Answer:
Approximately 3 photons
Explanation:
The energy of a photon at the peak of visual sensitivity is given by:
[tex]E=\frac{hc}{\lambda}[/tex]
where
h is the Planck constant
c is the speed of light
[tex]\lambda=550 nm = 5.5\cdot 10^{-7}m[/tex] is the wavelength of the photon
Substituting into the formula,
[tex]E_1=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^( m/s)}{5.50\cdot 10^{-7} m}=3.6\cdot 10^{-19} J[/tex]
This is the energy of one photon. The human eye can detect an amount of energy of
[tex]E=10^{-18} J[/tex]
So the amount of photons contained in this energy is
[tex]n=\frac{E}{E_1}=\frac{10^{-18} J}{3.6\cdot 10^{-19}J}=2.8 \sim 3[/tex]
so approximately 3 photons.
