Schrödinger's Equation and the Hydrogen Atom — Question 7

Section 39-8 · Physics pages 6564–6565

Question 7 Orbital angular momentum has a fixed magnitude and discrete axial projections Treat hydrogen with a fixed nucleus, a nonrelativistic Coulomb potential and no spin, fine structure or external field. Write the Bohr length as a. A state has n = 3, ℓ= 2 and mℓ= 2. Use L2 = ℓ(ℓ+ 1)ℏ2 and Lz = mℓℏ. z |L| Lz Schematic; labels, not drawing lengths, specify the data. (a) Find the angular-momentum magnitude and z component. (b) List all possible z components for this ℓand find the largest ratio |Lz|/|L|.
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Question 7 - Solution Orbital angular momentum has a fixed magnitude and discrete axial projections z |L| Lz (a) Method and calculation |L| = √ 6 ℏ, Lz = 2ℏ. |L|2 −L2 z = 2ℏ2 . (b) Interpretation and further calculation Lz ∈{−2, −1, 0, 1, 2}ℏ. (|Lz| |L| ) max = 2 √ 6 = √ 2/3 = .816497 . Physical check The vector-projection sketch represents constraints on angular-momentum quantum numbers. It does not assign simultaneous definite values to all components or define a classical electron orbit.

Question 7

Orbital angular momentum has a fixed magnitude and discrete axial projections

Treat hydrogen with a fixed nucleus, a nonrelativistic Coulomb potential and no spin, fine structure or external field. Write the Bohr length as a. A state has n=3n=3, ℓ=2\ell=2 and mℓ=2m_\ell=2. Use L2=ℓ(ℓ+1)ℏ2L^2=\ell(\ell+1)\hbar^2 and Lz=mℓℏL_z=m_\ell\hbar.

Diagram shown in the original SVG worksheet.

  1. Find the angular-momentum magnitude and z component.

  2. List all possible z components for this ℓ\ell and find the largest ratio |Lz|/|L||L_z|/|L|.

Question 7 - Solution

Orbital angular momentum has a fixed magnitude and discrete axial projections

Diagram shown in the original SVG worksheet.

(a) Method and calculation

|L|=6ℏ,Lz=2ℏ.\boxed{|L|=\sqrt6\,\hbar},\qquad\boxed{L_z=2\hbar}.|L|2−Lz2=2ℏ2.\boxed{|L|^2-L_z^2=2\hbar^2}.

(b) Interpretation and further calculation

Lz∈{−2,−1,0,1,2}ℏ.\boxed{L_z\in\{-2,-1,0,1,2\}\hbar}.(|Lz||L|)max=26=2/3=.816497.\boxed{\left(\frac{|L_z|}{|L|}\right)_{\max}=\frac2{\sqrt6}=\sqrt{2/3}=.816497}.

Physical check

The vector-projection sketch represents constraints on angular-momentum quantum numbers. It does not assign simultaneous definite values to all components or define a classical electron orbit.