1. A photon with an energy of 13,6eVknocks out an electron from a hydrogen atom at rest, which is in the ground state. The kinetic energy of an electron away from the nucleus is (eV) 1) -13,6 2) 10,2 3) 3,4 4) 0 5) 13,6

To solve this problem, we need to use the principles of conservation of energy and the Bohr model of the hydrogen atom.<br /><br />Given information:<br />- A photon with an energy of 13.6 eV knocks out an electron from a hydrogen atom at rest, which is in the ground state.<br /><br />Let's denote the kinetic energy of the electron as KE.<br /><br />According to the Bohr model, the energy of an electron in a hydrogen atom is given by:<br /><br />E = -13.6 eV / n^2<br /><br />where n is the principal quantum number (the energy level of the electron).<br /><br />In the ground state (n = 1), the energy of the electron is:<br /><br />E = -13.6 eV<br /><br />When the photon knocks out the electron, the electron is no longer bound to the nucleus, and its energy becomes 0 eV.<br /><br />The conservation of energy principle states that the total energy of the system must remain constant. Therefore, the energy of the photon must be equal to the kinetic energy of the electron.<br /><br />KE = 13.6 eV<br /><br />Therefore, the correct answer is:<br />5) 13.6 eV