Solar radiation falls on Earth’s surface at a rate of 1900 W/m^2 . Assuming that

Question

Solar radiation falls on Earth’s surface at a
rate of 1900 W/m^2
.
Assuming that the radiation has an average
wavelength of 560 nm, how many photons per
square meter per second fall on the surfaces?
The speed of light is 3 × 10^8 m/s and Planck’s
constant is 6.62607 × 10^34
J · s.
Answer in units of photon/(m^2· s)

Okay so using E = hf with h as planck's constant and f as frequency, I plugged in f = c/ with c = speed of light and being the wavelength.
Came out with E = hc/, some constant J per photon

I converted Watts into Joules per second
so I ended up with (1900 J/m^2·s)/(E J/photon), the units cancel to be photon per m^2 per second but the constant 1900/E seems to be incorrect. Am I using the right method?

Explanation / Answer

Solar power = 1900 W/m2 = 1900 J/m2.s

Wavelength = 560 nm = 560 x 10-9 m

Energy of one photon = E = hf = hc/wavelength (in meters)

= 6.62607 × 10-34 x 3 x 108/560 x 10-9

= 3.5497 x 10-19 J

Number of photons per square meter per second

= 1900/E = 1900/3.5497 x 10-19 = 5.353 x 1021

Answer: 5.353 x 1021 photon/(m2· s)

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