In a particular state of the hydrogen atom, the angle between the angular momentum vector L→\overrightarrow{L} and the zz-axis is u=26.6u = 26.6°. If this is the smallest angle for this particular value of the orbital quantum number ll, what is ll?

Ch 41: Quantum Mechanics II: Atomic Structure

Young & Freedman Calc - University Physics 14th Edition

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Young & Freedman Calc 14th Edition

Ch 41: Quantum Mechanics II: Atomic Structure

문제 11

41장, 문제 11

In a particular state of the hydrogen atom, the angle between the angular momentum vector $\vec{L}$ and the $z$ -axis is $u equals 26.6$ °. If this is the smallest angle for this particular value of the orbital quantum number $l$ , what is $l$ ?

검증된 단계별 안내

Step 1: Recall the relationship between the orbital quantum number (l) and the angular momentum vector magnitude. The magnitude of the angular momentum vector is given by $ L = \sqrt{l(l+1)} \hbar $, where $ \hbar $ is the reduced Planck's constant.

Step 2: Understand the quantization of the projection of angular momentum along the z-axis. The projection $ L_z $ is given by $ L_z = m_l \hbar $, where $ m_l $ is the magnetic quantum number and can take integer values from $ -l $ to $ +l $.

Step 3: The angle $ \theta $ between the angular momentum vector $ \overrightarrow{L} $ and the z-axis is determined by $ \cos \theta = \frac{L_z}{L} $. Substitute $ L_z = m_l \hbar $ and $ L = \sqrt{l(l+1)} \hbar $ into this equation to get $ \cos \theta = \frac{m_l}{\sqrt{l(l+1)}} $.

Step 4: Since the problem states that $ \theta = 26.6^\circ $ is the smallest angle, $ \cos \theta $ must be maximized. The maximum value of $ \cos \theta $ occurs when $ m_l = l $, which corresponds to the smallest angle. Substitute $ \cos \theta = \frac{l}{\sqrt{l(l+1)}} $ and $ \theta = 26.6^\circ $ into the equation.

Step 5: Solve for $ l $ by rearranging the equation $ \cos 26.6^\circ = \frac{l}{\sqrt{l(l+1)}} $. Use trigonometric values for $ \cos 26.6^\circ $ and algebraic manipulation to isolate $ l $.

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주요 개념

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Angular Momentum in Quantum Mechanics

In quantum mechanics, angular momentum is a fundamental property of particles, represented by the vector $ \overrightarrow{L} $. It is quantized, meaning it can only take on certain discrete values determined by the orbital quantum number $ l $. The magnitude of angular momentum is given by $ |\overrightarrow{L}| = \sqrt{l(l+1)}\hbar $, where $ \hbar $ is the reduced Planck's constant.

Orbital Quantum Number (l)

The orbital quantum number $ l $ defines the shape of an electron's orbital and is integral to determining the angular momentum of an electron in an atom. It can take on integer values from 0 to $ n-1 $, where $ n $ is the principal quantum number. Each value of $ l $ corresponds to a specific type of orbital (s, p, d, f) and influences the angular distribution of the electron's probability density.

Quantization of Angular Momentum

In quantum mechanics, the quantization of angular momentum implies that the angular momentum vector can only take specific orientations relative to an axis, such as the z-axis. The angle between the angular momentum vector and the z-axis is related to the magnetic quantum number $ m_l $, which can take values from $ -l $ to $ +l $. The smallest angle for a given $ l $ indicates the lowest energy state or configuration of the system.