Bohr model stability condition for hydrogen In a hydrogen atom, an electron orbits a proton. According to the Bohr (classical) picture, orbital stability requires equilibrium between the attractive Coulomb force and the required centripetal force. Is this statement correct?

Electronics and Communication Engineering Materials and Components Difficulty: Easy
Choose an option
  • A
    True
  • B
    False
  • C
    True only for n = 1
  • D
    False unless relativistic corrections are included
  • E
    True only at T = 0 K

Answer

Correct Answer: True

Explanation

Introduction / Context:The Bohr model, though superseded by quantum mechanics, provides an intuitive picture of hydrogenic orbits. It relates electrostatic attraction to centripetal requirements for circular motion, combined with angular-momentum quantization, to explain discrete energy levels and spectral lines.

Given Data / Assumptions:

  • Single electron moving in a circular orbit around a fixed proton.
  • Non-relativistic speeds; classical Coulomb force law.
  • Bohr quantization of angular momentum L = n h/(2π).

Concept / Approach:

For a circular orbit, the centripetal force m v^2 / r must be supplied by Coulomb attraction k e^2 / r^2 (with k = 1/(4π ε0), e the elementary charge). Equating these gives the relationship among m, v, and r. Combined with the angular momentum postulate, one solves for allowed radii r_n and energies E_n. This balance is the physical “stability” condition in the Bohr picture.

Step-by-Step Solution:

Set m v^2 / r = k e^2 / r^2.Rearrange: m v^2 = k e^2 / r.With L = m v r = n h/(2π), solve for r_n and E_n (results r_n ∝ n^2, E_n ∝ −1/n^2).

Verification / Alternative check:

In full quantum mechanics, the expectation values match Bohr predictions for hydrogen energy levels, validating the historical model’s results though not its orbital picture.

Why Other Options Are Wrong:

Limiting the validity to n = 1 or T = 0 is unnecessary in the Bohr framework; relativistic corrections are small and not required for the primary stability condition.

Common Pitfalls:

Confusing “stability” here with quantum stability; Bohr’s circular orbit analogy is a semi-classical construction leading to correct spectra despite its limitations.

Final Answer:

True

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