The de-Broglie wavelength of an electron moving with kinetic energy of 144 eV is nearly

The formula of de-Broglie wavelength, $\lambda =\frac{h}{\sqrt{2mK.E.}}$ and we know that, $K.E.=eV$. Recall the proper relation between wavelength and electric potential, in angstrom,  $\lambda =\frac{12.27}{\sqrt{V}}\mathrm{\AA }$$=\frac{12.27}{\sqrt{144}}\times {10}^{-10}$$=1.02\times {10}^{-10} \mathrm{m}$

it means the wavelength, $\lambda =102\times {10}^{-3} \mathrm{nm}$.