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

Energy of n photons $=\frac{nhc}{\lambda }$

Where, $h=6.625\times {10}^{-34}Joulesec$ , $c=3\times {10}^{8}mse{c}^{-1}$ , $\lambda =550nm$ and n = number of photons

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

Number of photons (n) $=\frac{{10}^{-17}}{Energyofaphoton}$

Energy of a photon $=0.0361\times {10}^{-17}Joule$

Number of photons $=27.668\approx 28photons$

Hence, number of photons are 28.