Given data

• Area of metal plate: $A = 1\times 10^{-4}\ \text{m}^2$
• Intensity of radiation: $I = 16\ \text{mW/m}^2 = 16\times 10^{-3}\ \text{W/m}^2$
• Work function: $\phi = 5\ \text{eV}$
• Energy of each incident photon: $E = 10\ \text{eV}$
• Only $10\%$ of the incident energy produces photoelectrons.

Power incident on the plate

$P = IA = (16\times 10^{-3})(1\times 10^{-4})$ $P = 16\times 10^{-7}\ \text{W} = 1.6\times 10^{-6}\ \text{J/s}$

Useful power for photoelectric emission

Since only $10\%$ of the incident energy is effective,

$P_{\text{eff}} = 0.1\times 1.6\times 10^{-6}$ $P_{\text{eff}} = 1.6\times 10^{-7}\ \text{J/s}$

Energy of one photon in joules

$1\ \text{eV} = 1.6\times 10^{-19}\ \text{J}$ Therefore, $E = 10\ \text{eV} = 10\times 1.6\times 10^{-19} = 1.6\times 10^{-18}\ \text{J}$

Number of photons effectively producing photoelectrons per second

$n = \frac{P_{\text{eff}}}{E} = \frac{1.6\times 10^{-7}}{1.6\times 10^{-18}}$ $n = 10^{11}\ \text{s}^{-1}$

Assuming one effective photon gives one photoelectron,

$\boxed{\text{Number of emitted photoelectrons per second} = 10^{11}}$

Maximum kinetic energy of emitted photoelectrons

By Einstein’s photoelectric equation,

$K_{\max} = E - \phi$ $K_{\max} = 10 - 5 = 5\ \text{eV}$

So, $\boxed{K_{\max} = 5\ \text{eV}}$

Match with options

The pair is: $\boxed{10^{11}\ \text{and}\ 5\ \text{eV}}$

This corresponds to Option C.