X rays of wavelength 0.0100 nm are directed in the positive direction of an axis onto a target containing loosely bound electrons. For Compton scattering from one of those electrons, at an angle of $\mathbf{180}\mathbf{°}$,what are (a) the Compton shift, (b) the corresponding change in photon energy, (c) the kinetic energy of the recoiling electron, and (d) the angle between the positive direction of the axis and the electron’s direction of motion?

(a)

Use the Compton shift formula is:

$$\begin{gathered} \Delta \lambda =\frac{h}{m_{e}c}\left(1-\cos \varphi \right) \\ =\left(2.43\text{pm}\right)\left(1-\cos 180°\right) \\ =+4.86\text{pm} \end{gathered}$$

Hence, the Compton shift is +4.86 pm.

(b)

Let the value of $hc=1240\text{eV}·\text{nm}$.

Use the above value to find the change in photon energy is;

$$\begin{gathered} \Delta E=\frac{hc}{\lambda \text{'}}-\frac{hc}{\lambda } \\ =1240\left(\frac{1}{0.01 \text{nm}+4.86 \text{nm}}-\frac{1}{0.01 \text{nm}}\right) \\ =-40.6 \text{keV} \end{gathered}$$

Hence, the change in photon energy is $-40.6 \text{keV}$.

(c)

Use the conservation of energy, that the energy neither be created nor be destroyed it only be transformed from one energy to another.

From the conservation of energy,

$$\begin{gathered} \Delta K=-\Delta E \\ =40.6 \text{keV} \end{gathered}$$

Hence, the kinetic energy of recoiling electron is 40.6 keV.

(d)

The electron will move straight ahead after the collision, since it has acquired some of the forward linear momentum from the photon. Thus, the angle between +x and the direction of the electron’s motion is zero.

Hence, the angle between positive direction of +x axis and the electron direction of motion is zero.