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NEET 2026 · Bohr Model: Orbits, Speed & Angular Momentum

Bohr Model: Orbits, Speed & Angular Momentum question from NEET 2026

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Bohr Model: Orbits, Speed & Angular Momentum

In the first excited state of hydrogen atom, the energy of its electron is -3.4 eV. The radial distance of the electron from the hydrogen nucleus in this case is approximately:

(Take 1 eV=1.6×10−19 J, e=1.6×10−19 C1\text{ eV} = 1.6 \times 10^{-19}\text{ J},\ e = 1.6 \times 10^{-19}\text{ C} and 14πε0=9×109 N m2/C2\frac{1}{4\pi\varepsilon_0} = 9 \times 10^{9}\text{ N m}^2/\text{C}^2)

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Answer: (c)

  1. Total energy E=−ke22r⇒r=ke22∣E∣E = -\frac{ke^2}{2r} \Rightarrow r = \frac{ke^2}{2|E|}
  2. r=9×109×(1.6×10−19)22×3.4×1.6×10−19=14.4×10−106.8r = \frac{9\times10^9 \times (1.6\times10^{-19})^2}{2\times3.4\times1.6\times10^{-19}} = \frac{14.4\times10^{-10}}{6.8}
  3. r≈2.1×10−10 mr \approx 2.1\times10^{-10}\text{ m}
  4. So the answer is (c) 2.1×10−10 m2.1\times10^{-10}\text{ m}.
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