An electron is accelerated through a potential difference $V$. According to the relation given in the notes, its de Broglie wavelength is

For an electron accelerated through potential difference $V$, the gained kinetic energy is $eV=\tfrac12 mv^2$. Using $\lambda=h/(mv)$ gives $\lambda=\dfrac{h}{\sqrt{2meV}}$. In the form quoted in the notes, this becomes $\lambda = 12.29/\sqrt{V}$ when suitable units are used.