Question

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

Answer 1

${\lambda }_1=400nm$ to ${\lambda }_2=780nm$
$h=6.63\times {10}^{-34}j-s,$ $c=3\times {10}^{8}m/s,$

The energy is given as:

$E=hv=\frac{hc}{\lambda }$

$E_1=\frac{6.63\times {10}^{-34}\times 3\times {10}^8}{400\times {10}^{-9}}=\frac{6.63\times 3}{4}\times {10}^{-19}=5\times {10}^{-19}J$

$E_2=\frac{6.63\times 3}{7.8}\times {10}^{-19}=2.55\times {10}^{-19}J$

So, the range is $5\times {10}^{-19}J$ to $2.55\times {10}^{-19}J.$