A photon emitted during the de-excitation of electron from a state n to the second excited state in a hydrogen atom, irradiates a metallic electrode of work function 0.5 ~eV , in a photocell, with a stopping voltage of 0.47 ~V . Obtain the value of quantum number of the state ' n '.

Step-by-step solution

The energy of a photon emitted during the transition from state n to the second excited state ( n_f = 3 ) in a hydrogen atom is given by the Rydberg formula: E = 13.6 ( 1 n_f^2 - 1 n_i^2 ) eV E = 13.6 ( 1 3^2 - 1 n^2 ) = 13.6 ( 1 9 - 1 n^2 ) eV According to Einstein's photoelectric equation, the maximum kinetic energy of the emitted photoelectrons is K_ max = E - , where is the work function. Given K_ max = e V_s = 0.47 eV and = 0.5 eV : 0.47 = E - 0.5 E = 0.47 + 0.5 = 0.97 eV Equating the two expressions for E : 1