Question

X-rays with an energy of 400 keV undergo Compton scattering with a target. If the scattered X-rays are detected at \theta = 30^{\circ}θ=30 ​∘ ​​ relative to the incident X-rays, what is the energy of the recoiling electron? [Assume that energy is conserved in this interaction]

Answer 1

Answer: 37.937 keV

Explanation:

Photons have momentum, this was proved by he American physicist Arthur H. Compton after his experiments related to the scattering of photons from electrons (Compton Effect or Compton Shift). In addition, energy and momentum are conserved in the process.

In this context, the Compton Shift $\Delta \lambda$ in wavelength when the photons are scattered is given by the following equation:

$\Delta \lambda=\lambda' - \lambda_{o}=\lambda_{c}(1-cos\theta)$     (1)

Where:

$\lambda_{c}=2.43(10)^{-12} m$ is a constant whose value is given by $\frac{h}{m_{e}.c}$, being $h=4.136(10)^{-15}eV.s$ the Planck constant, $m_{e}$ the mass of the electron and $c=3(10)^{8}m/s$ the speed of light in vacuum.

$\theta=30\°$ the angle between incident phhoton and the scatered photon.

We are told the scattered X-rays (photons) are detected at $30\°$:

$\Delta \lambda=\lambda' - \lambda_{o}=\lambda_{c}(1-cos(30\°))$   (2)

$\Delta \lambda=\lambda' - \lambda_{o}=3.2502(10)^{-13}m$   (3)

Now, the initial energy $E_{o}=400keV=400(10)^{3}eV$ of the photon is given by:

 $E_{o}=\frac{h.c}{\lambda_{o}}$    (4)

From this equation (4) we can find the value of $\lambda_{o}$:

$\lambda_{o}=\frac{h.c}{E_{o}}$    (5)

$\lambda_{o}=\frac{(4.136(10)^{-15}eV.s)(3(10)^{8}m/s)}{400(10)^{3}eV}$    

$\lambda_{o}=3.102(10)^{-12}m$    (6)

Knowing the value of $\Delta \lambda$ and $\lambda_{o}$, let's find $\lambda'$:

$\Delta \lambda=\lambda' - \lambda_{o}$

Then:

$\lambda'=\Delta \lambda+\lambda_{o}$  (7)

$\lambda'=3.2502(10)^{-13}m+3.102(10)^{-12}m$  

$\lambda'=3.427(10)^{-12}m$  (8)

Knowing the wavelength of the scattered photon $\lambda'$  , we can find its energy $E'$ :

$E'=\frac{h.c}{\lambda'}$    (9)

$E'=\frac{(4.136(10)^{-15}eV.s)(3(10)^{8}m/s)}{3.427(10)^{-12}m}$    

$E'=362.063keV$    (10) This is the energy of the scattered photon

So, if we want to know the energy of the recoiling electron $E_{e}$, we have to calculate all the energy lost by the photon, which is:

$E_{e}=E_{o}-E'$  (11)

$E_{e}=400keV-362.063keV$  

Finally we obtain the energy of the recoiling electron:

$E_{e}=37.937keV$