Visible light emitted from the sun that makes it through the Earth\'s atmosphere

Question

Visible light emitted from the sun that makes it through the Earth's atmosphere has a power of about 465 J/s per square meter. The roof of the Parker Building is approximately 5300 m^2. a. If this roof was completely covered with solar panels, what is the maximum amount of energy (in J) that can be harvested from the Sun during a two-hour class if it is assumed that the solar panels absorb visible light with 100% efficiency? b. If we assume that all of this visible light has an average wavelength of 510 nm, how many photons will be absorbed by the solar panels during this period of time?

Explanation / Answer

a)

1 hour = 3600 s

2hours = 7200 s

--

Amount of energy = Power (visible ligth emitted) x Area x Time x Efficiency

Amount of energy = 465 J/m^2s * 5300m^2 * 7200 s *1

Amount of energy = 17.74 x 10^9 J

b) n = E / Ep

E=17.74 x 10^9 J

Ep = Photon energy

Ep = hc /

Where h = plank's constant = 6.6 x 10^-34Js

          c = speed of light = 3 x 10^8 m / s

          = wavelength = 510 nm =510 x10^-9 m

Ep = hc /

Ep = (6.6 x 10^-34 J·s * 3 x 10^8 m / s) / 510 x10^-9 m

Ep = 3.88 x 10^-19 J

Finally

n=17.74 x 10^9 J/3.88 x 10^-19 J

n=4.57 x 10 ^28 photons

Related Questions

Navigate

• Browse (All)
• Browse V
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.