What are (a) the Compton shift $$\Delta \lambda$$, (b) the fractional Compton shift $$\Delta \lambda / \lambda,$$ and (c) the change $$\Delta E$$ in photon energy for light of wavelength $$\lambda=590$$ nm scattering from a free, initially stationary electron if the scattering is at $$90^{\circ}$$ to the direction of the incident beam? What are (d) $$\Delta \lambda,(\mathrm{e}) \Delta \lambda / \lambda,$$ and (f) $$\Delta E$$ for $$90^{\circ}$$ scattering for photon energy 50.0 keV (x-ray range)?

(a) As we know that,

$$\Delta \lambda=\dfrac{h}{m_{e} c}(1-\cos \phi)=(2.43 \mathrm{pm})\left(1-\cos 90^{\circ}\right)=2.43 \mathrm{pm}\\$$

(b) The dfractional shift should be interpreted as $$\Delta \lambda$$ divided by the original wavelength:

$$\dfrac{\Delta \lambda}{\lambda}=\dfrac{2.425 \mathrm{pm}}{590 \mathrm{nm}}=4.11 \times 10^{-6}\\$$

(c) The change in energy for a photon with $$\lambda=590 \mathrm{nm}$$ is given by

$$\Delta E_{\mathrm{ph}}=\Delta\left(\dfrac{h c}{\lambda}\right) \approx-\dfrac{h c \Delta \lambda}{\lambda^{2}}\\$$

$$\begin{array}{l}=-\dfrac{\left(4.14 \times 10^{-15} \mathrm{eV} \cdot \mathrm{s}\right)\left(2.998 \times 10^{8} \mathrm{m} / \mathrm{s}\right)(2.43 \mathrm{pm})}{(590 \mathrm{nm})^{2}} \\=-8.67 \times 10^{-6} \mathrm{eV}\end{array}\\$$

(d) For an x-ray photon of energy $$E_{\mathrm{ph}}=50 \mathrm{keV}, \Delta \lambda$$ remains the same $$(2.43 \mathrm{pm}),$$ since it is independent of $$E_{\mathrm{ph}}$$

(e) The dfractional change in wavelength is now

$$\dfrac{\Delta \lambda}{\lambda}=\dfrac{\Delta \lambda}{h c / E_{\mathrm{ph}}}=\dfrac{\left(50 \times 10^{3} \mathrm{eV}\right)(2.43 \mathrm{pm})}{\left(4.14 \times 10^{-15} \mathrm{eV} \cdot \mathrm{s}\right)\left(2.998 \times 10^{8} \mathrm{m} / \mathrm{s}\right)}=9.78 \times 10^{-2}\\$$

(f) The change in photon energy is now

$$\Delta E_{\mathrm{ph}}=h c\left(\dfrac{1}{\lambda+\Delta \lambda}-\dfrac{1}{\lambda}\right)=-\left(\dfrac{h c}{\lambda}\right) \dfrac{\Delta \lambda}{\lambda+\Delta \lambda}=-E_{\mathrm{ph}}\left(\dfrac{\alpha}{1+\alpha}\right)\\$$

where $$\alpha=\Delta \lambda / \lambda .$$ With $$E_{\mathrm{ph}}=50 \mathrm{keV}$$ and $$\alpha=9.78 \times 10^{-2},$$ we obtain $$\Delta E_{\mathrm{ph}}=-4.45 \mathrm{keV}$$

(Note that in this case $$\alpha \approx 0.1$$ is not close enough to zero so the approximation $$\Delta E_{\mathrm{ph}} \approx$$ $$h c \Delta \lambda / \lambda^{2}$$ is not as accurate as in the first case, in which $$\alpha=4.12 \times 10^{-6} .$$ In fact if one were to use this approximation here, one would get $$\Delta E_{\mathrm{ph}} \approx-4.89 \mathrm{keV},$$ which does not amount to a satisfactory approximation.)