Asked by Tonya

Sunlight is measured at 1m^2 solar collector with a power of 1216W. Peak intensity of this sunlight has a wavelength of 5.94x10^-7m how many photons arriving each second
Answered by DrBob222
1 watt = 1 J/sec; therefore, 1216 W = 1216 J/sec
E = hc/wavelength
E = 6.626E-34 J*sec x 3E8 m/s/5.94E-9 m = J/photon =approx 3.35E-17 J/photon
3.35E-17 J/photon x # photons = 1216 J/sec so
# photons/sec = ?
check my work
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