Light of wavelength 5000 mathring A and inte | vedclass

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Question

(D) Energy of a single photon $E = \frac{hc}{\lambda} = \frac{6.63 	imes 10^{-34} 	imes 3 	imes 10^8}{5000 	imes 10^{-10}} = 3.978 	imes 10^{-19}

Vedclass · Accepted Answer

$59 	imes 10^{17}$

Vedclass · Answer

(D) Energy of a single photon $E = \frac{hc}{\lambda} = \frac{6.63 	imes 10^{-34} 	imes 3 	imes 10^8}{5000 	imes 10^{-10}} = 3.978 	imes 10^{-19} \ J$. Intensity $I = 4.68 \ mW/cm^2 = 4.68 	imes 10^{-3} \ W / (10^{-4} \ m^2) = 46.8 \ W/m^2$. Number of incident photons per unit area per unit time $n_p = \frac{I}{E} = \frac{46.8}{3.978 	imes 10^{-19}} \approx 1.176 	imes 10^{20} 	ext{ photons}/(m^2 \cdot s)$. Number of photoelectrons emitted per unit area per unit time $n_e = n_p 	imes 	ext{efficiency} = 1.176 	imes 10^{20} 	imes 0.05 = 5.88 	imes 10^{18} 	ext{ electrons}/(m^2 \cdot s)$. Adjusting for the provided options,the closest value is $59 	imes 10^{17}$.

Light of wavelength $5000 \ \mathring A$ and intensity $4.68 \ mW/cm^2$ is incident on a photosensitive surface. If only $5\%$ of the incident photons produce photoelectrons,what is the number of photoelectrons emitted per unit area per unit time?

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