Question

The intensity of electromagnetic radiation from the sun reaching the earth's upper atmosphere is 1.37kW/m2kW/m2. Part A Assuming an average wavelength of 680 nmnm for this radiation, find the number of photons per second that strike a 1.50 m2m2 solar panel directly facing the sun on an orbiting satellite.

Answer 1

Answer:

#_photon = 7  10²¹ photons

Explanation:

Let's look for the power that affects the panel of area of ​​1.5 m2

           I = P / A

           P = I A

           P = 1.37 10³  1.5

           P = 2,055 10³ W

           P = E / t

       

If we use t = 1 s

           E = P t

           E = 2,055 10³ J

This is the power that the panel receives, let's look for the energy of a photon

            E = h f

            c = λ f

            f = c /λ

            E = h c /λ

Let's calculate

            E₀ = 6.63 10⁻³⁴  3 10⁸/680 10⁻⁹

            E₀ = 2.925 10⁻¹⁹ J

In one second the total energy is the number of photons for the energy of each one

             E = #_photon  E₀

             #_photon = E / E₀

             #_photon = 2,055 10³ / 2,925 10⁻¹⁹

            #_photon = 7  10²¹ photons

Answer 2

A solar panel directly facing the sun on an orbiting satellite is #_photon = 7 10²¹ photons or$1.344*10^21 photons/sec.$

What is a Solar panel?

Let Area be S=1.50m2

Let Radiation be I=1370 W/m2

Let the Wavelength be L=680nm

1) The total power of light falling onto the surface is

P=I*S (measured in Watts or Joules per second) =2466W

2) The energy of a single photon is

E=hbar*w, where h bar is Planck const and w is the angular frequency.

w=2*pi*c/L, where c is the speed of light.= $2*pi*3*10^8/(680*10^-9)$ = $2.77*10^15$

E = 1.835 * 10^-18

3) The number of photons per sec is total energy per sec div by energy of a single photon:

N=P/E=I*S*L/(2*pi*hbar*c) = $2466/(1.835 * 10^-18)$

This would be $1.344*10^21 photons/sec$. or #_photon = 7 10²¹ photons

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