A metal plate of area $$1 \times 10^{-4}$$ m$$^2$$ is illuminated by a radiation of intensity 16 m W/m$$^2$$. The work function of the metal is 5 eV. The energy of the incident photons is 10 eV and only 10% of it produces photo electrons. The number of emitted photo electrons per second and their maximum energy, respectively, will be: [$$1eV = 1.6 \times 10^{-19}$$ J]

We begin with the given data. The metal plate has an area $$A = 1 \times 10^{-4}\,{\rm m^2}$$ and the radiation that falls on it has an intensity $$I = 16\;{\rm milli\,W/m^2} = 16 \times 10^{-3}\,{\rm W/m^2} = 0.016\,{\rm W/m^2}$$. 1 watt is 1 joule per second, so intensity tells us how many joules reach one square metre each second.

The power (energy per second) actually falling on the small plate is obtained from the definition of intensity: $$P = I \times A$$. Substituting the values, we get

$$P = 0.016\,{\rm W/m^2} \times 1 \times 10^{-4}\,{\rm m^2} = 1.6 \times 10^{-6}\,{\rm W}.$$

Because a watt equals a joule per second, this also means that $$1.6 \times 10^{-6}\,{\rm J}$$ of radiant energy reaches the plate every second.

Now, each incident photon has an energy of $$10\,{\rm eV}$$. Converting this into joules with the relation $$1\,{\rm eV} = 1.6 \times 10^{-19}\,{\rm J}$$, we write

$$E_{\text{photon}} = 10\,{\rm eV} = 10 \times 1.6 \times 10^{-19}\,{\rm J} = 1.6 \times 10^{-18}\,{\rm J}.$$

The number of photons that hit the plate each second is simply the total incident energy per second divided by the energy of one photon:

$$N_{\text{photon/sec}} = \frac{P}{E_{\text{photon}}} = \frac{1.6 \times 10^{-6}\,{\rm J/s}}{1.6 \times 10^{-18}\,{\rm J}} = 1 \times 10^{12}\,{\rm photons/s}.$$

However, only 10 % of these photons actually eject photoelectrons. Therefore the rate of emission of photoelectrons is

$$N_{e^{-}/\text{sec}} = 0.10 \times 1 \times 10^{12} = 1 \times 10^{11}\,{\rm electrons/s}.$$

Next we calculate the maximum kinetic energy of those emitted electrons. According to Einstein’s photoelectric equation,

$$K_{\text{max}} = h\nu - \phi,$$

where $$h\nu$$ is the photon energy and $$\phi$$ is the work function of the metal. Both quantities are conveniently given in electron-volts:

$$K_{\text{max}} = 10\,{\rm eV} - 5\,{\rm eV} = 5\,{\rm eV}.$$

So, each second $$1 \times 10^{11}$$ electrons are emitted, and the most energetic among them possess $$5\,{\rm eV}$$ of kinetic energy.

Hence, the correct answer is Option C.