An electron in the hydrogen atom jumps from n^ th energy state to the ground state. The wavelength so emitted illuminates a photosensitive material having work function 2.65 eV. If the maximum kinetic energy of the emitted photoelectrons is 10.1 eV, then the value of ' n ' is

Step-by-step solution

Using Einstein's photoelectric equation: E = K_ max + Substituting the given values: E = 10.1 eV + 2.65 eV = 12.75 eV The energy of the photon emitted during the transition from the n^ th state to the ground state ( n=1 ) in a hydrogen atom is given by: E = 13.6 ( 1 - 1 n^2 ) eV Equating the two energies: 12.75 = 13.6 ( 1 - 1 n^2 ) 1 - 1 n^2 = 12.75 13.6 = 15 16 1 n^2 = 1 - 15 16 = 1 16 n^2 = 16 n = 4 Answer: 4