A parallel beam of light of wavelength $$900 \text{ nm}$$ and intensity $$100 \text{ W m}^{-2}$$ is incident on a surface perpendicular to the beam. The number of photons crossing $$1 \text{ cm}^2$$ area perpendicular to the beam in one second is

A parallel beam of light of wavelength $$\lambda = 900 \text{ nm} = 900 \times 10^{-9} \text{ m}$$ and intensity $$I = 100 \text{ W/m}^2$$ is incident on a surface perpendicular to the beam. We need to find the number of photons crossing $$1 \text{ cm}^2$$ area per second.

Since the energy of a single photon is $$E = \frac{hc}{\lambda} = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{900 \times 10^{-9}}$$, this gives $$E = \frac{19.8 \times 10^{-26}}{9 \times 10^{-7}} = 2.2 \times 10^{-19} \text{ J}$$.

Next, the area is $$1 \text{ cm}^2 = 10^{-4} \text{ m}^2$$ so $$P = I \times A = 100 \times 10^{-4} = 10^{-2} \text{ W} = 0.01 \text{ W}$$.

From this, the number of photons per second is $$n = \frac{P}{E} = \frac{10^{-2}}{2.2 \times 10^{-19}} = \frac{1}{2.2} \times 10^{17}$$ which gives $$n = 0.4545 \times 10^{17} = 4.545 \times 10^{16}$$ and thus $$n \approx 4.5 \times 10^{16}$$.

The correct answer is Option B: $$4.5 \times 10^{16}$$.