The formula for finding the energy of a photon is;
#E_gamma=(hc)/lamda#
Where #h# is Planck's constant, #c# is the speed of light in a vacuum, and #lamda# is the wavelength of the photon. The energy of photons is typically expressed in terms of electron volts, #eV#, where #1 "eV" = 1.602*10^-19 "J"# is the amount of energy required to accelerate an electron through one volt.
Planck's constant in terms of electron volts is #4.136xx10^-15 "eV s"#, and the speed of light in a vacuum is #2.998 xx 10^8 "m/s"#, so the energy for a single photon is;
#E_gamma = ((4.136 xx 10^-15"eV s")(2.998*10^8 "m/s"))/(1.670 * 10^-6 "m")#
#E_gamma = .7425 "eV"#
We are given the rate, #nu = 9xx10^7 "/s"#, at which photons are received, so the rate at which energy is received can be stated as;
#p = E_gamma*nu#
#p=(.7425 "eV")(9*10^7"/s")#
#p=6.682xx10^7 "eV/s"#
Now we can calculate the total energy received over an hour. There are #3600# seconds in an hour.
#E_T = p* t#
#E_T = (6.682 xx 10^7 "eV/s")(3600 "s")#
#2.406*10^11 eV#
For comparison, this is about 1 billionth of the energy given off by a 100 watt light bulb each second.
