$(a)2\times10^{-3}J/m^2/sec\qquad(b)1\times10^{-2}J/m^2/sec\qquad(c)2\times10J/m^2/sec\qquad(d)1\times10J/m^2/sec$

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Energy falling per square meter per second = No. of quanta falling/$m^2/sec\times$ energy of one quanta

$= 6.37\times10^{15}\times\large\frac{hc}{\lambda}$

$=6.37\times10^{15}\times\large\frac{6.625\times10^{-34}\times3.0\times10^8}{632.8\times10^{-9}}$

$=2\times10^{-3} J/m^2/sec$

Hence answer is (a)

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