Saved Bookmarks
| 1. |
A photon emitted during the de-excitation of electron from a state to the first excited state in a hydrogen atom, irradiates a metallic cathode of work function 2eV, in a photo cell, with a stopping potential of 0.55 V. Obtain the value of the quantum number of the state. |
|
Answer» Solution :When an electron in a hydrogen atom is DE excited from a staten to the firstexcited state (ie,. =2), the ENERGY of emitted photonis. `E = [-(13.6)/(n^(2))] - [- (13.6)/((2)^(2))] = [3.4 - (13.6)/(n^(2))]eV` As work function of photo cell `phi_(0) = 2 eV` and stopping potential `V_(0) =0.55 V`, the maximum kinetic energy of photoelectron `K_("max") - eV_(0) =0.55 eV`. As per Einstein.s photoelectric equation `=E = phi_(0) + k_("max")` `rArr "" [3.4 -(13.6)/(n^(2))] = 2+ 0.55 = 2.55 rArr (13.6)/(n^(2)) = 3.4 - 2.55 = 0.85rArr n = SQRT((13.6)/(0.85)) = 4` |
|