91影视

Use the formula then it gives;

$\begin{array}{c}螖位=\frac{h}{m_{e}c}(1鈭�\cos 蠒)\\ =2.43(1鈭�\cos {90}^{鈭�})\\ =2.43\text{鈥塸尘}\end{array}$

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

The dfractional shift should be interpreted as $螖位$divided by the original wavelength as follows:

$\begin{array}{c}\frac{螖位}{位}=\frac{2.425\text{鈥�}pm}{590\text{鈥�}nm}\\ =4.11脳{10}^{鈭�6}\end{array}$

Hence, the fractional shift is$4.11脳{10}^{鈭�6}$ .

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

$\begin{array}{c}螖E_{ph}=螖\left(\frac{hc}{位}\right)\\ 鈮�鈭�\frac{hc螖位}{{位}^2}\\ =鈭�\frac{(4.14脳{10}^{鈭�15}eV鈰�s)(2.998脳{10}^8\text{m/s})\left(2.43\text{鈥嬧赌�}pm\right)}{{\left(590\text{鈥�}nm\right)}^2}\\ =鈭�8.67脳{10}^{鈭�6}\text{鈥塭痴}\end{array}$

Hence, the change in photon energy is $鈭�8.67脳{10}^{鈭�6}\text{鈥塭痴}$.

For an x-ray photon of energy $螖E_{ph}=50keV$, $螖位$remains the same i.e, 2.43 pm, since it is independent of $E_{ph}$.

Hence, $螖位$remains the same as 2.43 pm.

The fractional change in wavelength is solved as follows:

$\begin{array}{c}\frac{螖位}{位}=\frac{螖位}{hc/E_{ph}}\\ =\frac{(50脳{10}^{3}eV)\left(2.43\text{鈥塸尘}\right)}{(4.14脳{10}^{鈭�15}eV鈰�s)(2.998脳{10}^8\text{m/s})}\\ =9.78脳{10}^{鈭�2}\end{array}$

Hence, the value of fractional change in wavelength is $9.78脳{10}^{鈭�2}$.

The change in photon energy is solved as follows:

$\begin{array}{c}螖E_{ph}=hc\left(\frac{1}{位+螖位}鈭�\frac{1}{位}\right)\\ =鈭�\left(\frac{hc}{位}\right)\left(\frac{螖位}{位+螖位}\right)\\ =鈭�E_{ph}\left(\frac{伪}{1+伪}\right)\end{array}$

Here,$伪=\frac{螖位}{位}$. With $E_{ph}=50keV$and $伪=9.78脳{10}^{鈭�2}$thus, it gives:

$螖E_{ph}=鈭�4.45kev$

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