Answer
The wavelength of the electron is $\lambda\approx10.59pm$
Work Step by Step
The kinetic energy of the electron is $E_K=13.4keV=1.34\times10^4eV$
$1eV\approx1.602\times10^{-19}J$
So, $E_K=(1.34\times10^4eV)\times(1.602\times10^{-19}J)\approx2.147\times10^{-15}J$
1) Calculate the velocity of the electron
The velocity of the electron is calculated according to the formula $$E_K=\frac{1}{2}mv^2$$$$v^2=\frac{2E_K}{m}$$
- Mass of the electron $m\approx9.109\times10^{-31}kg$
So, $$v^2=\frac{2\times(2.147\times10^{-15}J)}{9.109\times10^{-31}kg}\approx4.714\times10^{15}m/s$$$$v=6.866\times10^7m/s$$
2) Find the wavelength of the electron $(\lambda)$
The wavelength of the electron (matter) is calculated as follows $$\lambda=\frac{h}{mv}$$$$\lambda=\frac{6.626\times10^{-34}J.s}{(9.109\times10^{-31}kg)(6.866\times10^7m/s)}$$$$\lambda\approx1.059\times10^{-11}m\approx10.59pm$$
