JEE Main & Advanced
Physics
Atomic Physics
Question Bank
Atomic Structure
question_answer
Energy E of a hydrogen atom with principal quantum number n is given by \[E=\frac{-13.6}{{{n}^{2}}}eV\]. The energy of a photon ejected when the electron jumps from \[n=3\] state to \[n=2\] state of hydrogen is approximately [CBSE PMT 2004]