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JEE Main 2022 (30 Jun, Shift 1) · Bohr Model: Orbits, Speed & Angular Momentum

Bohr Model: Orbits, Speed & Angular Momentum question from JEE Main 2022 (30 Jun, Shift 1)

Q82JEE Main 2022 · 30 JUN · SHIFT 1Medium

Bohr Model: Orbits, Speed & Angular Momentum

A hydrogen atom in ground state absorbs 12.09 eV of energy. The orbital angular momentum of the electron is increased by:

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Answer: (b)

  1. Final energy =−13.6+12.09=−1.51 eV=−13.69= -13.6 + 12.09 = -1.51\text{ eV} = -\frac{13.6}{9}, so n=3n = 3.
  2. ΔL=(3−1)h2π=hπ\Delta L = (3 - 1)\frac{h}{2\pi} = \frac{h}{\pi}
  3. ΔL=6.63×10−343.14=2.11×10−34 Js\Delta L = \frac{6.63 \times 10^{-34}}{3.14} = 2.11 \times 10^{-34}\text{ Js}
  4. So the answer is (b) 2.11×10−342.11 \times 10^{-34} Js.
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