Energy of a photon with a wavelength of 450 nm is

According to Max Planck’s equation,

$\mathrm{E}=\frac{\mathrm{hc}}{\mathrm{\lambda }}\to \mathrm{eqn}\left(1\right)$

Here h is Planck’s constant and c is speed of light. 

The wave length of photon $\mathrm{\lambda }$ = 450nm

$\mathrm{h}=6.625\times {10}^{-27}\mathrm{ergs}.\sec$

$\mathrm{c}=3\times {10}^{10}\mathrm{cm}/\sec$

From equation 1, the energy of photon is

$\mathrm{E}=\frac{6.625\times {10}^{-27}\mathrm{ergs}.\sec \times 3\times {10}^{10}\mathrm{cm}.{\sec }^{-1}}{450\times {10}^{-7}\mathrm{cm}}$$\therefore \mathrm{E}=4.36\times {10}^{-12}\mathrm{ergs}$