Ultraviolet light of wavelength 300 nm and i | vedclass

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(C) The intensity of light $I$ is given by $I = \frac{n E}{A t}$,where $n$ is the number of photons,$E = \frac{hc}{\lambda}$ is the energy of one phot

Vedclass · Accepted Answer

$1.51 	imes 10^{12} 	ext{ per sec}$

Vedclass · Answer

(C) The intensity of light $I$ is given by $I = \frac{n E}{A t}$,where $n$ is the number of photons,$E = \frac{hc}{\lambda}$ is the energy of one photon,$A$ is the area,and $t$ is time.The number of incident photons per second is $N = \frac{I A}{E} = \frac{I A \lambda}{hc}$.Given:$I = 1.0 \ W/m^2$$A = 1.0 \ cm^2 = 1.0 	imes 10^{-4} \ m^2$$\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{1.0 	imes 10^{-4} 	imes 300 	imes 10^{-9}}{6.63 	imes 10^{-34} 	imes 3 	imes 10^8} = \frac{3 	imes 10^{-11}}{19.89 	imes 10^{-26}} \approx 1.508 	imes 10^{14} 	ext{ photons/sec}$.Since $1\%$ of the incident photons produce photoelectrons,the number of photoelectrons emitted per second is:$N_e = 0.01 	imes N = 0.01 	imes 1.508 	imes 10^{14} = 1.508 	imes 10^{12} 	ext{ per sec}$.Thus,the correct option is $C$.

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

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