Question

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 ______

Options

• 3 × 10¹⁶

• 4.5 × 10¹⁶

• 4.5 × 10¹⁷

• 4.5 × 10²⁰

Answer 1

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 4.5 × 10¹⁶.

Explanation:

Given that

Wavelength of light, λ = 900 nm = 900 × 10^-9 m

Intensity, I = 100 Wm^-2

Area, A = 1 cm² = 10^-4 m²

Time, t = 1 second

The number of photons in one second

= `"Energy of all photons(E')"/"Energy of one photon (E)"`             ..... (i) 

Energy of all photons, (E') = Power (P) × time (t)

= Intensity (I) × Area (A) × time (t)

= 100 × 10^-4 × 1 = 0.01 J 

Energy of each photon, (E) = `"hc"/lambda`

= `(6.6xx10^-34xx3xx10^8)/(900xx10^-9xx1.6xx10^-19)`

= 1.375 eV

Putting this value in equation (i), we have

The number of photons = `0.01/(1.375xx1.6xx10^-19)`

= 4.5 × 10¹⁶