Physics - Modern Physics Question with Solution | TestHub

Given, energy per unit area per second,$p=1.388\times {10}^3 \mathrm{W} {\mathrm{m}}^{-2}$

Let$n$be the number of photons incident on the earth per square metre. The wavelength of each photon =$550\times {10}^{-9} \mathrm{m}$

The energy of each photon,$\mathit{\text{E}}=\frac{\mathit{\text{hc}}}{\lambda }$

$=\frac{6\text{.}63\times 10^{-34}\times 3\times 10^8}{550\times 10^{-9}}=3\text{.}616\times 10^{-19}\text{ J}$

Number of photons incident on the earth's surface

$\mathit{\text{n}}=\frac{\mathit{\text{P}}}{\mathit{\text{E}}}=\frac{1\text{.}388\times 10^3}{3\text{.}616\times 10^{-19}}=3\text{.}838\times 10^{21}$

$=3.838\times {10}^{21} \mathrm{photon} {\mathrm{m}}^{-2} {\mathrm{s}}^{-1}$