Question

Suppose $3.1\times {10}^{-18}J$ energy is needed by the interior of the human eye to see an object. How many photons of light of $\lambda =400nm$ will be needed to see the object? $h=6.6\times {10}^{-34}Js$:

• A. seven
• B. six
• C. eight
• D. ten

Answer 1

Answer: (B)

six

The energy of one photon is $E=\frac{hc}{\lambda }=\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{400\times {10}^{-9}}=4.95\times {10}^{-19}$
The number of photons required is $\frac{3.1\times {10}^{-18}J}{4.95\times {10}^{-19}}=6.3\simeq 6$