Homework Answers
5) given
intensity, I = 0.85 kW/m^2 = 850 W/m^2
area, A = 1.5*3.5 = 5.25 m^2
lamda = 590 nm
a) the amount of energy stricking the area in one second, E = I*A*t
= 850*5.25*1
= 4462.5 J
Energy of each photon, E_photon = h*c/lamda
= 6.626*10^-34*3*10^8/(590*10^-9)
= 3.37*10^-19 J
the number of photons stricking the solar panel per unit time, N = E/E_photon
= 4462.5/(3.37*10^-19)
= 1.32*10^24 <<<<<<<---------------Answer
b) when the panel is tilted 23 degrees away,
N = 1.32*10^24*cos(23)
= 1.21*10^24 <<<<<<<---------------Answer
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