Question

Suppose ${10}^{-17}J$ of light energy is needed by the interior of the human eye to see an object. How many photons of green light $(\lambda =550nm)$ are needed to generate this minimum amount of energy?

Answer 1

$Explanation:$

From the quantum theory of radiation,

The energy of n photon $=\frac{nhc}{\lambda }$

where,

$n=$ No of photon

$h=$Planck's constant$=6.625\times {10}^{-34}Js$

$c=$Velocity of light$=3\times {10}^{8}m/s$

$\lambda =$Wavelength$=550nm$

$\frac{nhc}{\lambda }=\frac{n\times 6.625\times {10}^{-34}\times 3\times {10}^8}{550}=n\times 0.0361J$

$Numberofphotons\left(n\right)=\frac{Visiblelightenergy}{Energyofaphoton}$

$Energyofaphoton=0.0361\times {10}^{-17}$

$Visiblelightenergy={10}^{-17}\left(given\right)$

$Numberofphotons\left(n\right)=\frac{{10}^{-17}}{0.0361\times {10}^{-17}}$

$Hencethenumberofphotonsare28$