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JEE Main 2017 (8 Apr, Shift 1) · Bohr Model: Orbits, Speed & Angular Momentum

Bohr Model: Orbits, Speed & Angular Momentum question from JEE Main 2017 (8 Apr, Shift 1)

Q108JEE Main 2017 · 8 APR · SHIFT 1Easy

Bohr Model: Orbits, Speed & Angular Momentum

According to Bohr’s theory, the time averaged magnetic field at the centre (i.e. nucleus) of a hydrogen atom due to the motion of electrons in the nth orbit is proportional to: (n = principal quantum number)

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Answer: (b)

  1. B=μ0I2rB = \frac{\mu_0 I}{2r}, I=ev2πr∝n−1n2=n−3I = \frac{ev}{2\pi r} \propto \frac{n^{-1}}{n^2} = n^{-3}.
  2. B∝n−3n2=n−5B \propto \frac{n^{-3}}{n^2} = n^{-5}
  3. So the answer is (b) n−5n^{-5}.
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