A) give the energy (in units of keV) and the magnitude of the momentum (in units

Question

                    A) give the energy (in units of keV) and the magnitude of the momentum (in units of keV/c) of the photon before the collision.                 

                    B) give the (total relativistic) energy (in units of keV) and the magnitude of the momentum (in units of keV/c) of the electron before the collision.                 

                    C) give the wave length of the photon after the collosion.                 

                    D) give the kinetic energy and the total relativistic energy of the electron after the collision.

Explanation / Answer

A)

energy E1=h*c/lamda=52.8keV.


initial momentum is pi=E/c=52.8 keV/c...

here c is speed of photon.....

B)relativistic energy =52.8keV....

momentum is pe=0 kgm/s



C) lambda ' - lambda = h * (1 - cos(theta) ) / me * c

lamda'=23.5+0.328=23.828pm.....
E2=h*c/lamda'=52.131keV.....




D)K.E of electron is E..=E1-E2=0.67 kev=670eV.....

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