Light is shining perpendicularly on the surface of the earth with an intensity of 680 w/m^2. assuming that all the photons in the light have the same wavelength (in vacuum) of 678 nm, determine the number of photons per second per square meter that reach the earth.

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).

probably not the "magnets"in the magnet, so to speak, are thought to be electrons orbiting in shells. vurrent loops two sides to the loop