• # question_answer Light of frequency $8\times {{10}^{15}}Hz$ is incident on a substance of photoelectric work function $6.125\,eV.$ The maximum kinetic energy of the emitted photoelectrons is A) 17$eV$ B) 22$eV$ C) 27$eV$ D) 37$eV$

[c] $E=h\nu =6.6\times {{10}^{-34}}\times 8\times {{10}^{15}}=5.28\times {{10}^{-18}}J=33eV$ By using $E={{W}_{0}}+{{K}_{\max }}\Rightarrow {{K}_{\max }}=E-{{W}_{0}}$ $=33-6.125=27eV$