What is the difference in energy between two spin angular momentum states of a hydrogen atom in a 1s orbital, experiencing a magnetic field of 1 Tesla?

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

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

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

An atom with one outer electron having orbital angular momentum $$l$$ is placed in a weak magnetic field. The number of energy levels into which the higher total angular momentum state splits, is

A

2$$l$$ + 2

B

2$$l$$ + 1

C

2$$l$$

D

2$$l$$ - 1

If an atom is in the ³d₃ state, the angle between its orbital and spin angular momentum vectors ($$\overrightarrow {\bf{L}} $$ and $$\overrightarrow {\bf{S}} $$) is

A

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

B

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

C

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

D

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

A piece of paraffin is placed in a uniform magnetic field H₀. The sample contains hydrogen nuclei of mass m_p, which interact only with external magnetic field. An additional oscillating magnetic field is applied to observe resonance absorption. If g₁ is the g-factor of the hydrogen nucleus, the frequency at which resonance absorption takes place, is given by

A

$$\frac{{3{g_1}e{H_0}}}{{2\pi {m_p}}}$$

B

$$\frac{{3{g_1}e{H_0}}}{{4\pi {m_p}}}$$

C

$$\frac{{{g_1}e{H_0}}}{{2\pi {m_p}}}$$

D

$$\frac{{{g_1}e{H_0}}}{{4\pi {m_p}}}$$

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

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

A disc is a spinning with an angular velocity $$\omega $$ rad/s about the axis of spin. The couple applied to the disc causing precession will be (where $$I$$ = Mass moment of inertia of the disc and $$\omega {\text{P}}$$ = Angular velocity of precession of the axis of spin)

A

$$\frac{1}{2}I{\omega ^2}$$

B

$$I{\omega ^2}$$

C

$$\frac{1}{2}I\omega \omega {\text{P}}$$

D

$$I\omega \omega {\text{P}}$$

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

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

Superconducting tin has a critical temperature of 3.7K at zero magnetic field and a critical field at 0.0306 Tesla at 0K. Find the critical field at 2K.

A

0.0306 Tesla

B

7.4 Tesla

C

0.02166 Tesla

D

0 Tesla