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shutvik [7]
3 years ago
15

Light is shining perpendicularly on the surface of the earth with an intensity of 680 W/m^2. Assuming that all the photons in th

e light have the same wavelength (in vacuum) of 678 nm, determine the number of photons per second per square meter that reach the earth.
Physics
1 answer:
andrew11 [14]3 years ago
4 0

Answer:

3.066×10^21  photons/(s.m^2)

Explanation:

The power per area is:

Power/A = (# of photons /t /A)×(energy / photon)

E/photons = h×c/(λ)

photons /t /A = (Power/A)×λ /(h×c)  

photons /t /A = (P/A)×λ/(hc)

photons /t /A = (680)×(678×10^-9)/(6.63×10^-34)×(3×10^-8)

                      = 3.066×10^21

Therefore, the number of photons per second per square meter 3.066×10^21  photons/(s.m^2).

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