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(D) The power of the source is $P = 160 \,W$ and the wavelength is $\lambda = 50000 \,\mathring A = 5 	imes 10^{-6} \,m$.The energy of a single photo

Vedclass · Accepted Answer

$10^{20}$

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(D) The power of the source is $P = 160 \,W$ and the wavelength is $\lambda = 50000 \,\mathring A = 5 	imes 10^{-6} \,m$.The energy of a single photon is $E = \frac{hc}{\lambda} = \frac{6.63 	imes 10^{-34} 	imes 3 	imes 10^8}{5 	imes 10^{-6}} \approx 3.98 	imes 10^{-20} \,J$.The number of photons emitted per second by the source is $N = \frac{P}{E} = \frac{160}{3.98 	imes 10^{-20}} \approx 4.02 	imes 10^{21} \,s^{-1}$.The photon flux $n$ at a distance $r = 1.8 \,m$ is the number of photons passing through a unit area per second,given by $n = \frac{N}{4 \pi r^2}$.$n = \frac{4.02 	imes 10^{21}}{4 	imes 3.14 	imes (1.8)^2} = \frac{4.02 	imes 10^{21}}{40.69} \approx 0.0988 	imes 10^{21} \approx 10^{20} \,m^{-2} s^{-1}$.Thus,the order of magnitude is $10^{20} \,m^{-2} s^{-1}$.

$A$ $160 \,W$ infrared source is radiating light of wavelength $50000 \,\mathring A$ uniformly in all directions. The photon flux at a distance of $1.8 \,m$ is of the order of ............. $m^{-2} s^{-1}$.

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