An electron in a hydrogen atom is initially in an excited state where its kinetic energy is 1.51 eV . It then makes a transition to the ground state. What is the change in the electrostatic potential energy of the electron during this transition?

Step-by-step solution

For an electron in a hydrogen-like atom, the potential energy (PE) is related to the kinetic energy (KE) by PE = -2 KE . In the initial excited state, the kinetic energy is given as 1.51 eV . Initial potential energy, PE _ i = -2 1.51 eV = -3.02 eV . In the ground state of a hydrogen atom, the total energy (TE) is -13.6 eV . The potential energy is related to the total energy by PE = 2 TE . Final potential energy, PE _ f = 2 (-13.6 eV ) = -27.2 eV . The change in electrostatic potential energy is: PE = PE _ f - PE