1. The problem statement, all variables and given/known data What is the energy of H atom in the first excited state if the potential energy in the ground state is taken to be 0 ? 2. Relevant equations Usually the energy of H atom in the ground state is -13.6eV and in the 1st excited state is -10.2eV E(n) = πme2/8ε2h2 Bohr's radius = 0.529 angstrom The answer provided is : 23.8eV 3. The attempt at a solution So I went with how the energy of the electron is derived from ground up Knowing that Total energy = potential energy + Kinetic energy potential energy due to the positive charge on nucleus and negative on electron so if the potential energy is taken to be 0 then the only energy left is the kinetic one so the energy would just be 1/2 mvv then I substituted v = e2/2∏hn then I used the formula ΔE = 13.6(3/4) but everything is now messed up and I ended up with a gross looking equation , I know something doesn't makes sense above may be nothing does . Thanks in advance : P.S : the answer is 23.8 and 13.6 + 10.2 = 23.8 but how can I possibly do this ?