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The energy of the electron in the nth Bohr orbit of hydrogen is

A{'text': 'E_n = −13.6 / n² eV', 'label': 'A'}
B{'text': 'E_n = −13.6 × n² eV', 'label': 'B'}
C{'text': 'E_n = +13.6 / n² eV', 'label': 'C'}
D{'text': 'E_n = 13.6 × n eV', 'label': 'D'}
Answer & Solution
Correct answer: A. {'text': 'E_n = −13.6 / n² eV', 'label': 'A'}
1. Bohr found E_n = −(k Z² m e⁴ / (8 ε₀² h²)) × 1/n². 2. For H atom (Z = 1), evaluating gives E_n = −13.6 / n² eV. 3. Ground state (n = 1) energy is −13.6 eV. 4. Negative sign indicates bound state; energy → 0 as n → ∞. _Source: NCERT Class 11 Chemistry, Ch 2 "Structure of Atom", §2.4_
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