Answer:
a) 4.04*10^-12m
b) 0.0209nm
c) 0.253MeV
Explanation:
The formula for Compton's scattering is given by:
[tex]\Delta \lambda=\lambda_f-\lambda_i=\frac{h}{m_oc}(1-cos\theta)[/tex]
where h is the Planck's constant, m is the mass of the electron and c is the speed of light.
a) by replacing in the formula you obtain the Compton shift:
[tex]\Delta \lambda=\frac{6.62*10^{-34}Js}{(9.1*10^{-31}kg)(3*10^8m/s)}(1-cos132\°)=4.04*10^{-12}m[/tex]
b) The change in photon energy is given by:
[tex]\Delta E=E_f-E_i=h\frac{c}{\lambda_f}-h\frac{c}{\lambda_i}=hc(\frac{1}{\lambda_f}-\frac{1}{\lambda_i})\\\\\lambda_f=4.04*10^{-12}m +\lambda_i=4.04*10^{-12}m+(0.0169*10^{-9}m)=2.09*10^{-11}m=0.0209nm[/tex]
c) The electron Compton wavelength is 2.43 Ă— 10-12 m. Hence you can use the Broglie's relation to compute the momentum of the electron and then the kinetic energy.
[tex]P=\frac{h}{\lambda_e}=\frac{6.62*10^{-34}Js}{2.43*10^{-12}m}=2.72*10^{-22}kgm\\[/tex]
[tex]E_e=\frac{p^2}{2m_e}=\frac{(2.72*10^{-22}kgm)^2}{2(9.1*10^{-31}kg)}=4.06*10^{-14}J\\\\1J=6.242*10^{18}eV\\\\E_e=4.06*10^{-14}(6.242*10^{18}eV)=0.253MeV[/tex]
