Answer
$9.98\times 10^{21}~photons/s$
Work Step by Step
In part (a), we found that solar energy arrives at the panel at a rate of $~~3.61~kW$
We can find the photon energy:
$E = \frac{hc}{\lambda}$
$E = \frac{(6.63\times 10^{-34}~J\cdot s)(3.0\times 10^8~m/s)}{550\times 10^{-9}~m}$
$E = 3.616\times 10^{-19}~J$
We can find the rate at which photons are absorbed:
$\frac{3.61\times 10^3~W}{3.616\times 10^{-19}~J/photon} = 9.98\times 10^{21}~photons/s$
