Question

Visible light has wavelengths in the range of 400 nm to 780 nm. Calculate the range of energy of the photons of visible light.

Answer 1

Given:
Range of wavelengths, ${\lambda }_1$ = 400 nm to ${\lambda }_2$ = 780 nm
Planck's constant, h = 6.63$\times$10^$-$34 Js
Speed of light, c = 3$\times$10⁸ m/s
Energy of photon,
$E=hv$
$\nu =\frac{c}{\lambda }$
$\therefore E=h\nu =\frac{hc}{\lambda }$
Energy $\left(E_1\right)$ of a photon of wavelength $\left({\lambda }_1\right)$:
$$\begin{gathered} {E}_1=\frac{hc}{{\lambda }_1} \\ =\frac{6.63\times {10}^{-34}\times 3\times {10}^8}{400\times {10}^{-9}} \\ =\frac{6.63\times 3}{4}\times {10}^{-9} \\ =4.97725\times {10}^{-19} \\ =5\times {10}^{-19}\mathrm{J} \end{gathered}$$

Energy (E₂) of a photon of wavelength (${\lambda }_2$):
$$\begin{gathered} {E}_2=\frac{6.63\times 3}{7.8}\times {10}^{-19} \\ =2.55\times {10}^{-9}\mathrm{J} \end{gathered}$$
So, the range of energy is 2.55 × 10⁻¹⁹ J to 5 × 10⁻¹⁹ J.