Question

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

• A. 14
• B. 28
• C. 39
• D. 42

Answer 1

The correct option is B 28

Energy of n photons $=\frac{nhc}{\lambda }$
Where, $h=6.626\times {10}^{-34}Joulesec$ , $c=3\times {10}^{8}mse{c}^{-1}$ , $\lambda =550nm$ and n = number of photons
$\frac{nhc}{\lambda }=\frac{n\times 6.626\times {10}^{-34}\times 3\times {10}^8}{550\times {10}^{-9}}$ $=n\times 3.61\times {10}^{-19}Joule$
Number of photons (n) $=\frac{{10}^{-17}}{\text{Energy of a photon}}$
Energy of a photon $=0.0361\times {10}^{-17}Joule$
Number of photons $=27.668\approx 28photons$
Hence, number of photons are 28.