Bohr Model Basics

In the Bohr model, the electron orbits the proton in quantized circular paths. For the lowest energy orbit (n=1), two key conditions apply: the centripetal force equals the Coulomb attraction,

\[ \frac{mv^2}{r} = \frac{1}{4\pi\epsilon_0} \frac{e^2}{r^2} \]

and angular momentum is quantized, \( mvr = \hbar \) .

Solving these yields the ground state velocity \( v = \frac{1}{4\pi\epsilon_0} \frac{e^2}{\hbar} \). Dividing by c gives \( \frac{v}{c} = \frac{1}{4\pi\epsilon_0} \frac{e^2}{\hbar c} \), confirming option (A) .

Option Analysis

• (A) Correct, as derived above and numerically ≈0.0073 (1/137) .
• (B) Incorrect; yields ≈2.92×10²⁵, far exceeding 1 .
• (C) Incorrect; m/M ≈1/1836, unrelated to velocity ratio .
• (D) Numerically ≈0.0073 but structurally different; equals α only coincidentally via Bohr radius definition .

Derivation Steps

The velocity formula emerges from:

• Centripetal force: \( \frac{mv^2}{a_0} = \frac{e^2}{4\pi\epsilon_0 a_0^2} \)
• Quantization: \( mva_0 = \hbar \)

Solving gives \( v = \alpha c \), where \( \alpha \) matches option (A).

Exam Relevance

For competitive exams like CSIR NET, recognize α’s role and eliminate distractors: (B) dimensionally inconsistent, (C) mass ratio irrelevant, (D) equivalent but not the direct form .