The de Broglie wavelength of the most energetic photoelectrons emitted from a photosensitive metal of work function $\varphi$, when light frequency v incidents on it is $\lambda$. Then v=
(h-Planck's constant, m - mass of electron)

Since the de Broglie wavelength is given by $\lambda$, the momentum of the electrons will be given by $P_e=\frac{h}{\lambda }$.

So the kinetic energy of the most energetic photo-electrons is given by $K_e=\frac{P_e^2}{2m_e}=\frac{h^2}{2m_e{\lambda }^2}$.

From the photo electric equation $K_e=hv-\varphi$

$\Rightarrow \frac{h^2}{2m_e{\lambda }^2}=h\nu -\varphi$

$\Rightarrow \nu =\frac{\varphi }{\mathrm{h}}+\frac{\mathrm{h}}{2\mathrm{m}{\lambda }^2}$