From the theory of special relativity, \begin{align} E = \sqrt{p^2c^2 + m^2c^4}. \end{align} Since the photon is a massless particle, and \begin{align} E = pc. \end{align}

Then, \begin{align} \hbar &= \frac{h}{2\pi},\\ c &= \nu\lambda,\\ \omega &= 2\pi\nu,\\ \tilde{\nu} &= \frac{1}{\lambda} = \frac{\nu}{c},\\ k &= \frac{2\pi}{\lambda} = \frac{2\pi\nu}{c} = \frac{\omega}{c},\\ p &=h\tilde{\nu}={\hbar}k,\\ E &=h\nu=\hbar\omega=2\pi\hbar\nu=\frac{2\pi{\hbar}c}{\lambda},\\ E &=pc={\hbar}kc=\frac{2\pi{\hbar}c}{\lambda},\\ \end{align} where \begin{align} h &:\ \mathrm{Planck's\ constant},\\ \hbar &:\ \mathrm{Reduced\ Planck's\ constant},\\ c &:\ \mathrm{Speed\ of\ light},\\ \lambda &:\ \mathrm{Wavelength},\\ \nu &:\ \mathrm{Frequency},\\ \omega &:\ \mathrm{Angular\ frequency},\\ \tilde{\nu} &:\ \mathrm{Wave\ number},\\ k &:\ \mathrm{Angular\ wave\ number},\\ p &:\ \mathrm{Momentum},\\ E &:\ \mathrm{Energy}.\\ \end{align}

From the formula $E=\frac{2\pi{\hbar}c}{\lambda}$ and ${\hbar}c$ = 197.326 980 4… MeV fm (PDG - Physical Constants) \begin{align} E &=\frac{2\pi{\hbar}c}{\lambda}\\ &\sim\frac{2{\times}3.14159{\times}197.327\ \mathrm{eV\ nm}}{\lambda}\\ &\sim\frac{1240\ \mathrm{eV\ nm}}{\lambda}. \end{align}

• $\lambda$ = 700 nm –> $E$ = 1.8 eV