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(C) Given energy $E = 2 \ cal/cm^2 \cdot min$. Converting to $SI$ units: $E = 2 	imes 4.2 \ J/cm^2 \cdot min = 8.4 \ J/cm^2 \cdot min$.The energy of

Vedclass · Accepted Answer

$2.3 	imes 10^{19}$

Vedclass · Answer

(C) Given energy $E = 2 \ cal/cm^2 \cdot min$. Converting to $SI$ units: $E = 2 	imes 4.2 \ J/cm^2 \cdot min = 8.4 \ J/cm^2 \cdot min$.The energy of a single photon is given by $\epsilon = \frac{hc}{\lambda}$.The total energy $E$ incident on the surface is $E = n \epsilon = n \frac{hc}{\lambda}$,where $n$ is the number of photons.Rearranging for $n$: $n = \frac{E \lambda}{hc}$.Substituting the values: $n = \frac{8.4 	imes 5500 	imes 10^{-10}}{6.6 	imes 10^{-34} 	imes 3 	imes 10^8}$.$n = \frac{46200 	imes 10^{-10}}{19.8 	imes 10^{-26}} = \frac{4.62 	imes 10^{-6}}{1.98 	imes 10^{-25}} \approx 2.33 	imes 10^{19}$.Thus,the number of photons is $2.3 	imes 10^{19} \ photons/cm^2 \cdot min$.

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