Question

If the total energy of an electron in the 1st shell of H atom = 0.0 eV then its potential energy in the 1st
excited state would be

(a) +6.8 eV
(b) +20.4 eV
(c) -6.8 eV
(d) +3.4 eV​

Answer 1

Answer:

In 1st excited state,n=2. E=-13.6×Z^2÷n^2. E=-13.6×1^2÷2^2 (Z of H=1). E=-3.4ev. Ep=-2Ek. Ep=-2(-3.4)=+6.8ev