A beam of electromagnetic radiation of intensity $$6.4 \times 10^{-5}$$ W/cm$$^2$$ is comprised of wavelength, $$\lambda = 310$$ nm. It falls normally on a metal (work function $$\varphi = 2$$ eV) of surface area of 1 cm$$^2$$. If one in $$10^3$$ photons ejects an electron, total number of electrons ejected in 1s is $$10^x$$. ($$hc = 1240$$ eVnm, $$1$$ eV $$= 1.6 \times 10^{-19}$$ J), then $$x$$ is

We begin by noting that the incident beam has an intensity of $$6.4 \times 10^{-5}\;{\rm W/cm^2}$$ and the metal surface has an area of $$1\;{\rm cm^2}$$. Intensity is power per unit area, so the power $$P$$ falling on the plate is simply the product of intensity and area.

Hence, we have

$$P = \bigl(6.4 \times 10^{-5}\;{\rm W/cm^2}\bigr)\times \bigl(1\;{\rm cm^2}\bigr) = 6.4 \times 10^{-5}\;{\rm W}.$$

This power is the energy incident per second. Therefore the energy received by the plate in $$1{\rm \; s}$$ is

$$E_{\text{incident}} = P \times 1{\rm \; s} = 6.4 \times 10^{-5}\;{\rm J}.$$

Next, we calculate the energy of a single photon of the given wavelength. The photon energy is obtained from the relation $$E = \dfrac{hc}{\lambda}.$$

We are provided $$hc = 1240\;{\rm eV\,nm}$$ and $$\lambda = 310\;{\rm nm}.$$ Substituting these values, we get

$$E_{\text{photon}} = \dfrac{1240\;{\rm eV\,nm}}{310\;{\rm nm}} = 4.0\;{\rm eV}.$$

To keep all quantities in the same unit system, we convert this photon energy from electron-volts to joules. The conversion factor is $$1\;{\rm eV} = 1.6 \times 10^{-19}\;{\rm J}.$$ Hence,

$$E_{\text{photon}} = 4.0\;{\rm eV}\times \bigl(1.6 \times 10^{-19}\;{\rm J/eV}\bigr) = 6.4 \times 10^{-19}\;{\rm J}.$$

We now determine the number of photons striking the plate in one second. Because total incident energy in one second is $$E_{\text{incident}}$$ and each photon carries energy $$E_{\text{photon}},$$ the photon count $$N_{\text{incident}}$$ is given by

$$N_{\text{incident}} = \dfrac{E_{\text{incident}}}{E_{\text{photon}}} = \dfrac{6.4 \times 10^{-5}\;{\rm J}}{6.4 \times 10^{-19}\;{\rm J}}.$$

Both the numerator and the denominator contain the same coefficient $$6.4,$$ so they cancel, leaving powers of ten to handle:

$$N_{\text{incident}} = 10^{14}.$$

The problem states that only one photon out of every $$10^3$$ actually ejects an electron. Thus the fraction of photons effective in photo-emission is $$\dfrac{1}{10^3}.$$ Therefore the number of electrons emitted in one second, $$N_e,$$ is

$$N_e = N_{\text{incident}}\times \dfrac{1}{10^3} = 10^{14}\times 10^{-3} = 10^{11}.$$

We are told to write the answer in the form $$10^{x}$$ for the total electrons ejected per second, and we have discovered that $$x = 11.$$

Hence, the correct answer is Option C.