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(B) The intensity $I$ is given as $100 \ W \ m^{-2}$. The energy of a single photon is $E = \frac{hc}{\lambda}$.Number of photons incident per unit ar

Vedclass · Accepted Answer

$6.04 	imes 10^{14}$

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(B) The intensity $I$ is given as $100 \ W \ m^{-2}$. The energy of a single photon is $E = \frac{hc}{\lambda}$.Number of photons incident per unit area per unit time $(N)$ is given by $N = \frac{I}{E} = \frac{I \lambda}{hc}$.Substituting the values: $N = \frac{100 	imes 300 	imes 10^{-9}}{6.63 	imes 10^{-34} 	imes 3 	imes 10^8} \approx 1.51 	imes 10^{20} \ m^{-2} \ s^{-1}$.Given that $2 \%$ of incident photons produce photoelectrons,the number of photoelectrons emitted per unit area per unit time is $n' = 0.02 	imes N = 0.02 	imes 1.51 	imes 10^{20} = 3.02 	imes 10^{18} \ m^{-2} \ s^{-1}$.For an area $A = 2 \ cm^2 = 2 	imes 10^{-4} \ m^2$,the number of photoelectrons emitted per second is $n = n' 	imes A = 3.02 	imes 10^{18} 	imes 2 	imes 10^{-4} = 6.04 	imes 10^{14}$.

Radiation of wavelength $300 \ nm$ and intensity $100 \ W \ m^{-2}$ falls on the surface of a photosensitive material. If $2 \%$ of the incident photons produce photoelectrons,the number of photoelectrons emitted from an area of $2 \ cm^2$ of the surface per second is nearly:

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