The momentum of a photon is given by the following relationship:
[tex]p= \frac{h}{\lambda} [/tex]
where
h is the Planck constant
[tex]\lambda[/tex] is the photon wavelength
For the photon in our problem, [tex]\lambda=0.20 nm = 0.20 \cdot 10^{-9}m[/tex], so its momentum is
[tex]p= \frac{h}{\lambda}= \frac{6.6 \cdot 10^{-34} Js}{0.20 \cdot 10^{-9} m}=3.3 \cdot 10^{-24} kg m/s [/tex]
The electron must have the same momentum of this photon, and its momentum is given by (in the non-relativistic approximation)
[tex]p=mv[/tex]
where
m is the electron mass
v is its speed
Re-arranging this formula, we can calculate the electron speed:
[tex]v= \frac{p}{m}= \frac{3.3 \cdot 10^{-24} kg m/s}{9.1 \cdot 10^{-31} kg} =3.6 \cdot 10^6 m/s [/tex]
And this velocity is quite small compared to the speed of light ([tex]c=3 \cdot 10^8 m/s[/tex]), so the non-relativistic approximation that we used for the electron's momentum is valid.
