alleryThe energy of the electron in the nth Bohr orbit of hydrogen is
AE_n = −13.6 / n² eV
BE_n = −13.6 × n² eV
CE_n = +13.6 / n² eV
DE_n = 13.6 × n eV
Answer & Solution
Correct answer: A. E_n = −13.6 / n² eV
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: syllabus Chemistry, Ch 2 "Structure of Atom", §2.4_
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