To show that \(\lambda\) = \(\frac{\text{h}}{\sqrt{2meV_o}}\) The de broglie wavelength \(\lambda\), is given by \(\lambda\) = \(\frac{\text{h}}{\text{p}}\) where p = mv \(\lambda\) = \(\frac{\text{h}}{\text{mv}}\) When an electron is accelerated through a potential difference V\(_0\), it gains kinetic energy given K.E.= eV\(_0\) The kinetic energy can also be expressed in terms of the mass m of the electron and its speed v: K.E = \(\frac{1}{2}\)mv\(^2\) K.E = eV\(_0\) = \(\frac{1}{2}\)mv\(^2\) v = \(\sqrt{\frac{2eV_o}{m}}\) Substituting v = \(\sqrt{\frac{2eV_o}{m}}\) into the de Broglie equation \(\lambda\) = \(\frac{\text{h}}{m \times \sqrt{\frac{2eV_o}{m}}}\) = \(\frac{\text{h}}{\sqrt{\frac{2eV_o m^2}{m}}}\) \(\lambda\) = \(\frac{\text{h}}{\sqrt{2meV_o}}\) Note: use the fact that \(\sqrt{m}\) x \(\sqrt{m}\) = m