When a beam of photons each of energy 10.6 eV and of intensity 2.0 W/m² falls on a platinum surface of area   $1.0\times {10}^{-4}{\text{m}}^{\text{2}}$  and work function  $5.6\text{\hspace{0.17em}eV},0.53$    of the incident photons eject photoelectrons. The sum of minimum & maximum energies (in eV) is ______ . (Take  $1\text{\hspace{0.17em}eV}=1.6\times {10}^{-19}\text{J.}$ )

Energy incident on the platinum surface per second or  power  $=IA=2.0\times 1\times {10}^{-4}=2\times {10}^{-4}J/s$
The energy of each photon E  $=10.6eV$   $=10.6\times 1.6\times {10}^{-19}J$
The number of photons incident on the surface  $=\frac{2\times {10}^{-4}}{10.6\times 1.6\times {10}^{-19}}per\sec .$
As only 53% can eject electrons, and so photo electrons ejected per second  $=0.53\left[\frac{2\times {10}^{-6}}{10.6\times 1.6\times {10}^{-19}}\right]=6.25\times {10}^{11}$
According to Einstein photoelectric equation   $K_{\max }=E-W_0=10.6-56 =5eV.$
The minimum kinetic energy of photoelectrons will be zero.