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JEE Main 2021 (26 Feb, Shift 2) · Other / Mixed: Atoms & Bohr Links

Other / Mixed: Atoms & Bohr Links question from JEE Main 2021 (26 Feb, Shift 2)

Q7JEE Main 2021 · 26 FEB · SHIFT 2Difficult

Other / Mixed: Atoms & Bohr Links

The recoil speed of a hydrogen atom after it emits a photon in going from n = 5 state to n = 1 state will be:

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Answer: (d)

  1. Photon energy: E=13.6(1−125)=13.06 eV=2.09×10−18 JE = 13.6\left(1 - \frac{1}{25}\right) = 13.06\text{ eV} = 2.09 \times 10^{-18}\text{ J}
  2. Momentum conservation: Mv=EcMv = \frac{E}{c}
  3. v=2.09×10−181.67×10−27×3×108≈4.17 m/sv = \frac{2.09 \times 10^{-18}}{1.67 \times 10^{-27} \times 3 \times 10^{8}} \approx 4.17\text{ m/s}
  4. So the answer is (d) 4.17 m/s.
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