A parallel beam of light of wavelength 900 , | vedclass

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(B) Given:Wavelength $\lambda = 900 	imes 10^{-9} \, m$Intensity $I = 100 \, W/m^2$Area $A = 1 \, cm^2 = 10^{-4} \, m^2$Time $t = 1 \, s$The energy c

Vedclass · Accepted Answer

$4.5 	imes 10^{16}$

Vedclass · Answer

(B) Given:Wavelength $\lambda = 900 	imes 10^{-9} \, m$Intensity $I = 100 \, W/m^2$Area $A = 1 \, cm^2 = 10^{-4} \, m^2$Time $t = 1 \, s$The energy crossing the area per second is $P = I 	imes A = 100 	imes 10^{-4} = 10^{-2} \, J/s$.The energy of a single photon is $E = \frac{hc}{\lambda}$.The number of photons $n$ crossing the area per second is given by $n = \frac{P}{E} = \frac{P \lambda}{hc}$.Substituting the values:$n = \frac{10^{-2} 	imes 900 	imes 10^{-9}}{6.63 	imes 10^{-34} 	imes 3 	imes 10^8}$$n = \frac{9 	imes 10^{-9}}{19.89 	imes 10^{-26}} \approx 0.452 	imes 10^{17} = 4.52 	imes 10^{16}$.Rounding to the nearest option,we get $4.5 	imes 10^{16}$.

$A$ parallel beam of light of wavelength $900 \, nm$ and intensity $100 \, W/m^2$ is incident on a surface perpendicular to the beam. The number of photons crossing $1 \, cm^2$ area perpendicular to the beam in one second is:

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