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

The Compton Shift [tex]\Delta \lambda[/tex] in wavelength when the photons are scattered is given by the following equation:

[tex]\Delta \lambda=\lambda' - \lambda_{o}=\lambda_{c}(1-cos\theta)[/tex]     (1)

Where:

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

[tex]\theta=41\°[/tex] the angle between incident phhoton and the scatered photon.

We are told the scattered X-rays (photons) are deflected at [tex]41\°[/tex]:

[tex]\Delta \lambda=\lambda' - \lambda_{o}=\lambda_{c}(1-cos(41\°))[/tex]   (2)

[tex]\Delta \lambda=\lambda' - \lambda_{o}=5.950(10)^{-13}m[/tex]   (3)

But we are asked to express this in [tex]nm[/tex], so:

[tex]\Delta \lambda=5.950(10)^{-13}m.\frac{1nm}{(10)^{-9}m}[/tex]  

[tex]\Delta \lambda=0.000595nm[/tex]  (4)

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

 [tex]E_{o}=\frac{h.c}{\lambda_{o}}[/tex]    (5)

From this equation (5) we can find the value of [tex]\lambda_{o}[/tex]:

[tex]\lambda_{o}=\frac{h.c}{E_{o}}[/tex]    (6)

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

[tex]\lambda_{o}=4.682(10)^{-12}m[/tex]    (7)

Knowing the value of [tex]\Delta \lambda[/tex] and [tex]\lambda_{o}[/tex], let's find [tex]\lambda'[/tex]:

[tex]\Delta \lambda=\lambda' - \lambda_{o}[/tex]

Then:

[tex]\lambda'=\Delta \lambda+\lambda_{o}[/tex]  (8)

[tex]\lambda'=5.950(10)^{-13}m+4.682(10)^{-12}m[/tex]  

[tex]\lambda'=5.277(10)^{-12}m[/tex]  (9)

Knowing the wavelength of the scattered photon [tex]\lambda'[/tex]  , we can find its energy [tex]E'[/tex] :

[tex]E'=\frac{h.c}{\lambda'}[/tex]    (10)

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

[tex]E'=235.121keV[/tex]    (11) This is the energy of the scattered photon

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

[tex]K_{e}=E_{o}-E'[/tex]  (12)

[tex]K_{e}=265keV-235.121keV[/tex]  

Finally we obtain the kinetic energy of the recoiling electron:

[tex]E_{e}=29.878keV[/tex]  

Answer:

The first one:

the energy of the scattered x-ray

The answer for last on:

Kinetic energy of the recoiling electron

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