Respuesta :
Answer:
a) [tex] \Delta \lambda = \lambda' -\lambda_o = \frac{h}{m_e c} (1-cos \theta)[/tex]
For this case we know the following values:
[tex] h = 6.63 x10^{-34} Js [/tex]
[tex] m_e = 9.109 x10^{-31} kg [/tex]
[tex] c = 3x10^8 m/s[/tex]
[tex] \theta = 37[/tex]
So then if we replace we got:
[tex] \Delta \lambda = \frac{6.63x10^{-34} Js}{9.109 x10^{-31} kg *3x10^8 m/s} (1-cos 37) = 4.885x10^{-13} m * \frac{1m}{1x10^{-15} m}= 488.54 fm[/tex]
b) [tex] \lambda_0 = \frac{hc}{E_0}[/tex]
With [tex] E_0 = 300 k eV= 300000 eV[/tex]
And replacing we have:
[tex] \lambda_0 = \frac{1240 x10^{-9} eV m}{300000eV}=4.13 x10^{-12}m = 4.12 pm[/tex]
And then the scattered wavelength is given by:
[tex] \lambda ' = \lambda_0 + \Delta \lambda = 4.13 + 0.489 pm = 4.619 pm[/tex]
And the energy of the scattered photon is given by:
[tex] E' = \frac{hc}{\lambda'}= \frac{1240x10^{-9} eVm}{4.619x10^{-12} m}=268456.37 eV - 268.46 keV[/tex]
c) [tex] E_f = E_0 -E' = 300 -268.456 kev = 31.544 keV[/tex]
Explanation
Part a
For this case we can use the Compton shift equation given by:
[tex] \Delta \lambda = \lambda' -\lambda_0 = \frac{h}{m_e c} (1-cos \theta)[/tex]
For this case we know the following values:
[tex] h = 6.63 x10^{-34} Js [/tex]
[tex] m_e = 9.109 x10^{-31} kg [/tex]
[tex] c = 3x10^8 m/s[/tex]
[tex] \theta = 37[/tex]
So then if we replace we got:
[tex] \Delta \lambda = \frac{6.63x10^{-34} Js}{9.109 x10^{-31} kg *3x10^8 m/s} (1-cos 37) = 4.885x10^{-13} m * \frac{1m}{1x10^{-15} m}= 488.54 fm[/tex]
Part b
For this cas we can calculate the wavelength of the phton with this formula:
[tex] \lambda_0 = \frac{hc}{E_0}[/tex]
With [tex] E_0 = 300 k eV= 300000 eV[/tex]
And replacing we have:
[tex] \lambda_0 = \frac{1240 x10^{-9} eV m}{300000eV}=4.13 x10^{-12}m = 4.12 pm[/tex]
And then the scattered wavelength is given by:
[tex] \lambda ' = \lambda_0 + \Delta \lambda = 4.13 + 0.489 pm = 4.619 pm[/tex]
And the energy of the scattered photon is given by:
[tex] E' = \frac{hc}{\lambda'}= \frac{1240x10^{-9} eVm}{4.619x10^{-12} m}=268456.37 eV - 268.46 keV[/tex]
Part c
For this case we know that all the neergy lost by the photon neds to go into the recoiling electron so then we have this:
[tex] E_f = E_0 -E' = 300 -268.456 kev = 31.544 keV[/tex]
Answer:
When scattered rays are detected
(a) the Compton shift at this angle
(b) the energy of the scattered x-ray
(c) the energy of the recoiling electron.
