• question_answer A photon of energy 8 eV is incident on metal surface of threshold frequency $1.6\times {{10}^{15}}Hz.$ The maximum kinetic energy of the photoelectrons emitted (in eV) (Take $h=6\times {{10}^{-34}}Js)$.    A) 1.6 B) 6 C) 2 D) 1.2

[a] $K.E.=h\nu -h{{\nu }_{0}}$$=8\ eV-\left( \frac{6\times {{10}^{-34}}\times 1.6\times {{10}^{15}}}{1.6\times {{10}^{-19}}}eV \right)$ = $8\ eV-6\ eV=2\ eV$