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 :

Solution

Physics sir · 3 steps

Find the power through 1 cm², then divide by the energy of one photon. That count is the answer.

1. Energy per photon. One photon of wavelength λ = 900 nm = 9 × 10⁻⁷ m carries E = hc/λ = (6.63 × 10⁻³⁴ × 3 × 10⁸)/(9 × 10⁻⁷) = 2.21 × 10⁻¹⁹ J.
2. Power through area. Intensity is power per unit area. Since 1 cm² = 10⁻⁴ m², the power crossing it is P = 100 × 10⁻⁴ = 10⁻² W, which is 10⁻² joules every second.

   N = P/E = 10⁻²/(2.21 × 10⁻¹⁹) ≈ 4.5 × 10¹⁶ s⁻¹

3. So. About 4.5 × 10¹⁶ photons cross 1 cm² each second, which is option (b).

The idea to keep. Number of photons per second = (intensity × area) ÷ (hc/λ). Put everything in SI units first, especially cm² to m² and nm to m.

Answer from the NTA final answer key. Solution by PrepWiser.