[tex]7.1\cdot 10^{-11} J[/tex]
Explanation:
First of all, we have to calculate the energy of each photon, given by
[tex]E=\frac{hc}{\lambda}[/tex]
where
[tex]h=6.63\cdot 10^{-34} Js[/tex] is the Planck constant
[tex]c=3.0\cdot 10^8 m/s[/tex] is the speed of light
[tex]\lambda=3.35 mm=3.35\cdot 10^{-3} m[/tex] is the wavelength
Substitutitng,
[tex]E=\frac{(6.63\cdot 10^{-34})(3\cdot 10^8)}{3.35\cdot 10^{-3}}=5.94\cdot 10^{-23} J[/tex]
The problem says that the detector captures [tex]3.3\cdot 10^8[/tex] photons/second; since there are 3600 seconds in 1 hour, the number of photons detected per hour is
[tex]N=3.3\cdot 10^8 \cdot 3600 =1.19\cdot 10^{12}[/tex]
And so, the total energy of the photons detected in one hour is:
[tex]E_T=NE=(1.19\cdot 10^{12})(5.94\cdot 10^{-23})=7.1\cdot 10^{-11} J[/tex]
