Question

Calculate number of photons passing through a ring of unit area in unit time if light of intensity $100W{m}^2$ and of wavelength $400nm$ is falling normally on the ring.

Answer 1

Given : $\lambda =900nm=900\times {10}^{-9}m$

Energy of one photon $E=\frac{hc}{\lambda }$

$⟹E=\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{400\times {10}^{-9}}=4.95\times {10}^{-19}J$

Total energy of light falling per second per unit area $E_t=100J$

Number of photons $N=\frac{E_t}{E}=\frac{100}{4.95\times {10}^{-19}}=2\times {10}^{20}$ photons per second