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# Sun gives light at rate of $\;1400 Wm^{-2}\;$ of area perpendicular to direction of light . Assume $\;\lambda\;$ (sun light )= $\;6000 A^{0}\;.$ Then no.of Photons/sec arriving at $\;1m^2\;$ area at that part of earth is

$(a)\;4.2\times10^{20}\qquad(b)\;4.2\times10^{22}\qquad(c)\;4.2\times10^{21}\qquad(d)\;4.2\times10^{23}$

Answer : (b) $\;4.2\times10^{22}$
Explanation :
$\varepsilon =\;$ energy of photon =$h \nu=\large\frac{hC}{\lambda} \quad \; (C=3\times10^{8} m/s)$
Let n be no . of photon received/sec per unit area
$\eta=\large\frac{IA}{\large\frac{\varepsilon}{energy of photon}}=\large\frac{(1400\times1)\times(6000\times10^{-10})}{6.6\times10^{-34}\times3\times10^{8}}$
$=4.2\times10^{21}\;.$

edited Nov 11, 2014