Question

Calculate the number of the photons falling per second on each square meter of the earth's surface directly below the sun.

Answer 1

Number of photons falling per square meter of earth's surface $n=\frac{intensity\times \lambda }{hc}$

Where h= plank's constant=6.63$\times {10}^{-34}$,c=speed of light in air=3$\times {10}^8$, and intensity=1.4$\times {10}^3$

Therefore $n=\frac{1.4\times {10}^3\times 5\times {10}^{-7}}{6.63\times {10}^{-34}\times 3\times {10}^8}=3.51\times {10}^{21}{s}^{-1}$

here wavelength$\lambda$=5000$\mathring{A}$=5$\times {10}^{-7}$

Thus number of photons falling on each square meter of area per unit second$=3.51\times {10}^{21}\times 4\pi r^2$

$=9.9\times {10}^{44}$;here r=1.5$\times {10}^{11}$