The work functions for metals A, B and C are 1.92 eV , 2.0 eV and 5 eV respectively the metal (s) which will emit photoelectrons for incident radiation of wavelength 4100 ~A is /are [ h =6.63 10⁻¹⁹ ~J , e =1.6 10⁻¹⁹ C , c =3 10^8 ~m / s ]

Step-by-step solution

Energy of electron with an associated wavelength of 4100 is: hc =4.845 10⁻¹⁹ ~J =3.024 eV This incident electron would emit photon from metal whose work potential is less than its energy. Thus, it would emit photons from metals A and B.