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

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

$\eta=\large\frac{IA}{\large\frac{\varepsilon}{energy of photon}}=\large\frac{(1400\times1)\times(6000\times10^{-10})}{6.6\times10^{-34}\times3\times10^{8}}$

$=4.2\times10^{21}\;.$

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