91Ó°ÊÓ

Use the formula then it gives;

$\begin{array}{c}\mathrm{Δ}\mathrm{λ}=\frac{h}{m_{e}c}(1−\cos \mathrm{Ï•})\\ =2.43(1−\cos {90}^{∘})\\ =2.43\text{ p³¾}\end{array}$

Hence, the Compton shift is .$2.43\text{ p³¾}$

The dfractional shift should be interpreted as $\mathrm{Δ}\mathrm{λ}$divided by the original wavelength as follows:

$\begin{array}{c}\frac{\mathrm{Δ}\mathrm{λ}}{\mathrm{λ}}=\frac{2.425\text{ }pm}{590\text{ }nm}\\ =4.11×{10}^{−6}\end{array}$

Hence, the fractional shift is$4.11×{10}^{−6}$ .

The change in energy for a photon with $\mathrm{λ}=590\text{ }nm$is given by:

$\begin{array}{c}\mathrm{Δ}{E}_{ph}=\mathrm{Δ}\left(\frac{hc}{\mathrm{λ}}\right)\\ ≈−\frac{hc\mathrm{Δ}\mathrm{λ}}{{\mathrm{λ}}^2}\\ =−\frac{(4.14×{10}^{−15}eVâ‹\dots s)(2.998×{10}^8\text{m/s})\left(2.43\text{​â¶Ä‰}pm\right)}{{\left(590\text{ }nm\right)}^2}\\ =−8.67×{10}^{−6}\text{ e³Õ}\end{array}$

Hence, the change in photon energy is $−8.67×{10}^{−6}\text{ e³Õ}$.

For an x-ray photon of energy $\mathrm{Δ}{E}_{ph}=50keV$, $\mathrm{Δ}\mathrm{λ}$remains the same i.e, 2.43 pm, since it is independent of $E_{ph}$.

Hence, $\mathrm{Δ}\mathrm{λ}$remains the same as 2.43 pm.

The fractional change in wavelength is solved as follows:

$\begin{array}{c}\frac{\mathrm{Δ}\mathrm{λ}}{\mathrm{λ}}=\frac{\mathrm{Δ}\mathrm{λ}}{hc/E_{ph}}\\ =\frac{(50×{10}^{3}eV)\left(2.43\text{ p³¾}\right)}{(4.14×{10}^{−15}eVâ‹\dots s)(2.998×{10}^8\text{m/s})}\\ =9.78×{10}^{−2}\end{array}$

Hence, the value of fractional change in wavelength is $9.78×{10}^{−2}$.

The change in photon energy is solved as follows:

$\begin{array}{c}\mathrm{Δ}{E}_{ph}=hc\left(\frac{1}{\mathrm{λ}+\mathrm{Δ}\mathrm{λ}}−\frac{1}{\mathrm{λ}}\right)\\ =−\left(\frac{hc}{\mathrm{λ}}\right)\left(\frac{\mathrm{Δ}\mathrm{λ}}{\mathrm{λ}+\mathrm{Δ}\mathrm{λ}}\right)\\ =−E_{ph}\left(\frac{\mathrm{Î\pm }}{1+\mathrm{Î\pm }}\right)\end{array}$

Here,$\mathrm{Î\pm }=\frac{\mathrm{Δ}\mathrm{λ}}{\mathrm{λ}}$. With $E_{ph}=50keV$and $\mathrm{Î\pm }=9.78×{10}^{−2}$thus, it gives:

$\mathrm{Δ}{E}_{ph}=−4.45kev$

Hence, the change in photon energy is $−4.45kev$.