Energy E of a hydrogen atom with principal quantum number n is given E= -13.6 n^ 2 eV …
Physics · NEET · NTA Exams — Atoms and Nuclei
Energy E of a hydrogen atom with principal quantum number n is given \(E=\frac{-13.6}{n^{2}} \mathrm{eV}\). The energy of a photon ejected when an 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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