The De Broglie wavelength of the electron is equal to
[tex]\lambda= \frac{h}{p} [/tex]
where h is the Planck constant and p is the electron's momentum. Since this wavelength must be equal to that of the x-ray,
[tex]\lambda=0.185 nm=0.185 \cdot 10^{-9} m[/tex]
we can re-arrange the previous equation to find the momentum of the electron:
[tex]p= \frac{h}{\lambda}= \frac{6.6 \cdot 10^{-34}Js}{0.185 \cdot 10^{-9} m}=3.57 \cdot 10^{-24} kg m s^{-1} [/tex]
The kinetic energy of the electron is equal to the square of the momentum divided by twice its mass:
[tex]E= \frac{p^2}{2m}= \frac{(3.57 \cdot 10^{-24}kgms^{-1})^2}{2 (9.1 \cdot 10^{-31} kg)} =6.99 \cdot 10^{-18} J [/tex]
When the electron is accelerated by a potential difference [tex]\Delta V[/tex], the energy it gains is
[tex]E=q \Delta V[/tex]
where q is the electron charge. Re-arranging the formula, we find the potential difference:
[tex]\Delta V= \frac{E}{q}= \frac{6.99 \cdot 10^{-18} J}{1.6 \cdot 10^{-19} C}=43.7 V [/tex]
