Angular Momentum Vector Model Basics
For an electron in an atomic orbital, its angular momentum is described by quantum numbers.
- The principal quantum number $n$ indicates the energy level (here, $n=2$).
- The azimuthal or angular momentum quantum number $l$ determines the shape of the orbital. For a $p$ orbital, $l=1$.
- The magnetic quantum number $m_l$ determines the orientation of the orbital in space. For $l=1$, the possible values of $m_l$ are $-1, 0, +1$.
The vector model describes the orientation of the orbital angular momentum vector ($\vec{L}$) relative to an external axis, typically the $z$-axis.
Calculating Angle with Z-axis
The angle $\theta$ between the orbital angular momentum vector $\vec{L}$ and the positive $z$-axis is determined by the quantum numbers $l$ and $m_l$ using the following relationship:
$ \cos \theta = \frac{m_l}{\sqrt{l(l+1)}} $
For a $2p$ electron, we have $l=1$. We need to find the minimum angle $\theta$ (in degrees, integer) relative to the positive $z$-axis. This occurs for the non-zero values of $m_l$. Let's calculate $\cos \theta$ for $m_l = +1$ and $m_l = -1$:
- For $m_l = +1$: $ \cos \theta = \frac{+1}{\sqrt{1(1+1)}} = \frac{1}{\sqrt{2}} $
- For $m_l = -1$: $ \cos \theta = \frac{-1}{\sqrt{1(1+1)}} = -\frac{1}{\sqrt{2}} $
The value $m_l=0$ corresponds to $\cos \theta = 0$, which means $\theta = 90^\circ$. This is not the minimum angle.
Minimum Angle Determination
To find the minimum angle $\theta$, we look at the possible values of $\cos \theta$. A smaller angle corresponds to a larger positive value of $\cos \theta$. Comparing $\frac{1}{\sqrt{2}}$ and $-\frac{1}{\sqrt{2}}$, the larger value is $\frac{1}{\sqrt{2}}$.
We find the angle $\theta$ corresponding to $\cos \theta = \frac{1}{\sqrt{2}}$:
$ \theta = \arccos\left(\frac{1}{\sqrt{2}}\right) $ $ \theta = 45^\circ $The minimum angle made by the orbital angular momentum vector and the positive $z$ axis for a $2p$ electron is $45^\circ$. Since the question asks for an integer value, the answer is 45.
Important Questions from Angular Momentum Operators Eigenvalues Clebsch Gordan
- Consider two particles with angular momenta $j_1 = 2\hbar$ and $j_2 = \hbar/2$. If the expression
$$|j = 5/2, m = 3/2\rangle = \begin{cases} c_1|j_1 = 2, m_1 = 1\rangle|j_2 = 1/2, m_2 = 1/2\rangle + \\ c_2|j_1 = 2, m_1 = 2\rangle|j_2 = 1/2, m_2 = -1/2\rangle \end{cases}$$
gives an eigenstate of the total angular momentum of the two particles, using standard notation. Which of the following is true?
(Hint: $\hat{J}_{\pm}|j, m\rangle = \sqrt{j(j + 1) - m(m \pm 1)} |j, m \pm 1\rangle$) - A system of three non-identical spin $\frac{1}{2}$ particles has the Hamiltonian $H = \frac{A}{\hbar^2} (\vec{S}_1 + \vec{S}_2) \cdot \vec{S}_3$, where $\vec{S}_1, \vec{S}_2$ and $\vec{S}_3$ are the spin operators of particles labelled $1,2$ and $3$ respectively and $A$ is a constant with appropriate dimensions. The set of possible energy eigenvalues of the system is
- $H$ is the Hamiltonian, $\vec{L}$ the orbital angular momentum and $L_z$ is the $z$-component of $\vec{L}$. The $1s$ state of the hydrogen atom in the non-relativistic formalism is an eigen function of which one of the following sets of operators?
- An atom with non-zero magnetic moment has an angular momentum of magnitude $\sqrt{12}\hbar$. When a beam of such atoms is passed through a Stern-Gerlach apparatus, how many beams does it split into?
A particle has wavefunction
$\psi(x,y,z) = N ze^{-\alpha(x^2+y^2+z^2)}$,
where $N$ is a normalization constant and $\alpha$ is a positive constant. In this state, which one of the following options represents the eigenvalues of $L^2$ and $L_z$ respectively?
Some values of $Y_l^m$ are:
$Y_0^0 = \sqrt{\frac{1}{4\pi}}$, $Y_1^0 = \sqrt{\frac{3}{4\pi}} \cos\theta$, $Y_1^{\pm 1} = \mp \sqrt{\frac{3}{8\pi}} \sin\theta e^{\pm i\phi}$
