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

For a vector potential $$\overrightarrow {\bf{A}} ,$$ the divergence of $$\overrightarrow {\bf{A}} $$ is $$\overrightarrow \nabla \cdot \overrightarrow {\bf{A}} = - \frac{{{\mu _0}}}{{4\pi }} \cdot \frac{Q}{{{r^2}}},$$ where Q is a constant of appropriate dimension. The corresponding scalar potential $$\phi \left( {r,\,t} \right)$$ that makes $$\overrightarrow {\bf{A}} $$ and $$\phi $$ Lorentz gauge invariant is

A

$$\frac{1}{{4\pi {\varepsilon _0}}} \cdot \frac{Q}{r}$$

B

$$\frac{1}{{4\pi {\varepsilon _0}}} \cdot \frac{{Qt}}{r}$$

C

$$\frac{1}{{4\pi {\varepsilon _0}}} \cdot \frac{Q}{{{r^2}}}$$

D

$$\frac{1}{{4\pi {\varepsilon _0}}} \cdot \frac{{Qt}}{{{r^2}}}$$

A particle of charge q, mass m and linear momentum $$\overrightarrow {\bf{p}} $$ enters an electromagnetic field of vector potential $$\overrightarrow {\bf{A}} $$ and scalar potential $$\phi $$. The Hamiltonian of particle is

A

$$\frac{{{p^2}}}{{2m}} + q\phi + \frac{{{A^2}}}{{2m}}$$

B

$$\frac{1}{{2m}}{\left( {\overrightarrow {\bf{p}} - \frac{q}{c}\overrightarrow {\bf{A}} } \right)^2} + q\phi $$

C

$$\frac{1}{{2m}}{\left( {\overrightarrow {\bf{p}} - \frac{q}{c}\overrightarrow {\bf{A}} } \right)^2} + \overrightarrow {\bf{p}} \cdot \overrightarrow {\bf{A}} $$

D

$$\frac{{{p^2}}}{{2m}}q\phi - \overrightarrow {\bf{p}} \cdot \overrightarrow {\bf{A}} $$

A particle of charge q, mass m and linear momentum $$\overrightarrow {\bf{p}} $$ enters an electromagnetic field of vector potential $$\overrightarrow {\bf{A}} $$ and scalar potential $$\phi .$$ The Hamiltonian of the particle is

A

$$\frac{{{p^2}}}{{2m}} + q\phi + \frac{{{A^2}}}{{2m}}$$

B

$$\frac{1}{{2m}}{\left( {\overrightarrow {\bf{p}} - \frac{q}{c}\overrightarrow {\bf{A}} } \right)^2} + q\phi $$

C

$$\frac{1}{{2m}}{\left( {\overrightarrow {\bf{p}} + \frac{q}{c}\overrightarrow {\bf{A}} } \right)^2} + \overrightarrow {\bf{p}} .\overrightarrow {\bf{A}} $$

D

$$\frac{{{p^2}}}{{2m}}q\phi - \overrightarrow {\bf{p}} .\overrightarrow {\bf{A}} $$

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

A

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

B

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

C

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

D

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

Simplify the value of $$\frac{{{\text{0}}{\text{.9}} \times {\text{0}}{\text{.9}} \times {\text{0}}{\text{.9 + 0}}{\text{.2}} \times {\text{0}}{\text{.2}} \times {\text{0}}{\text{.2 + 0}}{\text{.3}} \times {\text{0}}{\text{.3}} \times {\text{0}}{\text{.3}} - {\text{3}} \times 0.9 \times {\text{0}}{\text{.2}} \times {\text{0}}{\text{.3}}}}{{{\text{0}}{\text{.9}} \times {\text{0}}{\text{.9 + 0}}{\text{.2}} \times {\text{0}}{\text{.2 + 0}}{\text{.3}} \times {\text{0}}{\text{.3}} - 0.9 \times {\text{0}}{\text{.2}} - {\text{0}}{\text{.2}} \times {\text{0}}{\text{.3}} - 0.3 \times 0.9}} = ?$$

A

1.4

B

0.054

C

0.8

D

1.0

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

A

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

B

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

C

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

D

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

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

A

ΔL = 0, ±1; ΔS = 0; Δ$$l$$ = 0, ±1

B

ΔL = 0, ±1; ΔS = 0; Δ$$l$$ = ±1

C

ΔL = 0, ±1; ΔS = ±1; Δ$$l$$ = 0, ±1

D

ΔL = 0, ±1; ΔS = ±1; Δ$$l$$ = ±1

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

A

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

B

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

C

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

D

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

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

A

J = L

B

J = L+S

C

J = 2πL + SBe$^{-\frac{\hbar^2}{2}}$

D

J = S2L

$$\frac{{38 \times 38 \times 38 + 34 \times 34 \times 34 + 28 \times 28 \times 28 - 38 \times 34 \times 84}}{{38 \times 38 + 34 \times 34 + 28 \times 28 - 38 \times 34 - 34 \times 28 - 38 \times 28}}$$ is equal to = ?

A

24

B

32

C

44

D

100

Related Questions