Question

What is the energy of a typical visible photon? How many photons enter the eye pre second when one looks at a weak source of light such as the moon who's intensity is about 4.0e-4 w/m 2? Assume a typical visible wavelength be 550nm and diameter of the eye pupil is 8.0 mm

Answer 1

The wavelength of visible light is given by = 550nm

The energy of a single photon is E = hf = hc /

  Since h has the dimen­sion energy * time, hc has the dimension energy *length and is conveniently expressed in eV×nm, as follows                                                            

hc = (6.63*10^-34 Js) * (3.00*10⁸ m/s)*(1 eV / 1.60*10^-19J)

hc = 1.24*10^-6 eV.m

or

hc = 1240 eV.nm

now we find the energy for a typical visible photon to be

                            E = hc / = 1240 eV.nm / 550 nm = 2.3 eV      

On the atomic level this energy is significant, but by everyday standards it is extremely small.

When we look at the moon, the energy entering our eye per second is given by IA,

where I is the intensity = 4.0*10^-4w/m 2 and

the diameter of the pupil to be about d = 8 mm = 8*10^-3m

area A = pi*(d/2)² = 5.02*10^-5 m²

A is the area of the pupil = 5.02*10^-5 m²,

Thus the number of photons entering our eye per second is = I*A / E

= (4.0*10^-4 w/m² * 5.02*10^-5 m²) / (2.3*1.6*10^-19 J)

= 5.45*10¹⁰ photons per second