Question:

The acceleration of an electron in the first orbit of the hydrogen atom (n = 1) is :

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The electron moves in a circle, so \(a = v^2/r\); get v from Bohr's quantisation \(mvr = nh/2\pi\) with n = 1.
Updated On: Nov 11, 2026
  • \(\frac{h^2}{\pi^2 m^2 r^3}\)
  • \(\frac{h^2}{8\pi^2 m^2 r^3}\)
  • \(\frac{h^2}{4\pi^2 m^2 r^3}\)
  • \(\frac{h^2}{4\pi m^2 r^3}\)
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The Correct Option is C

Solution and Explanation

Centripetal acceleration: \(a = \frac{v^2}{r}\).
Bohr's quantisation for n = 1: \(mvr = \frac{h}{2\pi}\), so \(v = \frac{h}{2\pi m r}\).
\(a = \frac{1}{r}\left(\frac{h}{2\pi m r}\right)^2 = \frac{h^2}{4\pi^2 m^2 r^3}\).
Answer: (3) \(\frac{h^2}{4\pi^2 m^2 r^3}\)
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