon as a part of its initial energy is transferred to the electron which is scattered. This effect is called Compton effect.