Question

Find energy of photon of wavelength $4000\stackrel{o}{A}$.

Answer 1

Given, $\lambda =4000\stackrel{o}{A}$
$=4000\times {10}^{-10}$m
$=4\times {10}^{-7}$m
We know that,
$E=hv$
But, $c=v=v=\frac{c}{\lambda }$
$\therefore E=\frac{c}{\lambda }$ $[\therefore h=6.6\times {10}^{-34}]$
$E=\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{4\times {10}^{-7}}$J and C$=3\times {10}^4$
E$=4.95\times {10}^{-19}$J
The energy of photon is $4.95\times {10}^{-19}$J.