Problem 57 - University Physics with Modern Physics

Question

(a) For an excited state of hydrogen, show that the smallest angle that the orbital angular momentum vector \(\vec{L}\) can have with the \(z\) -axis is $$\left(\theta_{L}\right)_{\min }=\arccos \left(\frac{n-1}{\sqrt{n(n-1)}}\right)$$ (b) What is the corresponding expression for \(\left(\theta_{L}\right)_{\max },\) the largest possible angle between \(\vec{L}\) and the \(z\) -axis?

Step-by-Step Solution

Answer

(a) \((\theta_{L})_{\min} = \arccos \left( \frac{n-1}{\sqrt{n(n-1)}} \right)\); (b) \((\theta_{L})_{\max} = \frac{\pi}{2}\).

Step 1: Understand Orbital Angular Momentum

The orbital angular momentum for hydrogen is characterized by the quantum number \(l\), where \(l = n-1\) for the excited state of hydrogen. The magnitude of the angular momentum vector, \(|\vec{L}|\), is given by \(\sqrt{l(l+1)\hbar^2}\). The projection of this vector on the z-axis is \(L_z = m_l \hbar\), where \(m_l\) is the magnetic quantum number and can take integer values from \(-l\) to \(l\).

Step 2: Use the Cosine Definition for Angles

The cosine of the angle \(\theta_L\) between \(\vec{L}\) and the z-axis is given by \(\cos(\theta_L) = \frac{L_z}{| \vec{L} |}\). Thus, we have \[\cos(\theta_L) = \frac{m_l \hbar}{\sqrt{l(l+1)}\hbar} = \frac{m_l}{\sqrt{l(l+1)}}\]

Step 3: Determine Smallest Angle \(\theta_{L_{\min}}\)

The smallest angle between \(\vec{L}\) and the z-axis occurs when \(m_l\) is maximized. For a given \(l = n-1\), the maximum \(m_l\) is \(l\). Plugging this into the equation, we get:\[\cos(\theta_{L_{\min}}) = \frac{l}{\sqrt{l(l+1)}} = \frac{n-1}{\sqrt{n(n-1)}}\]Thus, \[\theta_{L_{\min}} = \arccos \left( \frac{n-1}{\sqrt{n(n-1)}} \right)\]

Step 4: Determine Largest Angle \(\theta_{L_{\max}}\)

The largest angle between \(\vec{L}\) and the z-axis occurs when \(m_l\) is minimized (i.e., \(m_l = 0\)). This gives:\[\cos(\theta_{L_{\max}}) = \frac{0}{\sqrt{l(l+1)}} = 0 \\theta_{L_{\max}} = \arccos(0) = \frac{\pi}{2}\]which means the angle is 90 degrees or \(\frac{\pi}{2}\) radians.

Key Concepts

Quantum Numbers

Quantum numbers are essential for describing the unique quantum state of an electron in an atom. They serve as the "address" of an electron, detailing its energy level and position in relation to an atom's nucleus.

There are four primary quantum numbers:

• Principal Quantum Number (n): This number specifies the energy level of an electron and is a positive integer (1, 2, 3, ...). In the context of the hydrogen atom, it also largely determines the size of the orbital.
• Azimuthal Quantum Number (l): This number defines the shape of the orbital and ranges from 0 to (n-1). For example, if n=3, l can be 0, 1, or 2, describing s, p, or d orbitals, respectively.
• Magnetic Quantum Number (m_l): It determines the orientation of the orbital in space. The possible values for m_l are from -l to +l, including zero. This is crucial for understanding angular momentum projection as it defines the component of the orbital angular momentum along the z-axis.
• Spin Quantum Number (m_s): Intrinsic to the electron, it can be +1/2 or -1/2, indicating the two possible spin states of an electron.

Each of these numbers plays a crucial role in the behavior of electrons, particularly in complex atoms and while discussing atomic structure and spectral lines.

Hydrogen Atom

The hydrogen atom is one of the simplest systems to study in quantum mechanics and serves as a crucial model for understanding more complex atoms. It consists of a single electron orbiting a single proton, and its simplicity allows us to easily calculate properties like energy levels and angular momentum.

In a hydrogen atom:

• The energy levels are quantized, which means electrons can only exist at specific energy levels determined by the principal quantum number \( n \).
• Each of these levels further splits into specific shapes or subshells dictated by the azimuthal quantum number \( l \), with these subshells further oriented in space by the magnetic quantum number \( m_l \).
• The Schrödinger equation allows for the exact solution of the hydrogen atom's energy levels, principally defined by the Rydberg formula.

Learning about the hydrogen atom provides fundamental insights into the quantum mechanics principles governing atomic systems, making it a vital part of understanding atomic structure.

Angular Momentum Projection

Angular momentum projection in quantum mechanics relates to how the angular momentum vector of an electron's orbit is aligned with a chosen axis, generally the z-axis. This is described by the magnetic quantum number \(m_l\).

The angular momentum (\( \vec{L} \)) of an atom is a vector quantity and has both magnitude and direction. Its magnitude for an electron in a hydrogen atom is given by \(\sqrt{l(l+1)}\hbar\). The component of this vector along the z-axis is \(L_z = m_l \hbar\). This can be visualized if you consider the orbital as a cone aligned with the z-axis, where the tip of the vector traces a circle.

The angles the angular momentum vector makes with the z-axis are significant:

• Minimum Angle: Occurs when \(m_l\) is maximum, i.e., \(l = n-1\) (step 3 from the exercise), leading to the smallest angle that \(\vec{L}\) can have with the z-axis.
• Maximum Angle: Happens when \(m_l = 0\) (step 4 from the exercise), making it perpendicular to the z-axis, which corresponds to a right angle or \(\frac{\pi}{2}\).

Understanding these projections is crucial as they reveal how electrons will behave in magnetic fields, a concept exploited in technologies such as MRI.