Question

X-rays with an energy of 265 keV undergo Compton scattering from a target. If the scattered rays are deflected at 41.0° relative to the direction of the incident rays, find each of the following. (a) the Compton shift at this angle _________nm (b) the energy of the scattered x-ray __________keV (c) the kinetic energy of the recoiling electron ___________keV

Answer 1

Answers:

(a) Compton shift

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.

$\theta=41\°$ the angle between incident phhoton and the scatered photon.

We are told the scattered X-rays (photons) are deflected at $41\°$:

$\Delta \lambda=\lambda' - \lambda_{o}=\lambda_{c}(1-cos(41\°))$   (2)

$\Delta \lambda=\lambda' - \lambda_{o}=5.950(10)^{-13}m$   (3)

But we are asked to express this in $nm$, so:

$\Delta \lambda=5.950(10)^{-13}m.\frac{1nm}{(10)^{-9}m}$  

$\Delta \lambda=0.000595nm$  (4)

(b) the energy of the scattered x-ray

The initial energy $E_{o}=265keV=265(10)^{3}eV$ of the photon is given by:

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

From this equation (5) we can find the value of $\lambda_{o}$:

$\lambda_{o}=\frac{h.c}{E_{o}}$    (6)

$\lambda_{o}=\frac{(4.136(10)^{-15}eV.s)(3(10)^{8}m/s)}{265(10)^{3}eV}$    

$\lambda_{o}=4.682(10)^{-12}m$    (7)

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

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

Then:

$\lambda'=\Delta \lambda+\lambda_{o}$  (8)

$\lambda'=5.950(10)^{-13}m+4.682(10)^{-12}m$  

$\lambda'=5.277(10)^{-12}m$  (9)

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

$E'=\frac{h.c}{\lambda'}$    (10)

$E'=\frac{(4.136(10)^{-15}eV.s)(3(10)^{8}m/s)}{5.277(10)^{-12}m}$    

$E'=235.121keV$    (11) This is the energy of the scattered photon

(c) Kinetic energy of the recoiling electron

If we want to know the kinetic energy of the recoiling electron $E_{e}$, we have to calculate all the energy lost by the photon in the wavelength shift, which is:

$K_{e}=E_{o}-E'$  (12)

$K_{e}=265keV-235.121keV$  

Finally we obtain the kinetic energy of the recoiling electron:

$E_{e}=29.878keV$  

Answer 2

Answer:

The first one:

the energy of the scattered x-ray

The answer for last on:

Kinetic energy of the recoiling electron