Ultraviolet light of wavelength 300 nm and i | vedclass

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(C) The intensity of incident light is given by $I = \frac{E}{At} = \frac{nhc}{At\lambda}$,where $n$ is the number of photons incident per second.The

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$1.51 	imes 10^{12} \ s^{-1}$

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(C) The intensity of incident light is given by $I = \frac{E}{At} = \frac{nhc}{At\lambda}$,where $n$ is the number of photons incident per second.The number of photons incident per second is $n = \frac{IA\lambda}{hc}$.Given that $1 \%$ of incident photons emit photoelectrons,the number of photoelectrons emitted per second $N$ is:$N = \frac{1}{100} 	imes n = \frac{1}{100} 	imes \frac{IA\lambda}{hc}$.Substituting the given values:$I = 1.0 \ W/m^2$,$A = 1 \ cm^2 = 10^{-4} \ m^2$,$\lambda = 300 \ nm = 300 	imes 10^{-9} \ m$,$h = 6.63 	imes 10^{-34} \ J \cdot s$,$c = 3 	imes 10^8 \ m/s$.$N = \frac{1}{100} 	imes \frac{1.0 	imes 10^{-4} 	imes 300 	imes 10^{-9}}{6.63 	imes 10^{-34} 	imes 3 	imes 10^8}$.$N = \frac{3 	imes 10^{-11}}{100 	imes 19.89 	imes 10^{-26}} \approx 1.51 	imes 10^{12} \ s^{-1}$.

Ultraviolet light of wavelength $300 \ nm$ and intensity $1.0 \ W/m^2$ is incident on a photosensitive surface. If only $1 \%$ of the incident photons emit photoelectrons,calculate the number of photoelectrons emitted per second from a surface area of $1 \ cm^2$.

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