Answer:
Each second approximately [tex] 4.15\times10^{15} [/tex] photons hit the face
Explanation:
Emmited intensity of electromagentic waves is defined as:
[tex]I_{3m}=\frac{P}{A_{wave-front}} [/tex] (1)
with P the power and A_{wave-front} the surface of the sphere that defines the wave front at a given radial distance (r) from the source, this is:
[tex]A_{wave-front}=4\pi r^2 [/tex] (2)
Using (2) on (1):
[tex]I_{3m}=\frac{P}{4\pi d^2}=\frac{10}{4\pi 3^2} [/tex]
[tex]I_{3m}=0.088 \frac{W}{m^{2}} [/tex]
That is the intensity of ligth at 3 meters from the source so that it's the intensity the face absorbs so again using equation (1) but now for absorbed intensity:
[tex]I_{abs}=0.088 =\frac{P}{A_{face}} [/tex] (3)
Power is the energy over a perioid of time this is:
[tex]P=\frac{E}{t} [/tex] (4)
But energy of photons is:
[tex]E= \frac{nhc}{\lambda} [/tex] (5)
with n the number of photons, h Planck's constant,c velocity odf ligth and [tex] \lambda[/tex] the wavelength
Using (5) on (4) and (4) on (3):
[tex]0.088 =\frac{nhc}{\lambda A_{face} t} [/tex]
Solving for n
[tex]n=\frac{0.088 \lambda A_{face} t}{hc}=\frac{(0.088) (535\times10^{-9}) (0.0175)(1)}{(6.63\times10^{-34}) (3\times10^{8})} [/tex]
[tex] n=4.15\times10^{15} [/tex]
