Q.10 The orbital angular momentum of hydrogen atom in the ground state is ________.
The orbital angular momentum hydrogen atom ground state puzzle confuses many students preparing for exams like CSIR NET Life Sciences. In quantum mechanics, this value is zero, distinguishing it from the Bohr model’s prediction. This article breaks down the physics, analyzes MCQ options, and clarifies Bohr vs. quantum views for precise understanding.
The orbital angular momentum of a hydrogen atom in the ground state is 0.
Ground State Definition
The ground state of hydrogen corresponds to the principal quantum number n=1. In quantum mechanics, this is the 1s orbital where the azimuthal quantum number l=0.
Orbital Angular Momentum Formula
Orbital angular momentum magnitude is given by L = √[l(l+1)] ℏ, where ℏ = h / 2π is the reduced Planck’s constant. For l=0, L = √[0(0+1)] ℏ = 0.
Option Analysis
(A) 0: Correct. Quantum mechanics confirms zero orbital angular momentum in the 1s state due to l=0.
(B) h/2π: Incorrect for quantum mechanics; this is the Bohr model’s value for n=1.
(C) h/2: Incorrect; does not match any standard quantization for ground state.
(D) h: Incorrect; corresponds to neither model accurately for ground state.
Bohr vs Quantum Mechanics
Bohr model predicts L = n ℏ = ℏ for n=1, but quantum mechanics refines this to L=0 for l=0, resolving directional issues in Bohr’s approach.
Quantum Numbers in Ground State
Hydrogen’s ground state is the 1s orbital: n=1, l=0, m_l=0. The orbital angular momentum quantum number l determines L = √[l(l+1)] ℏ, yielding zero when l=0. This s-orbital’s spherical symmetry means no rotational motion.
Why Zero in Quantum Mechanics?
Substituting l=0 gives L=0, confirmed across sources. The wavefunction ψ_{100} lacks angular dependence, unlike p or d orbitals with l>0.
Bohr Model Misconception
Bohr’s postulate sets L = n ℏ, so for n=1, L = ℏ. This semi-classical view succeeds for energies but fails for angular momentum direction and ground state magnitude.
MCQ Options Explained
| Option | Value | Validity in QM Ground State | Reason |
|---|---|---|---|
| (A) | 0 | Correct | l=0 → L=0 |
| (B) | h/2π |
Incorrect | Bohr model only |
| (C) | h/2 |
Incorrect | No physical basis |
| (D) | h |
Incorrect | Overestimates magnitude |
For CSIR NET, select (A) as modern quantum mechanics prevails over Bohr.
Exam Relevance
CSIR NET questions test this distinction, emphasizing quantum numbers over Bohr limitations. Ground state has no orbital contribution; any angular momentum is spin-related (s=1/2).
