Question

In a hydrogen atom, if the energy of electron in the ground state is -x eV., then that in the 2nd excited state of He+ is

Answer 1

x=13.6 ev
enegy=13.6z^2/n^2

Answer 2

1. The problem statement, all variables and given/known data In a hydrogenlike ion with atomic number Z, the energies of the allowed states are given by E(n) = (-13.6eV) (Z^2/n^2) What is the wavelength asociated with the transition between first excited state and ground state of hydrogen-like helium? (He+) 2. Relevant equations E(n) = (-13.6eV) (Z^2/n^2) E = hf = hc/Lambda 3. The attempt at a solution Hydrogen like helium is He+ which has 2 protons and 1 electron. Atomic number Z is 2. So I'd sub in 2 for Z (4 for Z^2), but I don't know what to choose n as. I assume n=1 for ground state and n=2 for first excited state. An idea I had was to do it twice - sub in n=1 and complete the equation, sub in n=2 and complete the equation, and... take them away? I'm not sure :(