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JEE Main 2021 (31 Aug, Shift 2) · Other / Mixed: Atoms & Bohr Links

Other / Mixed: Atoms & Bohr Links question from JEE Main 2021 (31 Aug, Shift 2)

Q5JEE Main 2021 · 31 AUG · SHIFT 2Difficult

Other / Mixed: Atoms & Bohr Links

A free electron of 2.6 eV energy collides with a H+ ion. This results in the formation of a hydrogen atom in the first excited state and a photon is released. Find the frequency of the emitted photon. (h = 6.6 ×\times 10−-34 Js)

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Answer: (c)

  1. Energy of first excited state (n=2n = 2) of H: −3.4 eV-3.4\text{ eV}.
  2. Photon energy =2.6+3.4=6.0 eV=9.6×10−19 J= 2.6 + 3.4 = 6.0\text{ eV} = 9.6 \times 10^{-19}\text{ J}
  3. f=Eh=9.6×10−196.6×10−34=1.45×1015 Hzf = \frac{E}{h} = \frac{9.6 \times 10^{-19}}{6.6 \times 10^{-34}} = 1.45 \times 10^{15}\text{ Hz}
  4. =1.45×109 MHz= 1.45 \times 10^{9}\text{ MHz}
  5. So the answer is (c).
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