The light emitted during the transition from n = 3 to n = 2 in hydrogen is known as H_ light. Determine the maximum work function a metal can possess such that this H_ light is capable of emitting photoelectrons from it.

Step-by-step solution

The energy of an electron in the n -th orbit of a hydrogen atom is given by E_n = - 13.6 n^2 eV . For the H_ transition from n = 3 to n = 2 , the energy of the emitted photon is: E = E₃ - E₂ E = - 13.6 3^2 - (- 13.6 2^2 ) E = -1.51 + 3.40 = 1.89 eV For photoelectric emission to take place, the energy of the incident photon must be greater than or equal to the work function of the metal ( ). E Therefore, the maximum work function the metal can have is _ max = 1.89 eV .