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A 649keV x-ray photon scatters from an electron at rest, deflecting through 104 (DEGREES). What is the scattered photon's energy?
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The wavelength of the photon is given as follows:

$\lambda =\frac{hc}{E}$

Here, $h$ is the Planck's constant, $c$ is the speed of light, and $E$ is the energy of the photon.

$\lambda =\frac{\left (6.63\times 10^{-34} \textup{ J}\cdot \textup{ s} \right )\left (3\times 10^{8} \textup{ m/s} \right )}{\left (649 \textup{ keV} \right )\left ( \frac{1.6\times 10^{-16} \textup{ J}}{1 \textup{ keV}} \right )}$

$=1.91544\times 10^{-12} \textup{ m}$

Therefore, the wavelength of the incident rays is $1.91544\times 10^{-12} \textup{ m}$ .

From Compton scattering, the scattered photon energy is

$\lambda '=\lambda +\frac{h}{mc}\left ( 1-\cos \theta \right )$

Here, m is the mass of the electron and $\theta$ is the scattering angle.

$\lambda '=\left (1.91544\times 10^{-12} \textup{ m} \right ) +\frac{\left (6.63\times 10^{-34} \textup{ J}\cdot \textup{s} \right )}{\left (9.1\times 10^{-31} \textup{ kg} \right )\left (3\times 10^{8} \textup{ m/s} \right )}\left ( 1-\cos\left ( 104^{\textup{o}} \right ) \right )$

$=4.93153\times 10^{-12} \textup{ m}$

The energy of the scattered photon energy is,

$E'=\frac{hc}{\lambda '}$

$E'=\frac{\left (6.63\times 10^{-34} \textup{ J}\cdot \textup{ s} \right )\left (3\times 10^{8} \textup{ m/s} \right )}{4.93153\times 10^{-12} \textup{ m}}$

$=\left (4.03323\times 10^{-14}\textup{ J} \right )\left ( \frac{1 \textup{ keV}}{1.6\times 10^{-16} \textup{ J}} \right )$

$=252.08\textup{ keV}$

Therefore, the scattered photon energy is $252.08\textup{ keV}$ .

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