If an atom is in the <sup>3</sup>d<sub>3</sub> state, the angle between its orbital and spin angular momentum vectors ($$\overrightarrow {\bf{L}} $$ and $$\overrightarrow {\bf{S}} $$) is

$${\cos ^{ - 1}}\frac{1}{{\sqrt 3 }}$$

$${\cos ^{ - 1}}\frac{2}{{\sqrt 3 }}$$

$${\cos ^{ - 1}}\frac{1}{2}$$

$${\cos ^{ - 1}}\frac{{\sqrt 3 }}{2}$$

Correct Answer: $${\cos ^{ - 1}}\frac{1}{{\sqrt 3 }}$$

The Hamiltonian of a particle is given by $$H = \frac{{{p^2}}}{{2m}} + V\left( {\left| {\overrightarrow {\bf{r}} } \right|} \right) + \phi \left( { + \left| {\overrightarrow {\bf{r}} } \right|} \right)\overrightarrow {\bf{L}} .\overrightarrow {\bf{S}} ,$$ where $$\overrightarrow {\bf{S}} $$ is the spin, $$V\left( {\left| {\overrightarrow {\bf{r}} } \right|} \right)$$ and $$\phi \left( {\left| {\overrightarrow {\bf{r}} } \right|} \right)$$ are potential functions and $$\overrightarrow {\bf{L}} \left( { = \overrightarrow {\bf{r}} \times \overrightarrow {\bf{p}} } \right)$$ is the angular momentum. The Hamiltonian does not commute with

$$\overrightarrow {\bf{L}} + \overrightarrow {\bf{S}} $$

$$\overrightarrow {{{\bf{S}}^2}} $$

$${L_z}$$

$$\overrightarrow {{{\bf{L}}^2}} $$

In a cubic system with cell edge a, two phonons with wave vectors $${\overrightarrow {\bf{q}} _1}$$ and $${\overrightarrow {\bf{q}} _2}$$ collide and produce a third phonon with a wave. vector $${\overrightarrow {\bf{q}} _3}$$ such that $${\overrightarrow {\bf{q}} _1} + {\overrightarrow {\bf{q}} _2} = {\overrightarrow {\bf{q}} _3} + \overrightarrow {\bf{R}} $$ where, $$\overrightarrow {\bf{R}} $$ is a lattice vector. Such a collision process will lead to(a)

finite thermal resistance

zero thermal resistance

an infinite thermal resistance

a finite thermal resistance for certain $$\left| {\overrightarrow {\bf{R}} } \right|$$ only

For a multi-electron atom $$l$$, L and S specify the one electron orbital angular momentum, total orbital angular momentum and total spin angular momentum respectively. The selection rules for electric dipole transition between the two electronic energy levels, specified by $$l$$, L and S are

ΔL = 0, ±1; ΔS = 0; Δ$$l$$ = 0, ±1

ΔL = 0, ±1; ΔS = 0; Δ$$l$$ = ±1

ΔL = 0, ±1; ΔS = ±1; Δ$$l$$ = 0, ±1

ΔL = 0, ±1; ΔS = ±1; Δ$$l$$ = ±1

An atom with net magnetic moment $$\overrightarrow \mu $$ and net angular momentum $$\overrightarrow {\bf{L}} \left( {\overrightarrow \mu = \gamma \overrightarrow {\bf{L}} } \right)$$ is kept in a uniform magnetic induction $$\overrightarrow {\bf{B}} = {B_0}{\bf{\hat k}}.$$ The magnetic moment $$\overrightarrow \mu \left( { = {\mu _x}} \right)$$ is

$$\frac{{{d^2}{\mu _x}}}{{d{t^2}}} + \gamma {B_0}{\mu _x} = 0$$

$$\frac{{{d^2}{\mu _x}}}{{d{t^2}}} + \gamma {B_0}\frac{{d{\mu _x}}}{{dt}} + {\mu _x} = 0$$

$$\frac{{{d^2}{\mu _x}}}{{d{t^2}}} + {\gamma ^2}B_0^2{\mu _x} = 0$$

$$\frac{{{d^2}{\mu _x}}}{{d{t^2}}} + 2\gamma {B_0}{\mu _x} = 0$$

If for a system of N particles of different masses m₁, m₂, . . . m_N with position vectors $${\overrightarrow {\bf{r}} _1},\,{\overrightarrow {\bf{r}} _2},\,.\,.\,.\,{\overrightarrow {\bf{r}} _N}$$ and corresponding velocities $${\overrightarrow {\bf{v}} _1},\,{\overrightarrow {\bf{v}} _2},\,.\,.\,.\,{\overrightarrow {\bf{v}} _N}$$ respectively such that $$\sum\limits_i {\overrightarrow {{{\bf{v}}_i}} = 0,} $$ then

total momentum must be zero

total angular momentum must be independent of the choice of the origin

the total force on the system must be zero

the total torque on the system must be zero

A rigid body is rotating about its centre of mass; fixed at origin with an angular velocity $$\overrightarrow \omega $$ and angular acceleration $$\overrightarrow \alpha $$. If the torque acting on it is $$\overrightarrow \tau $$ and its angular momentum is $$\overrightarrow {\bf{L}} $$, then the rate of change of its kinetic energy is

$$\frac{1}{2}\overrightarrow \tau \cdot \overrightarrow \omega $$

$$\frac{1}{2}\overrightarrow {\bf{L}} \cdot \overrightarrow \omega $$

$$\frac{1}{2}\left( {\overrightarrow \tau \cdot \overrightarrow \omega + \overrightarrow {\bf{L}} \cdot \overrightarrow \alpha } \right)$$

$$\frac{1}{2}\overrightarrow {\bf{L}} \cdot \overrightarrow \alpha $$

The primitive translation vectors of the body centred cubic lattice are $$\overrightarrow {\bf{a}} = \frac{a}{2}\left( {{\bf{\hat x}} + {\bf{\hat y}} - {\bf{\hat z}}} \right),\,\overrightarrow {\bf{b}} = \frac{a}{2}\left( { - {\bf{\hat x}} + {\bf{\hat y}} + {\bf{\hat z}}} \right)$$ and $$\overrightarrow {\bf{c}} = \frac{a}{2}\left( {{\bf{\hat x}} - {\bf{\hat y}} + {\bf{\hat z}}} \right)$$ . The primitive translation vectors $$\overrightarrow {\bf{A}} ,\,\overrightarrow {\bf{B}} $$ and $$\overrightarrow {\bf{C}} $$ of the reciprocal lattice are

$$\overrightarrow {\bf{A}} = \frac{{2\pi }}{a}\left( {{\bf{\hat x}} - {\bf{\hat y}}} \right);\,\overrightarrow {\bf{B}} = \frac{{2\pi }}{a}\left( {{\bf{\hat y}} + {\bf{\hat z}}} \right);\,\overrightarrow {\bf{C}} = \frac{{2\pi }}{a}\left( {{\bf{\hat x}} + {\bf{\hat z}}} \right)$$

$$\overrightarrow {\bf{A}} = \frac{{2\pi }}{a}\left( {{\bf{\hat x}} - {\bf{\hat y}}} \right);\,\overrightarrow {\bf{B}} = \frac{{2\pi }}{a}\left( {{\bf{\hat y}} - {\bf{\hat z}}} \right);\,\overrightarrow {\bf{C}} = \frac{{2\pi }}{a}\left( {{\bf{\hat x}} + {\bf{\hat z}}} \right)$$

$$\overrightarrow {\bf{A}} = \frac{{2\pi }}{a}\left( {{\bf{\hat x}} + {\bf{\hat y}}} \right);\,\overrightarrow {\bf{B}} = \frac{{2\pi }}{a}\left( {{\bf{\hat y}} + {\bf{\hat z}}} \right);\,\overrightarrow {\bf{C}} = \frac{{2\pi }}{a}\left( {{\bf{\hat x}} - {\bf{\hat z}}} \right)$$

Deuteron in its ground state has a total angular momentum J = 1 and a positive parity. The corresponding orbital angular momentum L and spin angular momentum S combinations are

L = 0, S = 1 and L = 2, S = 0

L = 0, S = 1 and L = 1, S = 1

L = 0, S = 1 and L = 2, S = 1

L = 1, S = 1 and L = 2, S = 1

For two non-zero vectors $$\overrightarrow {\text{A}} $$ and $$\overrightarrow {\text{B}} $$, if $$\overrightarrow {\text{A}} $$ + $$\overrightarrow {\text{B}} $$ is perpendicular to $$\overrightarrow {\text{A}} $$ - $$\overrightarrow {\text{B}} $$ then,

Magnitude of $$\overrightarrow {\text{A}} $$ is twice magnitude of $$\overrightarrow {\text{B}} $$

Magnitude of $$\overrightarrow {\text{A}} $$ is half the magnitude of $$\overrightarrow {\text{B}} $$

$$\overrightarrow {\text{A}} $$ and $$\overrightarrow {\text{B}} $$ cannot be orthogonal

the magnitudes of $$\overrightarrow {\text{A}} $$ and $$\overrightarrow {\text{B}} $$ are equal

How is the total angular momentum of a system J described in terms of spin angular momentum S and orbital angular momentum L?

J = L

J = L+S

J = 2πL + SBe$^{-\frac{\hbar^2}{2}}$

J = S2L