Physics Problems & Solutions: Angular Momentum

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Quantum Mechanics - Spherical Harmonics

A diatomic molecules is initially in the state Ψ(Θ, Φ) = (5Y₁¹+ 3Y₅¹+ 2Y₅⁻¹) / (38)^½ where Y_l^m is a spherical harmonics. If measurements are made of the total angular momentum quantum number l and of azimuthal angular momentum quantum number m, what is the probability of obtaining the results l = 5?

A. 36/1444
B. 9/38
C. 13/38
D. 5/(38)^½
E. 34/38

(GR9677 #33)

Solution:

l = 5 → Y₅¹and Y₅⁻¹

Probability, P = ∑_{_i}|c_i|²

P = |3/√38|² + |2/√38|²= 9/38 + 4/38 = 13/38

Answer: C

Quantum Mechanics - Angular Momentum

At a given instant of time, a rigid rotator is in the state ψ(θ, ϕ) = √(¾π) sin θ sin ϕ, where θ is the polar angle relative to the z-axis and ϕ is the azimuthal angle. Measurement will find which of the following possible values of the z-component of the angular momentum L_z?

A. 0
B. ħ/2, −ħ/2
C. ħ, −ħ
D. 2ħ, −2ħ
E. ħ, 0, −ħ

(GR9677 #52)

Solution:

The eigenstates of L_z is L_zY_l^m= mħY_l^m
Possible values = eigen values of mħ.

L_z= −iħ ∂/∂ϕ → the information about m is contained in the sin ϕ term and it's proportional to e^imϕ:

sin ϕ = (e^iϕ− e⁻ⁱ^ϕ) / 2i

→ m = ±1

Thus, the eigenvalues are ±ħ

Answer: C

Nuclear & Particle Physics - Positronium

Positronium is the bound state of an electron and a positron. Consider only the states of zero orbital angular momentum l = 0. The most probable decay product of any such state of positronium with spin zero (singlet is)

A. 0 photons
B. 1 photons
C. 2 photons
D. 3 photons
E. 4 photons

(GR9677 #53)

Solution:

The singlet state, s = 0 of Positronium is known as para-Positronium, decays preferentially into two gamma rays.

It can decay into any even number of photons (2, 4, 6, ...), but the probability quickly decreases with the number.

Thus, the most probable decay product of singlet state of positronium is 2 photons.

Answer: C

Quantum Mechanics - Spin Angular Momentum

Two ions 1 and 2, at fixed separation, with spin angular momentum operators S₁ and S₂, have the interaction Hamiltonian H = −J S₁·S₂, where J > 0. The values of S₁² and S₂² are fixed at S₁(S₁+ 1) and S₂(S₂+ 1), respectively. Which of the following is the energy of the ground state of the system?

A. 0
B. –JS₁S₂
C. –J[S₁(S₁+ 1) – S₂(S₂+ 1)]
D. –(J/2)[(S₁+ S₂)(S₁+ S₂+ 1) – S₁(S₁+1) – S₂(S₂+1)]
E. –(J/2)[(S₁(S₁+ 1) + S₂(S₂+ 1))/(S₁+ S₂)(S₁+ S₂+ 1)]

(GR9677 #77)

Solution:

S₁² ψ = S₁(S₁+ 1) ψ
S₂² ψ = S₂(S₂+ 1) ψ
S_i² ψ = S_i(S_i+ 1) ψ

H = −J S₁·S₂

Using general arithmetic equation: ab = ½ [(a + b)² − a² − b²]

H = −(J/2)[(S₁ + S₂)² − S₁² − S₂²]

Since S_i² ψ = S_i(S_i+ 1) ψ
For (S₁ + S₂)² → replace S_i with S₁ + S₂
→ (S₁ + S₂)² ψ = (S₁+ S₂)(S₁+ S₂+ 1) ψ

H = −(J/2)[(S₁+ S₂)(S₁+ S₂+ 1) − S₁(S₁+ 1) − S₂(S₂+ 1)]

Answer: D

Electromagnetism - Angular Momentum

Two small pith balls, each carrying a charge q, are attached to the ends of a light rod of length d, which is suspended from the ceiling by a thin torsion-free fiber, as shown in the figure. There is a uniform magnetic field B, pointing straight down, in the cylindrical region of radius R around the fiber. The system is initially at rest. If the magnetic field is turned off, which of the following describes what happens to the system?

It rotates with angular momentum qBR².
It rotates with angular momentum ¼ qBd².
It rotates with angular momentum ½ qBRd.
It does not rotate because to do so would violate conservation of angular momentum.
It does not move because magnetic forces do no work.

(GR9677 #87)

Solution:

B and C are FALSE.
Angular momentum is finite, it cannot go to infinite.
If d → ∞ , angular momentum → ∞

D. FALSE
Since there is external torque, angular momentum is not conserved.

E. FALSE
The system will rotate due to B.

Answer: A:

Calculation:
Faraday's Law: $\varepsilon=-\frac{d \Phi _B}{dt}=\vec{E}\cdot dl$

with Φ = BπR² and dl = 2πd/2 = πd

$\pi R^2 \frac{dB}{dt}=E\pi d\rightarrow E=\frac{\dot{B}R^2}{d}$
$F=qE=\frac{q\dot{B}R^2}{d}$

Torque: $\tau =\sum\vec{r}\times \vec{F}$

Since there is a force contribution from each charge and by the right-hand-rule their cross products with the moment-arm point in the same direction,

$\tau =2 \frac{d}{2}\frac{q\dot{B}R^2}{d}=q\dot{B}R^2$

Torque and Angular momentum:

$\\\sum \tau = \dot{L} \\\rightarrow \dot{L} = q\dot{B}R^2 \\\rightarrow L= qBR^2$

Nuclear & Particle Physics - Spectroscopic Notation

In a ³S state of the helium atom, the possible values of the total electronic angular momentum quantum number are

A. 0 only
B. 1 only
C. 0 and 1 only
D. 0, 1/2, and 1
E. 0, 1, and 2

(GR9277 #31)

Solution:

Spectroscopic notation: N^2s+1 L_j=l+s

N = the principal quantum number and will often be omitted
s = the total spin quantum number
L = the orbital angular momentum quantum number, l but is written as S, P, D, F, … for l = 0, 1, 2, 3,⋯
j = the total angular momentum quantum number

For ³S
S → l = 0
3 → 3 = 2s + 1 → s = 1
j = l + s = 0 + 1 = 1

Answer: B

Nuclear & Particle Physics - Spectroscopic Notation

The configuration of three electrons is 1s2p3p has which of the following as the value of its maximum possible total angular momentum quantum number?

A. 7/2
B. 3
C. 5/2
D. 2
E. 3/2

(GR9277 #76)

Solution:

In Spectroscopic Notation, the total angular momentum quantum number is j = l + s, where
s = the total spin quantum number → 2n+1= multiplicity, the number of spin states
l = the orbital angular momentum quantum number

Since none of the electron sub-shells are filled, we add the individual angular momentum quantum number:
1s → l₁ = 0, s₁= 1/2 → j₁= 1/2
2p → l₂ = 1, s₂= 1/2 → j₂= 1+1/2 = 3/2
3p → l₃ = 1, s₃ = 1/2 → j₃ = 1 + 1/2 = 3/2

The maximum possible total angular momentum quantum number j = j₁+ j₂ + j₃ = 7/2

Answer: A

Nuclear & Particle Physics - Hydrogen

The spacing of the rotational energy levels for the hydrogen molecule H₂ is most nearly

A. 10⁻⁹ eV
B. 10⁻³ eV
C. 10 eV
D. 10 MeV
E. 100 MeV

(GR9277 #90)

Solution:

Rotational kinetic energy: E = ^L²∕_2I
Angular momentum: L² = l(l+1)ħ² with l = 0,1,2,⋯
Moment Inertia: I = mr²

For hydrogen atom:
r = 0.529 × 10⁻¹⁰ m
m =10⁻²⁷ kg (mass of proton)
→ I = 10⁻²⁷(0.529×10⁻¹⁰)² ≈ 10⁻⁴⁷ kgm²
ħ = ^h∕_2π = ^{6.63×10⁻³⁴}∕ _2(3.14) ≈ 10⁻³⁴

For l = 0 → L = 0 → E = 0
For l = 1 → L = 2ħ² → E = ^ħ²∕_I

The spacing of the rotational energy levels:
∆E = ^ħ²∕_I − 0 = ^{(10⁻³⁴)²}∕ _10⁻⁴⁷ = 10⁻⁶⁸⁺⁴⁷ = 10⁻²¹ J

Convert to eV: 1 eV = 1.6 × 10⁻¹⁹ J
→ ∆E = ^10⁻²¹∕ _{1.6 × 10⁻¹⁹} ≈ 10⁻³eV

Answer: B

Nuclear & Particle Physics - Selection Rules

A transition in which one photon is radiated by the electron in a hydrogen atom when the electron's wave function changes from ψ₁ to ψ₂ is forbidden if ψ₁ and ψ₂

A. have opposite parity
B. are orthogonal to each other
C. are zero at the center of the atomic nucleus
D. are both spherically symmetrical
E. are associated with different angular momenta

(GR8677 #48)

Solution:

Selection rules:

1.	Principal quantum number	:	∆n = anything
2.	Orbital angular momentum	:	∆l = ±1
3.	Magnetic quantum number	:	∆m_l = 0, ±1
4.	Spin	:	∆s = 0
5.	Total angular momentum	:	∆j = 0, ±1, but j = 0 ↛j = 0

A. FALSE
It’s not related to the selection rules.

B. FALSE
In any transition, eigenstates should always be mutually orthogonal.

C. FALSE.
Eigenstates zero at the center → l

0 could change, for example from 3 to 2. This is allowed.

D. TRUE.
If both initial and final states have spherically symmetrical wave functions, then they have the same angular momentum. l = 0 → l = 0 is forbidden.

E. FALSE.
The selection rules require l to change.

Answer: D

Classical Mechanics - Rotational Motion

A hoop of mass M and radius R is at rest at the top of an inclined plane. The hoop rolls down the plane without slipping. When the hoop reached the bottom, its angular momentum around its center of mass is

A. MR√(gh)
B. ½ MR√(gh)
C. M√(2gh)
D. Mgh
E. ½ Mgh

(GR8677 #76)

Solution:

Conservation of energy: Total energy at the top = Total energy at the bottom

$Mgh= \frac{1}{2} Mv^2_{\text{CM}}+\frac{1}{2} I_\text{CM} \omega^2$

Velocity: v_CM = ωR
Moment inertia of the hoop: I_CM = MR²

$\\Mgh= \frac{1}{2} MR^2\omega^2+\frac{1}{2} MR^2 \omega^2=MR^2 \omega^2\\\\\rightarrow \omega^2 = \frac{gh}{R^2}\rightarrow \omega= \frac{\sqrt{gh}}{R}$

Angular Momentum: $L=I\omega=MR^2 \frac{\sqrt{gh}}{R} = MR\sqrt{gh}$

Answer: A

Quantum Mechanics - Commutation Relations

The components of the orbital angular momentum operator $\vec{L}=(L_x,L_y,L_z )$ satisfy the following commutation relations.

$\\\left[L_x,L_y \right ]=i \hbar L_z \\\left[L_y,L_z \right ]=i \hbar L_x \\\left[L_z,L_x \right ]=i \hbar L_y$

What is the value of the commutator $\left[L_x L_y,L_z \right ]$ ?

A. $2i\hbar L_x L_y$
B. $i\hbar (L_x^2+L_y^2 )$
C. $-i\hbar (L_x^2+L_y^2 )$
D. $i\hbar (L_x^2-L_y^2 )$
E. $-i\hbar (L_x^2-L_y^2 )$

(GR0177 #43)

Solution:

Commutator identities: [A, B] = − [B, A]

and, [B, AC] = A[B, C] + [B, A]C

Therefore,
$\left[\underset{A}{\underbrace{L_x L_y}},\underset{B}{\underbrace{L_z}}\right] = - \left[\underset{B}{\underbrace{L_z}}, \underset{A}{\underbrace{L_x L_y}}\right]$

$\left[\underset{B}{\underbrace{L_z}}, \underset{A}{\underbrace{L_x}}\underset{C}{\underbrace{L_y}}\right]=L_x [L_z,L_y ]+[L_z,L_x ] L_y$

$\begin{aligned} -[L_z,L_x L_y ]&=-L_x [L_z,L_y ]-[L_z,L_x ] L_y \\&=-L_x (-i\hbar L_x)-i\hbar L_y L_y \\&=i\hbar L_x^2-i\hbar L_y^2 \\&=i\hbar (L_x^2-L_y^2 ) \end{aligned}$

Answer: D

Quantum Mechanics - Orbital Angular Momentum

Which is the following is the orbital angular momentum eigenfuction $Y_l^m (\theta,\phi )$ in a state for which the operators L^2

and

have eigenvalues $6\hbar^2$ and $-\hbar$ respectively?

$\\$A.$ \hspace{5 mm} Y_2^1 (\theta,\phi) \\\\$B.$ \hspace{5 mm} Y_2^{-1} (\theta,\phi) \\\\$C.$ \hspace{5 mm} \frac{1}{\sqrt{2}}[Y_2^1 (\theta,\phi)+Y_2^{-1} (\theta,\phi)] \\\\$D.$ \hspace{5 mm} Y_2^3 (\theta,\phi) \\\\$E.$ \hspace{5 mm} Y_3^{-1} (\theta,\phi)$

(GR0177 #81)

Solution:

Orbital angular momentum: $L=\sqrt{l(l+1)} \hbar$

$\begin{aligned} L^2=l(l+1) \hbar^2&=6\hbar^2 \\l(l+1)=l^2+l&=6 \\l^2+l-6&=0 \\(l+3)(l-2)&=0 \\l&=-3;2 \end{aligned}$

Allowed values of l:

$l=\left\{\begin{matrix} $0 1 2 3 4 ...$ \\ $s p d f g ...$ \end{matrix}\right.$

l cannot be negative. So in this case, l=2

The z-component of L: $L_z=m_l \hbar=-\hbar\rightarrow m=-1$

Thus, $Y_l^m (\theta,\phi)=Y_2^{-1} (\theta,\phi )$

Answer: B

Nuclear & Particle Physics - Selection Rules

An energy-level diagram of the n = 1 and n = 2 levels of atomic hydrogen (including the effect of spin-orbit coupling and relativity) is shown in the figure. Three transitions are labeled A, B, and C. Which of the transitions will be possible electric-dipole transition?

A. B only
B. C only
C. A and C only
D. B and C only
E. A, B, and C

(GR0177 #84)

Solution:

Selection rules:

1.	Principal quantum number	:	∆n = anything
2.	Orbital angular momentum	:	∆l = ±1
3.	Magnetic quantum number	:	∆m_l = 0, ±1
4.	Spin	:	∆s = 0
5.	Total angular momentum	:	∆j = 0, ±1, but j = 0 ↛j = 0

Transition A:
l = 0 to l = 0 → ∆l = 0 → NOT ALLOWED

Transition B:
l = 1 to l = 0 → ∆l = −1 → ALLOWED
j = 3/2 to j = 1/2 → ∆j = −1 → ALLOWED

Transition C:
j = 1/2 to j = 1/2 → ∆j = 0 → ALLOWED

Answer: D

Classical Mechanics - Rotational Motion

A child is standing on the edge of a merry-go-around that has the shape of a solid disk, as shown in the figure. The mass of the child is 40 kilograms. The merry-go-around has a mass 200 kilograms and a radius of 2.5 meters and it is rotating with an angular velocity of 2.0 radians per second. The child then walk slowly toward the center of the merry-go-around. What will the final angular velocity of the merry-go-around when the child reaches the center? (The size of the child can be neglected).

A. 2.0 rad/s
B. 2.2 rad/s
C. 2.4 rad/s
D. 2.6 rad/s
E. 2.8 rad/s

(GR0177 #89)

Solution:

Angular Momentum: L = Iω

Conservation of Angular Momentum:

∑ L_i = ∑ L_f

(I_m+ I_c)ω = (I_m'+ I_c')ω'
ω' = ω(I_m+ I_c)/(I_m'+ I_c')

Moment inertia of the uniform disk is the same before and after the child moves to the center:

I_m= I_m' = I_m= ½ m_mr²= ½ (200)(2.5)²
ω= 2 rad/s

Initally, moment inertia of the child:

I_c= m_cr²= (40)(2.5)²

When the child reaches the center, r= 0:

I_c' = 0

Thus,

ω' = ω(I_m+ I_c)/(I_m'+ I_c')
= (2) [½(200)(2.5)²+ (40)(2.5)²] / [½ (200)(2.5)²]
= (2) [½(200)+ (40)] / [½ (200)]
= 280/100
= 2.8 rad/s

Answer: E

Quantum Mechanics – Selection Rules

Which of the following is NOT compatible with the selection rule that controls electric dipole emission of photons by excited states of atoms?

A. ∆n may have any negative integral value
B. ∆l = ±1
C. ∆m_l = 0, ±1
D. ∆s = ±1
E. ∆j = ±1

(GR8677 #92)

Solution:

Selection rules:

1.	Principal quantum number	:	∆n = anything
2.	Orbital angular momentum	:	∆l = ±1
3.	Magnetic quantum number	:	∆m_l = 0, ±1
4.	Spin	:	∆s = 0
5.	Total angular momentum	:	∆j = 0, ±1, but j = 0 ↛j = 0

Answer: D

Classical Mechanics - Rotational Motion

Two uniform cylindrical disks of identical mass M, radius R, and moment inertia ½MR² collide on a frictionless, horizontal surface. Disk I, having an initial counterclockwise angular velocity ω₀ and a center-of-mass velocity v₀ =½ω₀R to the right, makes a grazing collision with disk II initially at rest. If after the collision the two disks stick together, the magnitude of the total angular momentum about the point P is

A. Zero
B. ½MR²ω₀
C. ½MR²v₀
D. MRv₀
E. Dependent on the time of the collision

(GR8677 #97)

Solution:

L_total = L_trans + L_rot

L_trans = r × p
r = distance from the center of disk I to point P.
At point P, R = r.

L_trans = R × p = R × Mv₀ = M(R × v₀)

From the problem, ω₀ is counterclockwise and v₀is to the right. Thus, the crossproduct (R × v₀) is negative. Also, v₀ =½ω₀R

L_trans = −½MR²ω₀

L_rot= Iω₀

Moment inertia, I = ½MR²

L_rot= ½MR²ω₀

L_total = −½MR²ω_{0 +}½MR²ω₀= 0

Answer: A

Quantum Mechanics - Commutation Relations

Let $\hat{\mathbf{J}}$ be a quantum mechanical angular momentum operator. The commutator [ $\hat{J}_x$ $\hat{J}_y$ , $\hat{J}_x$ ] is equivalent to which of the following?

A. 0
B. iħ $\hat{J}_z$
C. iħ $\hat{J}_z$ $\hat{J}_x$
D. −iħ $\hat{J}_x$ $\hat{J}_z$
E. iħ $\hat{J}_x$ $\hat{J}_y$

(GR0877 #95)

Solution:

Commutator identities: [A, B] = − [B, A]

and, [B, AC] = A[B, C] + [B, A]C

[ $\hat{J}_x$ $\hat{J}_y$ , $\hat{J}_x$ ] = − [ $\hat{J}_x$ , $\hat{J}_x$ $\hat{J}_y$ ]

A = $\hat{J}_x$
B = $\hat{J}_x$
C = $\hat{J}_y$

[ $\hat{J}_x$ , $\hat{J}_x$ $\hat{J}_y$ ] = $\hat{J}_x$ [ $\hat{J}_x$ , $\hat{J}_y$ ] + [ $\hat{J}_x$ , $\hat{J}_x$ ] $\hat{J}_y$
= $\hat{J}_x$ [ $\hat{J}_x$ , $\hat{J}_y$ ] + 0
= $\hat{J}_x$ [iħ $\hat{J}_z$ ] + 0

− [ $\hat{J}_x$ , $\hat{J}_x$ $\hat{J}_y$ ] = −iħ $\hat{J}_x$ $\hat{J}_z$

Answer: D

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