First we need to convert the electron's energy in Joule. Keeping in mind that
[tex]1 eV = 1.6 \cdot 10^{-19} J[/tex]
3 eV corresponds to
[tex]K=3 eV = 3 \cdot 1.6 \cdot 10^{-19}J=4.8 \cdot 10^{-19} J[/tex]
The kinetic energy of an electron is related to its momentum by the formula
[tex]K= \frac{p^2}{2m} [/tex]
where p is the electron's momentum, while m is its mass. Re-arranging the formula, we find
[tex]p= \sqrt{2Km}= \sqrt{2 (4.8 \cdot 10^{-19}J)(9.1 \cdot 10^{-31} kg)}=9.35 \cdot 10^{-25} kg m/s [/tex]
And then we can use De Broglie relationship to find its wavelength:
[tex]\lambda= \frac{h}{p}= \frac{6.6 \cdot 10^{-34}Js}{9.35 \cdot 10^{-25} kg m/s}=7.06 \cdot 10^{-10} m [/tex]
