A beam of light has two wavelengths of 4972 $$\text{Å}$$ and 6216 $$\text{Å}$$ with a total intensity of $$3.6 \times 10^{-3}$$ Wm$$^{-2}$$ equally distributed among the two wavelengths. The beam falls normally on an area of 1 cm$$^2$$ of a clean metallic surface of work function 2.3 eV. Assume that there is no loss of light by reflection and that each capable photon ejects one electron. The number of photoelectrons liberated in 2 s is approximately:

A beam of light has two wavelengths: 4972 Å and 6216 Å. The total intensity is $$3.6 \times 10^{-3}$$ W/m², equally distributed between the two wavelengths. This means each wavelength has an intensity of half the total, so intensity for 4972 Å is $$1.8 \times 10^{-3}$$ W/m² and for 6216 Å is also $$1.8 \times 10^{-3}$$ W/m².

The beam falls normally on an area of 1 cm². First, convert area to square meters: 1 cm² = $$1 \times 10^{-4}$$ m². The power (energy per second) for each wavelength is intensity multiplied by area.

For 4972 Å: power = $$(1.8 \times 10^{-3}) \times (1 \times 10^{-4}) = 1.8 \times 10^{-7}$$ W.

For 6216 Å: power = $$(1.8 \times 10^{-3}) \times (1 \times 10^{-4}) = 1.8 \times 10^{-7}$$ W.

We need the number of photoelectrons liberated in 2 seconds. The energy delivered by each wavelength in 2 seconds is power multiplied by time.

For 4972 Å: energy = $$(1.8 \times 10^{-7}) \times 2 = 3.6 \times 10^{-7}$$ J.

For 6216 Å: energy = $$(1.8 \times 10^{-7}) \times 2 = 3.6 \times 10^{-7}$$ J.

Each photon has energy $$E_{\text{photon}} = \frac{hc}{\lambda}$$, where $$h$$ is Planck's constant ($$6.626 \times 10^{-34}$$ J s), $$c$$ is the speed of light ($$3 \times 10^8$$ m/s), and $$\lambda$$ is the wavelength in meters.

Convert wavelengths to meters:

4972 Å = $$4972 \times 10^{-10}$$ m = $$4.972 \times 10^{-7}$$ m.

6216 Å = $$6216 \times 10^{-10}$$ m = $$6.216 \times 10^{-7}$$ m.

Calculate photon energy for 4972 Å:

$$E_{\text{ph1}} = \frac{(6.626 \times 10^{-34}) \times (3 \times 10^8)}{4.972 \times 10^{-7}} = \frac{1.9878 \times 10^{-25}}{4.972 \times 10^{-7}} = \frac{1.9878 \times 10^{-25}}{4.972 \times 10^{-7}} = 3.9976 \times 10^{-19} \text{ J}.$$

Calculate photon energy for 6216 Å:

$$E_{\text{ph2}} = \frac{(6.626 \times 10^{-34}) \times (3 \times 10^8)}{6.216 \times 10^{-7}} = \frac{1.9878 \times 10^{-25}}{6.216 \times 10^{-7}} = \frac{1.9878 \times 10^{-25}}{6.216 \times 10^{-7}} = 3.1987 \times 10^{-19} \text{ J}.$$

The work function of the metal is 2.3 eV. Convert to joules using $$1 \text{ eV} = 1.602 \times 10^{-19}$$ J:

Work function $$\phi = 2.3 \times 1.602 \times 10^{-19} = 3.6846 \times 10^{-19}$$ J.

Compare photon energies with work function:

For 4972 Å: $$E_{\text{ph1}} = 3.9976 \times 10^{-19}$$ J > $$3.6846 \times 10^{-19}$$ J, so photons can eject electrons.

For 6216 Å: $$E_{\text{ph2}} = 3.1987 \times 10^{-19}$$ J < $$3.6846 \times 10^{-19}$$ J, so photons cannot eject electrons.

Only photons of 4972 Å contribute to photoelectrons. The number of photons for 4972 Å is the energy delivered divided by energy per photon:

$$N_{\text{ph1}} = \frac{3.6 \times 10^{-7}}{3.9976 \times 10^{-19}} = \frac{3.6 \times 10^{-7}}{3.9976 \times 10^{-19}} = 9.003 \times 10^{11}.$$

Since each capable photon ejects one electron, the number of photoelectrons is $$9.003 \times 10^{11}$$, approximately $$9 \times 10^{11}$$.

Hence, the correct answer is Option B.