Answer
The kinetic energy of each electron is $~390~eV$
Work Step by Step
Let $E_x = 20,000~eV$
Note that the momentum of each electron $p_e = \sqrt{2~m_e~E_e}$
To produce the same diffraction pattern, the de Broglie wavelength of the electron should be the same as the wavelength of the x-ray photons. We can find the kinetic energy of each electron:
$\lambda = \frac{hc}{E_x} = \frac{h}{p_e}$
$\frac{hc}{E_x} = \frac{h}{\sqrt{2~m_e~E_e}}$
$\sqrt{2~m_e~E_e} = \frac{E_x}{c}$
$2~m_e~E_e = \frac{E_x^2}{c^2}$
$E_e = \frac{E_x^2}{2~m_e~c^2}$
$E_e = \frac{[(20,000~eV)(1.6\times 10^{-19}~J/eV)]^2}{(2)~(9.1\times 10^{-31}~kg)~(3.0\times 10^8~m/s)^2}$
$E_e = (6.25\times 10^{-17}~J)(\frac{1~eV}{1.6\times 10^{-19}~J})$
$E_e = 390~eV$
The kinetic energy of each electron is $~390~eV$.
