Physics Problems & Solutions: #76

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Quantum Mechanics - Gaussian Wave Packet

A Gaussian wave packet travels through free space. Which of the following statement about the wave packet are correct for all such wave packets?

The average momentum of the wave packet is zero
The width of the wave packet increases with time, as t → ∞.
The amplitude of the wave packet remains constant with time.
The narrower the wave packet is in momentum space, the wider it is in coordinate space.

A. I and III only
B. II and IV only
C. I, II, and IV only
D. II,III, and IV only
E. I, II,III, and IV only

(GR9677 #76)

Solution:

The Gaussian wave packet satisfies the Heisenberg uncertainty principle: ΔxΔp ≥ ħ/2

→ Statement IV is TRUE.
→ Statement I is FALSE, Δp cannot be 0
→ Answer A, C, E are FALSE.

Answer D suggests that both II and III are true.
If II and III are true, the area of gaussian wave packet can go to ∞
This is not allowed by the Heisenberg uncertainty principle.
→ D is FALSE

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

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

Thermal Physics – Conduction Electrons

The mean kinetic energy of the conduction electrons in metals is ordinarily much higher than kT because

A. electrons have many more degrees of freedom than atoms do
B. the electrons and the lattice are not in thermal equilibrium
C. the electrons form a degenerate Fermi gas
D. electrons in metals are highly relativistic
E. electron interact strongly with phonons

(GR0177 #76)

Solution:

The mean kinetic energy of electrons in metals at room temperature is usually many times the thermal energy kT due to Pauli’s exclusion principle.

According to classical physics, the mean thermal energy of the electrons is ³⁄₂kT.

According to quantum physics, the mean energy of electrons (fermion energy) is ³⁄₅E_F
where
$E_F= $ Fermi energy $ = \frac{\pi^2 \hbar^2}{2m_e} \left(\frac{3n}{\pi}\right)^{2/3}$
n = N/a³→ the number of electrons per unit volume

If E_F ≪ kT : electrons are hot (high T) → electrons are non-degenerate → classic (Maxwell-Boltzman) statistic.

In reality, E_F ≫ kT: electrons are cold (room T) → electrons are degenerate → quantum (Fermi-Dirac) statistic.

Answer: C