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 5.20 cm^2 area per second?

Answer 1

Answer:

The number of photons per second that strike the given area is 2.668 x 10⁸ photons/second

Explanation:

Given;

intensity of the sunlight, I = 2.00 kJ·s−1·m^−2

area of incident, A = 5.2 cm² = 5.2 x 10⁻⁴ m²

Energy of incident photons per second on the given area;

E = IA

E = (2000)( 5.2 x 10⁻⁴)

E = 1.04 J/s

Energy of a photon is given is by;

$E = \frac{hc}{\lambda} \\\\E = \frac{(6.626*10^{-34})(3*10^8)}{(510*10^{-9})}\\\\E = 3.898*10^{-19} \ J/photon$

The number of photons per second that strike the given area is;

$n = \frac{1.04 \ J/s}{3.898*10^{-19} \ J/photon} \\\\n = 2.668*10^{18} \ photons/second$

Therefore, the number of photons per second that strike the given area is 2.668 x 10⁸ photons/second