Quantum Numbers

Question

Energy difference (ΔE) between n=2 and n=4 levels of hydrogen atom is approximately (in eV):

RankGuru · Accepted Answer

1. **Bohr's Energy Formula**: For Hydrogen, $E_n = -\frac{13.6}{n^2}$ eV.
2. **Identify States**:
   - $n_1 = 2 \implies E_2 = -\frac{13.6}{4} = -3.40$ eV.
   - $n_2 = 4 \implies E_4 = -\frac{13.6}{16} = -0.85$ eV.
3. **Calculate Difference**:
   - $\Delta E = |E_4 - E_2|$
   - $\Delta E = |-0.85 - (-3.40)| = 2.55$ eV.
4. **Conclusion**: The energy required for this transition (or emitted during decay) is **$2.55$ eV**.

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