The ratio of the energy of a photon with $\lambda = 150$ nm to that with $\lambda = 300$ nm is

Question

Dr. Abubacker · Accepted Answer

(a) According to Einstein, the energy of photon is given by
$E = hv = \frac{hc}{\lambda}$
Where $h$ is Planck’s constant, $c$ the speed of light and $\lambda\ the\ wavelength.$
$	herefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{E_{1}}{E_{2}} = \frac{\lambda_{2}}{\lambda_{1}}$
Given, $\lambda_{1} = 150\ nm\ ,\ \lambda = 300$nm
$	herefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{E_{1}}{E_{2}} = \frac{300}{150} = \frac{2}{1}\ $

Q. The ratio of the energy of a photon with $\lambda = 150$ nm to that with $\lambda = 300$ nm is

Explanation

Reviewed by Dr. Abubacker

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