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JEE Main 2023 (1 Feb, Shift 1) · Bohr Model: Orbits, Speed & Angular Momentum

Bohr Model: Orbits, Speed & Angular Momentum question from JEE Main 2023 (1 Feb, Shift 1)

Q69JEE Main 2023 · 1 FEB · SHIFT 1Medium

Bohr Model: Orbits, Speed & Angular Momentum

A light of energy 12.75 eV12.75\text{ eV} is incident on a hydrogen atom in its ground state. The atom absorbs the radiation and reaches to one of its excited states. The angular momentum of the atom in the excited state is xπ×10−17 eVs\frac{x}{\pi} \times 10^{-17}\text{ eVs}. The value of xx is ___________ (use h=4.14×10−15 eVs, c=3×108 ms−1h = 4.14 \times 10^{-15}\text{ eVs},\ c = 3 \times 10^{8}\text{ ms}^{-1} ).

Numerical answer type. Work it out, then open the answer.

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Answer: 828

  1. En=−13.6+12.75=−0.85 eV=−13.6n2⇒n=4E_n = -13.6 + 12.75 = -0.85\text{ eV} = -\frac{13.6}{n^2} \Rightarrow n = 4
  2. L=nh2π=4h2π=2hπL = \frac{nh}{2\pi} = \frac{4h}{2\pi} = \frac{2h}{\pi}
  3. L=2×4.14×10−15π=828π×10−17 eVsL = \frac{2\times4.14\times10^{-15}}{\pi} = \frac{828}{\pi}\times10^{-17}\text{ eVs}
  4. So x=828x = 828.
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