Question

At noon on a clear day, sunlight reaches the earth\'s surface at Madison, Wisconsin, with an average intensity of approximately 2.00 kJ·s–1·m–2. If the sunlight consists of photons with an average wavelength of 510.0 nm, how many photons strike a 4.80 cm2 area per second?

Answer 1

Intensity of sunlight at given position is defined as power received per unit area

so here we can say

$I = 2 kJ/s*m^2$

area  on which photons are received is given as

$A = 4.80 cm^2 = 4.80 * 10^-4 m^2$

now we can find the power received due to sunlight

$P = I*A$

$P = 2* 10^3 * 4.80 * 10^-4$

$P = 0.96 Watt$

now we can say this power is due to photons that strikes on surface of earth

so here we can say

$P = N\frac{hc}{\lambda}$

given here that

$\lambda = 510 nm$

$0.96 = N\frac{6.6 * 10^{-34}* 3 * 10^8}{510*10^{-9}}$

$0.96 = N * 3.88 * 10^{-19}$

$N = \frac{0.96}{3.88*10^{-19}}$

$N = 2.47 * 10^{18}$

so it will strike 2.47 * 10^18 photons on given area per second