A parallel beam of light of intensity 33.15 W m ⁻² is incident normally on a surface. If the number of photons crossing an area of 2 cm ^2 perpendicular to the beam is 2 10¹⁶ per second, the wavelength of the light beam is (Take h = 6.63 10⁻³⁴ J s and c = 3 10^8 m s ⁻¹ )

Step-by-step solution

The total power P incident on the given area is the product of intensity and area. P = I A = 33.15 (2 10⁻⁴) = 66.3 10⁻⁴ W Let E be the energy of a single photon. The total power is also equal to the number of photons per second n multiplied by E . P = n E E = P n E = 66.3 10⁻⁴ 2 10¹⁶ = 33.15 10⁻²⁰ J The energy of a photon is given by E = hc . Therefore, the wavelength is: = hc E = 6.63 10⁻³⁴ 3 10^8 33.15 10⁻²⁰ = 19.89 10⁻²⁶ 33.15 10⁻²⁰ = 0.6 10⁻⁶ m = 600 10⁻⁹ m = 600 nm . Answer: 600 nm