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JEE Main 2019 (12 Apr, Shift 2) · Bohr Model: Orbits, Speed & Angular Momentum

Bohr Model: Orbits, Speed & Angular Momentum question from JEE Main 2019 (12 Apr, Shift 2)

Q102JEE Main 2019 · 12 APR · SHIFT 2Medium

Bohr Model: Orbits, Speed & Angular Momentum

Consider an electron in a hydrogen atom revolving in its second excited state (having radius 4.65 A˚\mathring{A}). The de-Broglie wavelength of this electron is:

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Answer: (c)

  1. Second excited state: n=3n = 3.
  2. λ=2πrn=2×3.14×4.653 A˚\lambda = \frac{2\pi r}{n} = \frac{2\times3.14\times4.65}{3}\ \mathring{A}
  3. λ≈9.7 A˚\lambda \approx 9.7\ \mathring{A}
  4. So the answer is (c) 9.7 A˚9.7\ \mathring{A}.
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