08 Mar '09 14:07>
Originally posted by sonhouse1 eV = 1.60217653(14) E-19 Joules - you have to divide by the charge on an electron. To get the energy of a photon in electon Volts we have E = (hc/e)/lambda, Since we´ve got lambda in nm we should multiply by a billion. So replace (hc) with (hc*1E+09/e) = 1239.84 which is how he gets his 1240 figure. So a 635nm photon has energy 1239.84/635 = 1.95 eV.
I took that quote to mean you are in a dark room and how much energy does it take to cause a visible flash of light. There wasn't much in the way of citations though. I'll have to see where I saw that one. I couldn't find the 2E-17 reference but did find this one, first talks about light in terms of electron volts then has a conversion chart at the end.
ht ...[text shortened]... e packet and inverting that you get 3.1E18 as the # of such packets to = one joule of energy.
Direct sunlight has an intensity of 120 W/m² so assuming the pupil is about 12 mm² we´ve got about 10 microwatts hitting the retina. Based on your figure of 3.1E18 photons for a joule this gives us of the order of 10^13 photons per second in direct sunlight. There are of the order of 150,000,000 receptors (wiki page on retina) so that gives us a few tens of thousands of photons per receptor per second in direct sunlight.