First, we need to convert electron's kinetic energy into Joules. Keeping in mind that
[tex]1 eV=1.6 \cdot 10^{-19}J[/tex]
we have
[tex]E=3 eV \cdot 1.6 \cdot 10^{-19} J/eV=4.8 \cdot 10^{-19} J[/tex]
The kinetic energy of the electron is equal to:
[tex]E= \frac{1}{2}mv^2 [/tex]
where m is the electron mass and v its speed. If we re-arrange this equation, we can find the electron speed:
[tex]v= \sqrt{ \frac{2E}{m} }= \sqrt{ \frac{2 \cdot 4.8 \cdot 10^{-19} J}{9.1 \cdot 10^{-31} kg} } =1.03 \cdot 10^6 m/s [/tex]
And now we can use De Broglie's relationship to find the electron's wavelength:
[tex]\lambda= \frac{h}{p} [/tex]
where h is the Planck constant and p=mv is the electron momentum. Substituting numbers, we get
[tex]\lambda= \frac{h}{mv}= \frac{6.6 \cdot 10^{-34} Js}{(9.1 \cdot 10^{-31}kg)(1.03 \cdot 10^6 m/s)}=7.04 \cdot 10^{-10} m [/tex]
