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Q.

A parallel beam of light of wavelength 900 nm and intensity 100 Wm^–2 is incident on a surface perpendicular to the beam. The number of photons crossing 1 cm² area perpendicular to the beam in one second is

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a

4.5 × 10¹⁶

b

4.5 × 10²⁰

c

3 × 10¹⁶

d

4.5 × 10¹⁷

answer is B.

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Detailed Solution

$\begin{array}{l} λ = 900 nm \\ I = 100 W / m^{2} \\ A = 10^{- 4} m^{2} \\ \Rightarrow P = 10^{- 2} W \end{array}$
$\Rightarrow$ Number of photons incident per second
$\begin{array}{l} = \frac{10^{- 2} λ}{hc} \\ = \frac{9 \times 10^{- 11} \times 10^{2}}{6 .63 \times 10^{- 34} \times 3 \times 10^{8}} = 4 .5 \times 10^{16} \end{array}$

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