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(C) The photon flux is defined as the number of photons passing through a unit area per unit time.The energy of a single photon is given by $E = \frac

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$10^{19}$

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(C) The photon flux is defined as the number of photons passing through a unit area per unit time.The energy of a single photon is given by $E = \frac{hc}{\lambda}$.The intensity $I$ at a distance $r$ from a source of power $P$ is $I = \frac{P}{4 \pi r^2}$.The photon flux $\Phi$ is given by $\Phi = \frac{I}{E} = \frac{P}{4 \pi r^2} 	imes \frac{\lambda}{hc}$.Substituting the given values: $P = 160 \,W$,$\lambda = 6200 	imes 10^{-10} \,m$,$r = 1.8 \,m$,$h = 6.63 	imes 10^{-34} \,J \cdot s$,and $c = 3 	imes 10^8 \,m/s$.$\Phi = \frac{160 	imes 6200 	imes 10^{-10}}{4 	imes 3.14 	imes (1.8)^2 	imes 6.63 	imes 10^{-34} 	imes 3 	imes 10^8}$.Calculating the value: $\Phi \approx 1.22 	imes 10^{19} \,m^{-2} s^{-1}$.Thus,the order of magnitude is $10^{19} \,m^{-2} s^{-1}$.

$A$ $160 \,W$ light source is radiating light of wavelength $6200 \,\mathring A$ uniformly in all directions. The photon flux at a distance of $1.8 \,m$ is of the order of .......... $m^{-2} s^{-1}$ (Planck's constant $= 6.63 \times 10^{-34} \,J \cdot s$)

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