Answer:
a) [tex]=4.84*10^{-12}[/tex]
b)[tex]= -2.76*10^{-14} J[/tex]
c)[tex]i.e -2.76*10^{-14} J[/tex]
d)= 0 and the direction of motion is equal to zero
Explanation:
a) compton shift
[tex]\Delta\lambda = \frac{h}{mc} (1-cos\theta)[/tex]
[tex]\Delta\lambda = \frac{6.626*10^{-34}}{9.11*10^{-11}3*10^8} (1-cos180)[/tex]
[tex]=4.84*10^{-12}[/tex]
b) the new wavelength
[tex]\lambda' = 10.0*10^{-12} +4.84^10^{-12}[/tex]
[tex]=14.84*10^{-12} m[/tex]
[tex]\Delta E = E' - E[/tex]
[tex]=hc[\frac{1}{\lambda'}-\frac{1}{\lambda}][/tex]
[tex]\Delta E = 6.626*10^{-34}*(3*10^8)[\frac{1}{14.84*10^{-12}}-\frac{1}{4.8*10^{-12}}][/tex]
[tex]= -2.76*10^{-14} J[/tex]
C)By conservation of energy, the kinetic energy of recoiling electron is equal to the magnitude of energy between the photon energy
[tex]i.e -2.76*10^{-14} J[/tex]
d) the angle between the positive direction of motion
[tex]sin\phy = \frac{\lambda_t sin\theta}{\lambda'}[/tex]
[tex] =\frac{2.43*10^{-12}sin180}{14.84*10^{-12}}[/tex]
= 0
the direction of motion is equal to zero.