Ultraviolet light of wavelength 300 nm and i | vedclass

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Question

(C) Given: Wavelength $\lambda = 300 \ nm = 300 	imes 10^{-9} \ m$,Intensity $I = 1.0 \ W/m^2$,Area $A = 1.0 \ cm^2 = 1.0 	imes 10^{-4} \ m^2$,Effic

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

Vedclass · Answer

(C) Given: Wavelength $\lambda = 300 \ nm = 300 	imes 10^{-9} \ m$,Intensity $I = 1.0 \ W/m^2$,Area $A = 1.0 \ cm^2 = 1.0 	imes 10^{-4} \ m^2$,Efficiency $\eta = 1\% = 0.01$.The power incident on the surface is $P = I 	imes A = 1.0 	imes 1.0 	imes 10^{-4} = 1.0 	imes 10^{-4} \ W$.The energy of a single photon is $E = \frac{hc}{\lambda} = \frac{6.63 	imes 10^{-34} 	imes 3 	imes 10^8}{300 	imes 10^{-9}} = 6.63 	imes 10^{-19} \ J$.The number of incident photons per second is $N_{ph} = \frac{P}{E} = \frac{1.0 	imes 10^{-4}}{6.63 	imes 10^{-19}} \approx 1.508 	imes 10^{14} \ s^{-1}$.The number of photoelectrons emitted per second is $N_e = N_{ph} 	imes \eta = 1.508 	imes 10^{14} 	imes 0.01 = 1.508 	imes 10^{12} \ s^{-1}$.Thus,the number of photoelectrons emitted is approximately $1.51 	imes 10^{12} \ s^{-1}$.

Ultraviolet light of wavelength $300 \ nm$ and intensity $1.0 \ W/m^2$ is incident on the surface of a photosensitive material. If $1\%$ of the incident photons produce photoelectrons,the number of photoelectrons emitted from an area of $1.0 \ cm^2$ of the surface is:

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