Energy E of a hydrogen atom with principal quantum number n is given by E= -13.6 n^ 2 eV…
Physics · Class 12 · CBSE — Atoms
Energy E of a hydrogen atom with principal quantum number n is given by \(E=\frac{-13.6}{n^{2}} \mathrm{eV}\) The energy of a photon ejected when the electron jumps from n = 3 state to n = 2 state of hydrogen is approximately
- 1.5 eV
- 0.85 eV
- 3.4 eV
- 1.9 eV
Answer
(D) 1.9 eV
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