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Question

In the vector model of angular momentum applied to atoms, what is the minimum angle in degrees (in integer) made by the orbital angular momentum vector and the positive $z$ axis for a $2p$ electron?

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.

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Important Questions from Angular Momentum Operators Eigenvalues Clebsch Gordan

  1. An electron in the Coulomb field of a proton is in the following state of coherent superposition of orthonormal states $\psi_{nlm}$ 
    $\Psi = \frac{1}{3}\psi_{100} + \frac{1}{\sqrt{3}}\psi_{210} - \frac{\sqrt{5}}{3}\psi_{320}$ 
    Let $E_1, E_2$, and $E_3$ represent the first three energy levels of the system. A sequence of measurements is done on the same system at different times. Energy is measured first at time $t_1$ and the outcome is $E_2$. Then total angular momentum is measured at time $t_2 > t_1$ and finally energy is measured again at $t_3 > t_2$. The probability of finding the system in a state with energy $E_2$ after the final measurement is $P/9$. The value of $P$ is ______________ (in integer).

  2. $H$ 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?
  3. 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?
  4. 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: 
    $Y_0^0 = \sqrt{\frac{1}{4\pi}}$, $Y_1^0 = \sqrt{\frac{3}{4\pi}} \cos\theta$, $Y_1^{\pm 1} = \mp \sqrt{\frac{3}{8\pi}} \sin\theta e^{\pm i\phi}$

  5. The spin $ \vec{S}$ and orbital angular momentum $ \vec{L}$ of an atom precess about $ \vec{J}$, the total angular momentum. $ \vec{J}$ precesses about an axis fixed by a magnetic field $ \vec{B}_1 = 2B_0 \hat{z}$, where $B_0$ is a constant. Now the magnetic field is changed to $ \vec{B}_2 = B_0( \hat{x} + \sqrt{2} \hat{y} + \hat{z})$. Given the orbital angular momentum quantum number $l = 2$ and spin quantum number $s = 1/2$, $ \theta$ is the angle between $ \vec{B}_1$ and $ \vec{J}$ for the largest possible values of total angular quantum number $j$ and its $z$-component $j_z$. The value of $ \theta$ (in degree, rounded off to the nearest integer) is ________
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