When light of intensity 1 W/m^2 and waveleng | vedclass

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(B) Given: Intensity $I = 1 \ W/m^2$, Wavelength $\lambda = 5 	imes 10^{-7} \ m$, Area $A = 1 \ cm^2 = 10^{-4} \ m^2$.Power $P = I 	imes A = 1 	ime

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$0.4 \ \mu A$

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(B) Given: Intensity $I = 1 \ W/m^2$, Wavelength $\lambda = 5 	imes 10^{-7} \ m$, Area $A = 1 \ cm^2 = 10^{-4} \ m^2$.Power $P = I 	imes A = 1 	imes 10^{-4} \ W = 10^{-4} \ W$.Energy of one photon $E = \frac{hc}{\lambda} = \frac{6.63 	imes 10^{-34} 	imes 3 	imes 10^8}{5 	imes 10^{-7}} \approx 3.978 	imes 10^{-19} \ J$.Number of photons incident per second $n_p = \frac{P}{E} = \frac{10^{-4}}{3.978 	imes 10^{-19}} \approx 2.51 	imes 10^{14} \ s^{-1}$.Since $100$ photons emit $1$ electron, the number of electrons emitted per second $n_e = \frac{n_p}{100} = 2.51 	imes 10^{12} \ s^{-1}$.Photoelectric current $I_p = n_e 	imes e = 2.51 	imes 10^{12} 	imes 1.6 	imes 10^{-19} \ C/s \approx 4.0 	imes 10^{-7} \ A = 0.4 \ \mu A$.

When light of intensity $1 \ W/m^2$ and wavelength $5 \times 10^{-7} \ m$ is incident on a surface, it is completely absorbed. If $100$ photons emit one electron and the surface area is $1 \ cm^2$, what is the photoelectric current?

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