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A parallel beam of light of wavelength $900 \mathrm{~nm}$ and intensity $100 \; \mathrm{Wm}^{-2}$ is incident on a surface perpendicular to the beam. The number of photons crossing $1 \mathrm{~cm}^{2}$ area perpendicular to the beam in one second is :
Option: 1
$3 \times 10^{16}$

Option: 2
$4.5 \times 10^{16}$

Option: 3
$4.5 \times 10^{17}$

Option: 4
$4.5 \times 10^{20}$

Answers (1)

best_answer

$\mathrm{\lambda=900 \mathrm{~nm}}$

$\mathrm{I=100 \mathrm{~W} / \mathrm{m}^2}$

$\mathrm{I=\frac{\text { Energy }}{\text { (Area)(time) }}=\frac{n(n f)}{(\text { Area)(time) }}}$

$\mathrm{100 \frac{w}{m^2}=\left(\frac{n}{t}\right) \times\left(\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{900 \times 10^{-9} \times 10^{-4}}\right)}$

$\mathrm{\frac{n}{t}= no. of \: photons \: per\: unit \: time\: crossing\: area\: 1cm^{2}}$

$\mathrm{\frac{n}{t} =\frac{9 \times 10^{-9}}{19.89 \times 10^{-26}} }$

$\mathrm{=\frac{9}{19.89} \times 10^{17} }$ $\mathrm{=\frac{9}{19.89}\times 10^{17}}$

$\mathrm{\frac{n}{t} =4.5 \times 10^{16}}$

Hence (2) is correct option

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