JEE Main · 2022 · 26 Jul 2022 · Shift 1 · Modern Physics

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 cm^{2}

area perpendicular to the beam in one second is :

(A) $3 \times 1 0^{16}$
(B) $4.5 \times 1 0^{16}$ ✓ Correct
(C) $4.5 \times 1 0^{17}$
(D) $4.5 \times 1 0^{20}$

Answer: (B)

Solution

Number of photons per second is

n = \frac{I A λ}{h c}

.

Given:

I = 100 W/m^{2}

A = 1 cm^{2} = 1 0^{- 4} m^{2}

λ = 900 \times 1 0^{- 9} m

n = \frac{100 \times 1 0 ^{- 4} \times 900 \times 1 0 ^{- 9}}{6.62 \times 1 0 ^{- 34} \times 3 \times 1 0 ^{8}}

n = \frac{9 \times 1 0 ^{- 9}}{19.86 \times 1 0 ^{- 26}} \approx 4.5 \times 1 0^{16}

JEE Main · 2022 · 26 Jul 2022 · Shift 1 · Single correct option

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