## College Physics (4th Edition)

We can find the energy of each photon: $E_p = \frac{hc}{\lambda}$ $E_p = \frac{(6.626\times 10^{-34}~J~s)(3.0\times 10^8~m/s)}{510\times 10^{-9}~m}$ $E_p = 3.9\times 10^{-19}~J$ We can find the minimum number of photons $N$: $N = \frac{I~t~A}{E_p}$ $N = \frac{I~t~\pi~r^2}{E_p}$ $N = \frac{(5.0\times 10^{-13}~W/m^2)(1~s)(\pi)(4.25\times 10^{-3}~m)^2}{3.9\times 10^{-19}~J}$ $N = 73$ The minimum number of photons per second that an owl eye can detect is 73.