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.
1 answer:
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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